text:
Kröger-Vink notation for defect reactions
In the Kröger-Vink notation the effective charge $c$ is marked as superscript while the site location $s$ is marked as subscript on the atom itself $\ce{A}$:
$$\ce{A^{c}_{s}}$$
As far as I have learned the effective charge is the charge difference from the perfect lattice situation. The Wikipedia definition agrees:

$c$ corresponds to the electronic charge of the species relative to the site that it occupies

While neutral charge (no difference) is usually an $x$, I have always worked with a slash $/$ or similar for one negative charge and a dot $\bullet$ for one positive charge.
I have run across a notation in a teacher's lecture material that says:
$$\ce{Zn^{/}_{Zn}} \quad \text{ and } \quad \ce{Zn^{\bullet}_{Zn}}$$
As I understand it, adding a $\ce{Zn}$ to it's own natural position makes no charge change, so it should be $\ce{Zn^{x}_{Zn}}$. How is the above notation then possible? Does it imply that the $\ce{Zn}$ has lost or gained one electron, respectivily, and thus has a net charge?

Yes, the Kröger-Vink notation you are asking about, implies that the metal has lost or gained one electron, relative to the case of the perfect crystal. This situation can be explained by non-stoichiometry.
A non-stoichiometric compound is one of which the proportions of the atoms can not be represented by integers.
Consider the case of $\ce{Fe}_{0.95}\ce{O}$. Indicated by the subscript 0.95, there are some metal atoms missing, which is expressed as the defect $\ce{$V$^{''}_{Fe}}$. This has a double negative charge, which can be compensated by 'replacing' $3~\ce{Fe^2+}$ by $2~\ce{Fe^3+}$ ions. That is, 2 iron ions need to be oxidized and 1 needs to disappear.
This can be accomplished by dissolving $\ce{Fe_2O_3(s)}$ into $\ce{FeO(s)}$, as in:
$$\ce{Fe_2O_3(s) ->[FeO] 2Fe^{.}_{Fe} + 3O_{O} + $V$^{''}_{Fe}}$$
Alternatively:
$$\ce{2Fe_{Fe} + 2O_{O} + 1/2 O_2(g) ->[FeO] 2Fe^{.}_{Fe} +3O_{O} + $V$^{''}_{Fe}}$$
More examples are described in my source, chapter 11 of the book 'Ceramic Materials' by Carter & Norton.


text:
My text seems to say that the high(est) energy orbitals are not valence orbitals - everything else says otherwise. What's true?
A paragraph in my text reads

Bonding involves the valence orbitals almost exclusively because these orbitals have the appropriate energies to interact strongly. Examine the electron energy-level diagram of a fluorine atom. The core $1 \mathrm{s}$ orbitals on a fluorine atom are at much lower energies than any of the $n = 2$ orbitals and do not participate effectively in bonding with other atoms. On the other hand, orbitals with $n > 2$ lie at considerably higher energy than the $n = 2$ orbitals. These orbitals can interact, but the resulting orbitals are too unstable to form strong covalent bonds. Therefore, the only orbitals of fluorine that form chemical bonds are the valence orbitals, those with $n = 2$. Typically the highest energy occupied orbitals are the ones used to bond with other atoms.

In contrast this simple paragraph from SparkNotes expresses what I had believed to be the case

The outermost orbital shell of an atom is called its valence shell, and the electrons in the valence shell are valence electrons. Valence electrons are the highest energy electrons in an atom and are therefore the most reactive.

I had thought that the outermost shell, the valence shell, the most energetic shell, and the shell involved in molecular bonds were all the same thing. However, the quoted paragraph of my text says otherwise. Where have I gone wrong?

Ooof, I have a little problem understanding your problem. None of the things you quoted cancel each other. Note that for a Neon atom, for instance, there is an n=4; it's just that it's vacant and "not occupied" by electrons in the base state.

@M.A.Ramezani What about 'orbitals with n>2 lie at considerably higher energy than the n=2 orbitals... but the resulting orbitals are too unstable to form covalent bonds.' Does that not preclude the largest and most energetic shell from being the shell involved in bonding? –

Well, the shell involved in bonding is the most energetic shell that is occupied by electrons.

Hal anytime! I'm pretty sure others will come and write better and more explained answers, and I'm going to sleep now, so I'm not changing that into an answer.

The problem is the words used to describe energy. I am not native english speaker so I am not the best person to answer your question but I'll try.
As smaller $n$, smaller energy. Problems arises when the words higher lower are used to compare numeric values (as those of the energy). They must say greater or smaller.
Without accurate words all of us intuitively agree in that:
"2 is higher that 1", and we should say that 2 is greater than 1 (2>1)
When negative numbers are involved it is not so clear. For some authors "-2 is higher than -1" and "-2 is higher than 1", for others "-2 is lower than -1" and "-2 is lower than 1". In the later cases the correct word is "smaller" (-2<-1).
In this context energies are negatives because of the definition of the reference states. So the energy of an $1s$ orbital is smaller than the energy of a $2s$ orbital.
I have some discrepancies about the meaning and usage of orbital energies but I adopted the common usage to avoid extra confusions.

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text:
What is the "rate factor" or "second-step rate constant" in the reaction rate equation?
The reaction rate equation can be written as:
$$r=r_0 K \prod_n a_n$$
where the $a_n$s are the activities of each reactant in the reaction equation (in the case where we find the reaction rate from reactants to products, of course. Else the other way around). These activities are multiplied together. (This is often written as concentrations instead, I have seen, which seems to be an idealization.)
$K$ is the equillibrium constant, $K=\exp\left(\frac{-\Delta G}{RT}\right)$.
I am curious as to what exactly the $r_0$ is? In the wikipedia article it is denoted $k_2$ and called the "rate constant for the second step." The second step here refers to the intermediate step at the actived-complex.
So far so good, but what is this quantity at this intermediate step? Is it a reaction rate from the activated-complex to the final product (in that case, how can this be different from the reaction rate from the reactants to this state)?

$k_2$ is used in a particular example (reaction of hydrogen and nitric oxide) in the Wikipedia article, not "in general" for a formula derivation.

You seem to be talking about Michaelis-Menten kinetics? If so, it's the rate constant for the production of the product and the release of the enzyme from the enzyme-substrate complex. It may have many steps embedded into it though.

If not, then you can take that comment about MM kinetics as just an example; we have:
$$E + S ⇌ ES → E + P$$
And $k_2$ would be for the non-elementary, potentially multi-step process for going from $ES$ complex to free enzyme $E$ and product $P$.

The implication is that the reaction mechanism for any particular reaction, if written like this:
$$A + B → C$$
might not proceed as one step. Having a $k_2$ indicates that there is a second step to the mechanism, and potentially even several more. $k_{obs}$ is probably a better name for it, because sometimes you see it written as a simplified "one-step" process, such as in Michaelis-Menten Kinetics, when really, it might be $1$ more step, or it might even be $20$, depending on the reaction (I'm exaggerating a bit, but you get the point).
If we take Michaelis-Menten enzyme kinetics as an example:
$$E+S⇌ES→E+P$$
The first step is an equilibrium in which the concentration of the intermediate is assumed to not change because the intermediate is being produced about as fast as it is being consumed. This would be the Steady State Approximation. In other words:
$\frac{d[ES]}{dt} = 0 = k_1[E][S] - k_{-1}[ES] - k_2[ES]$
Eventually...
$[ES] = \frac{k_1}{k_{-1} + k_2}[E][S]$
The second step is where the $ES$ complex does something in such a way that the enzyme unbinds somehow, and the product is made... somehow. Depending on the reactants and products, the mechanism is different.
Anyways, the point is that $k_2$ describes this second step in this type of reaction (which is what you were wondering---why the "second step"?), hence the $2$.
In any case, $k_2$ corresponds to the reaction step(s) that proceed after an intermediate product is formed on the first step for a reaction mechanism.

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text:
Why does hybridization produce a more stable configuration?
My text, and the other sources I've checked, include information on the effects, and kinds, of hybridized orbitals; however, they do not explain what properties of hybridized bonds conduce greater stability.
Why does hybridization produce a (lower energy) more stable configuration?

What causes orbital hybridization?

Hybridization only occurs in response to bond formation, that is an atom will not just spontaneously hybridize on its own.  Further, an atom will only hybridize if hybridization leads to the formation of a more stable (lower energy) molecule.
Nature is very efficient at exploring the energy surface and finding whatever energy minimum is available.  For example, ammonia ($\ce{NH3}$) is roughly a tetrahedron with $\ce{H-N-H}$ bond angles around 109.5° (the tetrahedral angle) using (roughly) $\mathrm{sp^3}$ hybridized orbitals. For ammonia, mixing (hybridizing) it's $\mathrm{2s}$ and three $\mathrm{2p}$ orbitals to form four $\mathrm{sp^3}$ hybridized orbitals that are used to form three $\ce{N-H}$ bonds and one lone pair orbital produces a more stable molecule than if it were not hybridized.  On the other hand, the analogue phosphine ($\ce{PH3}$) chooses to remain roughly unhybridized, using p-orbitals to form its $\ce{P-H}$ bonds with a resultant $\ce{H-P-H}$ bond angle around 90°.  Bond strengths, steric factors, etc. all play into determining whether a hybridized or unhybridized structure will be the more stable.

I'm sorry, I should have been more clear in my question. I (unjustifiably) presumed readers would regard the disposition of atoms to configure themselves most stably as common knowledge. I've edited my question.

As I tried to explain in my answer, hybridization does not always produce a more stable configuration - some atoms, when they combine to form a bond, do not hybridize ($\ce{PH3}$).

Whichever configuration (hybridized or unhybridized) has the strongest bonds, the smallest steric repulsion, etc. (e.g. the lowest energy) is the configuration that will be adopted.

In sp2 configuration does the repulsion occurs between hybridized and unhybridized orbitals @ron too ?

Or just the repulsion is considered between the three hybridized orbitals and hence thats the reason bond angles are 120° otherwise it could be less if considered the unhybridized orbital too ?

@Orion_Pax  Repulsion occurs between all orbitals, hybridized and unhybridized, that contain electrons.  The repulsions are minimized when the molecule attains the lowest energy geometry possible.


text:
Calculate specific heat of a substance given other information

What is the specific heat of a $\pu{20 g}$ substance that releases $\pu{979 J}$ of heat when changing from $\pu{70 ^\circ C}$ to $\pu{20 ^\circ C}$?

I started out with the following equation: $c = \frac{Q}{m\Delta T}$ and $Q = 979\ \mathrm{J}$ and $m = 20\ \mathrm{g}$ and $\Delta T = 45\ \mathrm{^\circ C}$.
My final answer was $2202.75\ \mathrm{J}$
Did I use the right equations?

you got to convert g to kg.

I rolled back your edit, as it shows effort and your question.

possible duplicate of How to calculate the enthalpy change of water?

Your equation is correct, as are most of your measurements.
The unit for $c$ is $\ce{Jg^{-1}K^{-1}}$ if using grams as mass, ($\ce{Jkg^{-1}K^{-1}}$ if you convert the mass to kilograms).
Using your formula, however, does not give the answer you gave. When calculating using a calculator, consider that a parenthesis needs to be put around the denominator, as such:
$$c = \frac{q}{(m\Delta T)}$$
or else the calculator takes it as being $\ce{979J / 20g \times 45K}$ which gives the value you got.
It needs to be $\ce{979J / (20g \times 45K)}$ which gives an answer that is much more reasonable.


text:
How to calculate the enthalpy change of water?

Calculate the $\Delta H$ of the reaction where $2.6\:\mathrm{g}$ of water, $C_\mathrm{s}= 4.184 \mathrm{\frac{J}{g\:K}}$ is heated, raising the temperature increases from $298\:\mathrm{K}$ to $303\:\mathrm{K}$.

I used the following equation: $\Delta H=mC_\mathrm{s}\Delta T$
$m = 2.6~\mathrm{g}$, $C_\mathrm{s} = 4.184~\mathrm{J/g}$, $\Delta T= 5$
What I got: $(2.6)(4.184)(5)= 54.392~\mathrm{J/g}$
Did I solve this correctly?

Yes, but your answer should be just in joules not $\mathrm{J/g}$, and your units are off.
$$(2.6~\mathrm{\color{\red}{g}})\left(4.184~\mathrm{\frac{J}{\color{\red}{g}\ K}}\right)(5~\mathrm{K})= 50~\mathrm{J}$$

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text:
How to calculate the pH of a 0.325 M solution of pyridine hydrofloride?
How to calculate the $\mathrm{pH}$ of a $0.325~\mathrm{M}$ solution of $\ce{C5H5NHF}$? The $K_\mathrm{b}$ for $\ce{C5H5N}$ and $\ce{C5H5NH+}$ is $1.7\times 10^{-9}$.
Here's my work, but it's wrong (at least I tried):
$$\ce{C5H5NHF -> C5H5NH+ + F- }$$
$$K_\mathrm{a}(\ce{HF})= 4.0\cdot10^{-4}$$
$$K=\frac{K_\mathrm{w}}{K_\mathrm{a}}$$
$$K=\frac{1\cdot10^{-14}}{4.0\cdot10^{-4}}=2.5\cdot10^{-11}$$
Construct an ice table:
$$
\begin{array}{|c|c|c|c|}\hline
&\ce{F-}&\ce{+H2O(l)->}&\ce{HF}&+\ce{OH-}\\\hline
\text{I}&\pu{0.325M}&&0&0 \\\hline
\text{C}&-x&&x&x\\\hline
\text{E}&\pu{0.325M}-x&&x&x\\\hline
\end{array}
$$
Set up the equation (small x-approximation):
$$2.5\cdot10^{-11} = \frac{x^2}{0.325-x}$$
Now, cross multiply:
$$x^2 = 5.87\cdot10^{-12}$$
Take the square root of both sides:
$$[\ce{OH-}] = x = 2.42\cdot10^{-6}$$
You want to get from $[\ce{OH-}]$ to $\mathrm{pOH}$:
$$\mathrm{pOH} = -\log\left(2.42\cdot10^{-6}\right) = 5.616$$
Then, from $\mathrm{pOH}$ to $\mathrm{pH}$:
$$\mathrm{pH} = 14 - 5.616 = 8.384$$
Which is the wrong answer.
What am I doing wrong? How do I even use the $K_\mathrm{b}$ value in my calculation?

You didn't take into account the ionization of $\ce{C5H5NH^{+}}$ anywhere. That equilibrium will affect the final pH. You have a problem asking about the pH of a salt solution made of both a weak acid and a weak base. This will make it trickier than usual.

So, my K equation should actually look like 1/Ka * 1/Kb * Kw to take in the account of the Kb value?

The fact that the pyridinium cation $\ce{C5H5NH+}$ is more acidic (pKa = 5.25) than $\ce{F-}$ is basic (pKb = 10.83) somewhat supports this idea that the dissociation of $\ce{F-}$ is supposed to be ignored. Although I definitely think that the question should have given a non-basic anion such as $\ce{Cl-}$ in this case.

The intro of this old ACS paper might answer your question.

I think part of the issue is that you are looking at the wrong reaction. I believe the reaction should be $$\ce{C5H5NHF_{(s)} ->C5H5NH+_{(aq)} + F-_{(aq)}->C5H5N_{(aq)} + HF_{(aq)}}$$ When looking at this, we can start by noting that $\ce{HF}$ is a stronger acid than $\ce{C5H5NH+}$, so the reaction will favor the reactants. Then, we can compare the reactants and note that $\ce{C5H5NH+}$ has a greater acid strength than the base strength of $\ce{F-}$  (ie $\mathrm{K_a} $ of $\ce{C5H5NH+} >\mathrm{K_b} \text{ of } \ce{F-} $) so the solution will be acidic. Its always useful to have this sort of sanity to check to make sure your answer makes sense. (For more details on this portion, Cengage has a chapter discussing acid/base mixtures)
A good approximation for the pH can be found in this paper, as shown here: $$[\ce{H3O+}]=\mathrm{K_a}*\sqrt{\frac{\mathrm{K_w}}{\mathrm{K_a * K_b}}}$$ where $\mathrm{K_a}$ is for $\ce{C5H5NH+}$ and $\mathrm{K_b}$ is for
$\ce{F-}$. This leads to $[\ce{H3O+}]=4.85*10^{-5}\to\mathrm{pK_a}=4.314$
As a comparison, the $\text{p}\ce{H}$ of $.325$M pyridinium is $2.86$, which it makes sense that the $\text{p}\ce{H}$ is higher when it is mixed with a base.


text:
Oxymercuration and demercuration mechanism
There are two proposed mechanisms for Oxymercuration and demercuration.
The first mechanism is concerted (i.e. single transition state connecting reactants and products) characterized by an attack of the $\pi$-bond on the $\ce{Hg(OAc)2}$.
Why does mercuric acetate act as an electrophile in the first mechanism? It is in its most common 2+ oxidation state. Does this happen because mercury has empty $6d$ orbitals.
The second mechanism is a stepwise mechanism (i.e. contains at least one intermediate species along the reaction pathway between reactants and products) that is characterized by incorporating the first ionization of $\ce{Hg(OAc)2}$ followed by an attack of the nucleophilic $\pi$-bond.
$\ce{Hg(OAc)2}$ dissociates in water as follows:
$$\ce{Hg(OAc)2 + H2O<=>Hg+OAc + OAc-}$$
Which one is the actual mechanism?

The mechanism you are looking for is probably not known and might well be never known. According to Brown et. al. there is a rapid and reversible equilibrium between associated and dissociated reactants.[1] They tested this for cyclohexene, norbonene and apobornylene, but I believe it is fair to generalise in this matter.
$$\ce{R2C=CR2 + Hg(OOCCF3)2 <=> (F3CCOO)R2C=CR2Hg(OOCCF3)}$$
Due to this, even the formation of a mercurinium ion seems to be sometimes not observed. For more information please have a look at the mechanistic studies of Pasto and Gontarz,[2] where they focussed on the stereo selectivity of substituted cyclohexenes, and references within.
Whatever molecule you might assume acting as the electrophile, it is quite obvious, that the bond mercury oxygen bond is highly ionic, with most of the electron density located in the acetyl moieties, leaving a rather substantial positive charge at the mercury itself. The following structures were optimised at DF-BP86/def2-SVP and the charges displayed are natural charges from the Natural Bond Orbital analysis to illustrate the above point.

The lower structure of $\ce{Hg(OAc)2}$ is slightly more stable.

Herbert Charles Brown, Min-Hon Rei, Kwang Ting Liu, J. Am. Chem. Soc., 1970, 92 (6), 1760–1761.
Daniel J. Pasto, John A. Gontarz, J. Am. Chem. Soc., 1971, 93 (25), 6902–6908.

I don't know the answer (and I thought for a while that it was not knowable), but I can at least tell you how you would find out whether $\ce{Hg(OAc)2}$, $\ce{Hg(OAc)+},$ or (more likely) some combination is the active electrophile.
If $\ce{Hg(OAc)2}$ dissociates, then both possible electrophiles are in solution. However, if we can effectively remove one electrophile, and see if the reaction works, then we know something.

Easy: shift equilibrium to favor $\ce{Hg(OAc)2}$

We can shift the equilibrium to heavily favor $\ce{Hg(OAc)2}$ by adding a large excess of $\ce{OAc-}$ ions, for example as $\ce{NaOAc}$.
There are two possible outcomes:

The reaction does not work (or goes very slowly). If this is the case, then $\ce{Hg(OAc)+}$ is the active electrophile.
The reaction does work. We know that $\ce{Hg(OAc)2}$ can be the electrophile, but this does not rule out $\ce{Hg(OAc)+}$ as an electrophile.

Hard: Shift equilibrium to favor $\ce{Hg(OAc)+}$

I am not coming up wit ha good way to do this, but if we could, then we could test whether $\ce{Hg(OAc)+}$ was necessary.

It's very likely that nobody knows the answer to this, because nobody was interested enough to find out. +1 for the proposed experiments.

Perhaps one of the acetate ions can be protonated at a low pH, but not low enough that both acetate ions are removed?

In the solution both
$\ce{HgOAc2} $ and $\ce{HgOAc+} $ exists.But why does $\ce{HgOAc2} $ acts as an electrophile here. Is it just because it has empty 6d orbitals.

When $\ce{Hg(OAc)2}$ acts as an electrophile, one acetate ion gets displace. I would not say that this displacement occurs by an associative or intermediate ligand substitution mechanism (because it is very challenging to tell the difference). If you had to identify some "orbitals" to be involved, why would you not consider the probably more accessible $6p$ orbitals?

All i am asking is why would even HgOAc2 behave as an electrophile here.It is just in its stable state, and higher oxidation states of mercury are not known except with compounds of flouride leading to +4 oxidation state.

@yasir - See Martin's great answer! It looks like even in $\ce{Hg(OAc)2}$, the mercury atom still bears almost a full positive charge balanced by substantial negative charge on the carboxy oxygen atoms. Mercuric acetate is ionic and mercury is big. An alkene can get the the mercury atom and will be attracted to the positive charge. The acetate groups are free to leave.


text:
Decomposition of axial vector and vector representions of C4v group
Let $R$ be the orthogonal matrix corresponding to an operation in $O(3)$. If
R is a proper rotation, then both vectors $\vec{V}$ and axial vectors $\vec{A}$ are transformed in the same way
$$ \vec{V} \rightarrow  \vec{V}' = R\vec{V}$$
$$ \vec{A} \rightarrow  \vec{A}' = R\vec{A}$$
if $R$ is an improper rotations, then
$$ \vec{V} \rightarrow  \vec{V}' = R\vec{V}$$
$$ \vec{A} \rightarrow  \vec{A}' = -R\vec{A}$$
And let $\mathcal{V}$ be the vector representation of $O(3)$ and $\mathcal{P}$ be the axial-vector representation of $O(3)$. Thus we can write,
$$\vec{V} \rightarrow  \vec{V}' = \mathcal{V}(R)\vec{V}$$
$$\vec{A} \rightarrow  \vec{A}' = \mathcal{P}(R)\vec{A}$$
My question is, considering $\mathcal{V}$ and $\mathcal{P}$ as representations of $C_\mathrm{4v}$ group, how can I find their composition in terms of irreps ?
$$\begin{array}{c|ccccc|cc} \hline
C_\mathrm{4v} & E & 2C_4 & C_2 & 2\sigma_\mathrm{v} & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_1} & 1 & 1 & 1 & 1 & 1 & z & x^2+y^2, z^2 \\
\mathrm{A_2} & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_1} & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 \\
\mathrm{B_2} & 1 & -1 & 1 & -1 & 1 & & xy \\
\mathrm{E} & 2 & 0 & -2 & 0 & 0 & (x,y),(R_x,R_y) & (xz,yz) \\ \hline
\end{array}$$


text:
How to perform oxidation of keto group in presence of hydroxyl group?
How to perform oxidation of keto group in presence of hydroxyl group. I fear there may be halogenation at $\ce{OH}$ group. Is there a way to protect the $\ce{OH}$ group ?

Protect the hydroxyl group, e.g. as an ether. Have a look here.


text:
Why is electron gain generally exothermic?

I understand that we need to supply energy to counter the nuclear attraction when we remove electrons, and that is the reason why ionization energy is endothermic. However, why does an atom release energy when we add an electron to it?
With respect to ionization energy, why is energy required to remove electrons of elements like $\ce{Be}$, $\ce{Li}$ that naturally want to lose their electrons?
My teacher told me that elements with fully filled and half filled highest occupied orbitals have positive electron affinity (endothermic). Why is this true?

The energy of an electron in a vacuum is zero. If it can attach (however weakly), the energy is gained and the process is therefore exothermic. It is not an atom that releases energy, it is the whole system.

Exactly opposite of previous. The atoms you mentioned gladly release electrons to other atoms, but not to vacuum (which is how ionization energy is defined).

In some cases, adding of extra electron to neutral atom is not an exothermic process. See Periodic table of electron affinities. For example, a helium atom has filled shell with $n=1$. The next electron would be accepted to the shell $n=2$ which has much higher energy, and therefore is not stable. For the half-filled orbitals, similar reasoning applies, but spicing it with Hund's rule.

I don't get it. 1) If an electron in vacuum attaches itself to the atom, energy is gained by the electron?  why? and doesn't "gain" mean endothermic? are you saying the energy lost by the atom goes into the electron that attaches itself?3) don't you mean "exothermic" instead of "endothermic"?  just to clarify endothermic- gain and exothermic -loss right? everything else you said makes sense.

Few words of thermodynamics... you describe the reaction $\ce{A + e^- -> A^-}$. If the system gains energy, the reaction is exotermic. Lowering the energy corresponds to more stable system.  Not atom, not electron, the whole system.
I hope I do not use the energy stuff too vaguely to confuse you.

i am rather confused.  Please do forgive me but i am new to this.

So the reaction is exothermic and so energy is gained by the whole system? But why does the reaction have to be exothermic? How is it exothermic? What happens for it to be exothermic? And does the system become stable again by  releasing the energy it got from the reaction?

Let's call classical mechanics to help. You drop 1kg weight from 1m to the ground. The system is the Earth and the 1kg weight. System goes to the state with lowest energy (guess which), process is spontaneous, energy is released, as long as we think only about gravitational interaction. The system gained these 10J which would finally change to heat. So the process is exothermic.

So comparing this to an atom and the  system, when the electron is pulled in, it drops from a higher electric potential to a lower one. And so energy is released and the system gains that energy.And that's how it's exothermic?

Looking at the comment thread above, I believe a bit more elaboration can help, so here it is
Why is electron gain generally exothermic?
Think of a unit positive sphere and a unit negative sphere. When is their system most stable? It is when the spheres are closest to each other. Why is that position stable? Because if we slightly disturb those spheres to take them farther from each other, they'll again come back to their original position, due to their attractive forces.
This stable position is an energy minima for this sphere system. When a system goes from a higher energy state to a lower energy state, it loses energy.
The same is the case with the system of the negative electron and the positive nucleus pair. The energy of an electron in vacuum (isolation from the nucleus) is zero (because it cannot feel the nucleus' attractive pull). But when it occupies a definite energy level in the atom, the energy of the system reduces. The extra energy is then released, which is why you find that the electron gain is generally exothermic.
Why is energy required to remove electrons of elements like $\ce{Be, Li}$ that naturally want to lose their electrons?
I assume you mean to say that $\ce{Na}$ easily loses electrons in presence of, say $\ce{Br}$ (assume $\ce{NaBr}$ ionic), even though $\ce{Na}$'s  ionization energy is positive. That is because the formation of $\ce{NaBr}$, or any ionic compound in general, involves a number of steps, each involving a definite energy change. Ionization of the sodium atom is just one of those several steps!
This is called the Born-Haber cycle. The general idea is that you start with solid sodium and bromine liquid, and eventually form solid sodium bromide. The net enthalpy change for this entire process is negative, or exothermic, hence it is favorable. The positive ionization enthalpy is compensated by the other processes which release energy.
However, in vacuum, the only process that occurs with the isolated (other ions are absent) gaseous sodium atom is to remove an electron. That process - which defines ionization energy - is still endothermic, because you need to oppose the nucleus' attractive pull.
Elements with fully filled and half filled highest occupied orbitals have positive electron affinity (endothermic). Why is this true?
That is true because elements with fully filled and half filled highest occupied orbitals are stable (you can read why that's true here). Adding an electron to this atom would make it unstable, thereby increasing the atom's energy. Now, don't get me wrong, when you add an electron to the atom: the energy of the nucleus-electron pair still decreases, but the rise in energy due to the instability is much more, causing a net increase in the energy of the atom.
Hence, in such cases, the electron gain process is endothermic.

When two oppositely charged particles are brought closer together under a Coulombic potential, the potential energy of the pair decreases. However, conservation of energy requires that this potential energy be converted into some other form (total energy is conserved). In a simple classical description we say that it is converted into kinetic energy, ie $-\Delta E_{Coulomb} = \Delta E_{kin}$, but the total energy remains constant, unless some of it is transferred to some other body. In the quantum mechanical description of atoms, electrons occupy discrete states, each state characterized by a total electronic energy (with potential and kinetic contributions) and a distance distribution between electron and nucleus (which can be used to compute an average distance). Transitions can occur between states through the exchange of discrete amounts of energy (quanta) with other bodies.
An ion formed by a neutral atom and an electron bound at a large separation is in a high energy state relative to the ground state, the relative energy being nearly the ionization energy of the ion. This is a highly unstable state because the slightest perturbation can drive electron and nucleus apart into an unbound (free) state. For the ion to settle into a lower energy (more stable) bound electronic state, corresponding to a smaller average distance between electron and nucleus, it has to release energy. It might do this radiatively (emitting photons) or through collisions with other atoms, dissipating the energy as heat. Another way of seeing this is that the newly formed ion has an excess of energy that it can give away to colder atoms in order to relax into a more stable lower energy state. QM dictates what are the most stable arrangements of electrons about nuclei, and only particular arrangements are possible. It is largely a geometric problem involving a balance between electron-electron repulsions, electron-nuclear attraction, the wave nature of matter evident particularly at small scales, and odd effects such as Pauli exclusion (the impossibility of two electrons having identical properties).

Considering now your list in more detail:

It doesn't always do so. Sometimes it's impossible. Not all combinations are stable. Only when the electron affinity is consistent with electron attraction is a stable combination possible. This is largely addressed above. Assuming it is possible to form a stable atom or ion, the combined particles have to shed energy to the surroundings, otherwise they risk falling apart again. When it gives off energy the combination relaxes from an excited state to a lower energy state. It might only shed some energy and remain in a highly excited (reactive) state, however.

This is addressed in another answer to your post. The argument is somewhat circular. Higher energy is less stable. Basically you are creating an unstable (excited) system when you attempt to remove an electron from a stable atom, so you have to add energy to do so.

This is addressed by the hand-wavy QM argument at the end of my dissertation. The combination of charge attraction and repulsion with the wave nature of subatomic particles and the Pauli exclusion principle lead to particular particle combinations and geometric arrangements being stable. This is the subject of a more rigorous QM course.


text:
CuCl2 dissolved in water
We dissolve $10\ \mathrm{g}$ of $\ce{CuCl2}$ in $0.1\ \mathrm{L}$ of water.
Given that:
$M(\ce{Cl})=35.5\ \mathrm{g/mol}$
$M(\ce{Cu})=63.5\ \mathrm{g/mol}$
How can one calculate the molar concentration of $\ce{Cu^2+}$ and $\ce{Cl-}$?
First of all I calculated the the quantity of matter of $\ce{CuCl2}$:
$n(\ce{CuCl2})=0.074\ \mathrm{mol}$
The the molar concentration:
$c=\frac{0.074\ \mathrm{mol}}{0.1\ \mathrm{L}}=0.74\ \mathrm{mol/L}$
Then I don’t know how can I calculate the molar concentration of $\ce{Cu^2+}$ and $\ce{Cl-}$ because I don’t know the quantity of matter of them!
But here im blocked!

You can assume that CuCl2 will be fully dissociated.

@JaroslavKotowski yes can this change something ?

Jaroslav already gave you a good hint: Assume full dissociation, which leads to the following equation: $$\ce{CuCl2 <=> Cu^2+ + 2Cl^-}$$
Since you know the initial concentration of copper chloride, $c(\ce{CuCl2})=0.74~\mathrm{mol/L}$, what can you say about the ratio of $n(\ce{CuCl2})/n(\ce{Cu^2+})$ and $n(\ce{CuCl2})/n(\ce{Cl^-})$?
Hint: In one mole of of a salt of the composition $\ce{AB}$, you will find one mole of $\ce{A+}$ and one mole of $\ce{B-}$, according to $\ce{AB <=> A+ + B-}$.
Therefore:

 $n(\ce{CuCl2})/n(\ce{Cu^2+}) = 1$, thus $c(\ce{CuCl2})=n(\ce{Cu^2+})=0.74~\mathrm{mol/L}$
 and
$n(\ce{CuCl2})/n(\ce{Cl^-}) =\frac12$, thus $c(\ce{CuCl2})=2\cdot c(\ce{Cl^-})=1.48~\mathrm{mol/L}$

I think that $n(\ce{CuCl2})/n(\ce{Cu^2+})$ is the concentration of Cu^2+ ?!

@user233658 That ratio is just a number, check the part in the spoiler tags, I included the solution.

How did you know that his ratio is 1 ? Is it because there is 2 molecule of Cu^2+ ? n(Cu^2+)=0.074mol ? How did you know it :/ ? Oh im feeling stupid :(

The formula gives you the ratio already. If you assume a molecule $\ce{CuCl2}$ there is one copper and two chlorine. So for each mole of copper chloride, full dissociation will give you one mole of copper ions.

When a salt is fully dissociated, than the atoms in a solution exist separately:
$$\ce{CuCl2<=>Cu^{2+}}+2\ce{Cl-}$$
Amount of substance is 0.074 mol. The molecule is made of 1 atom of Cu and 2 atoms of Cl.
$$ n=\frac{m}{M}$$
$$ n \cdot M =m$$
Amount of copper in the mixture then is: $$0.074~\mathrm{mol} \cdot 63.5~\mathrm{g/mol} = 4.7~\mathrm{g} $$ and the rest is formed by chlorine. $$ 10~\mathrm{g}- 4.7~\mathrm{g} = 5.3~\mathrm{g}$$


text:
How to add hydrogens to a crystal structure?
PDB protein crystal structures most of the time lack hydrogen atoms. Usually I use PyMOL (h_add) but it does not always seem to be giving the correct result. What can I use for accurate hydrogen addition?

I've used Reduce before, which is excellent. I'm not sure what you mean by ' does not always seem to be giving the correct result', though? Perhaps you could file a bug report with the PyMOL team if you think there is a bug in h_add.

Indeed, I think the PyMOL team would be highly interested in knowing about bugs in hydrogen addition from a PDB crystal structure.

@GeoffHutchison : True. Actually, it wouldn't be so hard to run h_add and Reduce over some subset of the pdb (small, high-res structures?) to see what differences there are - if any.

Olex2 (free, available for Windows, MacOS, Linux) can also be used to add hydrogens to any structure with the similar to PyMOL's command hadd. Example: hemoglobin PDB without H-atoms loaded in Olex2:

Executing hadd command adds hydrogens to all atoms where it is appropriate (taking hybridization into account), chemdraw assigns proper bonds order:

Alternatively, one can only populate certain atom types with hydrogens by selecting them first, e.g. for carbons: sel $C, then hadd. Also, one can hide/show added hydrogens by pressing Ctrl+H.
To save the file, type file name_of_protein.xyz. Saving in PDB also works, but the metadata/header is going to be lost, and sometimes the PDB file cannot be read properly by third-party software. So XYZ is the most safe option here.

There are many software packages that do what you request. PHENIX, Schrodinger, SCWRL4. It depends on how accurate you want, and options for pH etc.


text:
Haloform reaction in acidic condition
I want to know about the reaction mechanism of haloform reaction in acidic condition. I am following a persons doctoral thesis where he used acid medium haloform reaction but unfortunately he has not given any references.

Do you have a link to the thesis?

Please give necessary references, it took me for a surprise and I'm interested to know more.

In the book by Clayden et al. you can find a very nice description of the mechanism of the halogenation of enols and the haloform reaction on pp. 533-536. Especially on p. 536 the authors give the mechanistic details of the acid catalysed halogenation of enols and some reasons why multiple halogenation steps on the same carbon atom (as happens in the haloform reaction) occur easily under base catalysis but do usually not occur under acidic conditions. Since I lack the time to write it up myself (and since the authors already did a marvelous job at describing it) I simply post the relevant passage here:


text:
Vapour absorption in 'normal conditions'
When is described: "In normal conditions, viscose absorbs 11 to 14% of water vapour. In liquid water, they swell and can absorb 80 to 120% of water." (Clothing Technology)
What do they mean with 'normal conditions'? In open air?

Normal or standard conditions are usually ment pressure of 101.325 kPa and temperature of 298.15 K or 293.15 K. Please see the link for reference.


text:
Mechanism of enol ether formation

I assumed the the $\ce{TsOH}$ would dissociated to produce hydronium ions that would protonate the carbonyl. This would make the carbon more delta + and so susceptible to attack by $\ce{MeOH}$. After proton exchange I thought the $\ce{H2O}$ group would leave and we would have a carbocation left over. I can't work out how to resolve this carbocation with the changes in double bonds.

Based on what you have in the question, you are good up until formation of the carbocation by loss of a water leaving group. It looks like you are missing one of two things:

The carbocation is resonance stabilized.
One of the carbon-carbon double bonds in the product is new.

The scheme below has two of the major resonance contributors to that carbocation. Perhaps one of give you some insight on how to show the formation of the second alkene.

so we have an Elimination of H by MeOH on the bottom carbon of the RH ring?

That's correct as I see it.


text:
Law of mass action
If you have the forward reaction
$$\ce{2X ->[K] P}$$
which of the following systems of differential equations would model the reaction's kinetics?
$$\begin{array}{rl} \dfrac{\mathrm{d}[\ce{X}]}{\mathrm{d}t} &= -2K[\ce{X}]^2\\\\ \dfrac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} &= K[\ce{X}]^2\end{array} \tag{system 1}$$
or
$$\begin{array}{rl}  \dfrac{\mathrm{d}[\ce{X}]}{\mathrm{d}t} &= -K[\ce{X}]^2\\\\ \dfrac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} &= K[\ce{X}]^2\end{array} \tag{system 2}$$

The first answer is correct.  If the reaction is $\ce{2X->P}$, then two units of $\ce{X}$ should disappear for every unit of $\ce{P}$ formed. Only the first possibility meets this criterion.
The rate law for the reaction could be anything.  In your example you have apparently assumed that it is a second order reaction in $\ce{X}$, but we could make any other assumption.  For example, product formation rate could be $\frac{dP}{dt}=k~X^{1.2}$ or $\frac{dP}{dt}=k~\frac{X}{K_m + X^{2.4}}$.  But if the reaction you want to model is really $\ce{2X->P}$, then these rate laws would imply that $\frac{dX}{dt}=-2~k~X^{1.2}$ or $\frac{dX}{dt}=-2~k~\frac{X}{K_m + X^{2.4}}$.
Thus, if $\frac{dP}{dt}=k X^2$, then the rate of X depletion must be -2 times that, or $\frac{dX}{dt}=-2k X^2$.
Here is another way to see the problem with the rate law without the $-2$: suppose there was a reaction $\ce{X->P}$, i.e. only one molecule of $\ce{X}$ was needed to form $\ce{P}$.  Suppose also that this reaction followed second-order kinetics in $\ce{X}$, i.e. $\frac{dP}{dt}\propto X^2$.  What would the rate laws for this alternate reaction be?

Do you have a citation for this? I though the molar coefficients are taken into account by the multiplication of the reactant side. IE, that's where the squaring comes from.

Don't confuse stoichiometry with mechanism.  Mechanism determines the order of a reaction, i.e. the exponents in the rate law.  "Elementary" reactions like 2X -> P with two reactant molecules will indeed have a second-order rate law.  But that's just the exponent.  The stoichiometry is another matter.  If two molecules of X disappear to make one molecule of P, then the rate laws you write down better reflect that.  A citation for this could be any book on chemical kinetics.  I recommend Fogler.  Extra thought question: what would the rate laws look like for an "elementary" reaction 2X -> 2P?

another way to look at it is that if the reaction is $\ce{A + B \rightarrow P}$ then we would have say 1 mol A and 1 mol of B. But in your case if we have 1 mol of A it behaves as 1/2 mol because its A + A reacting, so we need to double up.

@CurtF. Thanks .Correct me for the following :1-i.e. $\frac{dP}{dt}\propto X^2$.  What would the rate laws for this alternate reaction be($\ce{X->P}$)?
$\frac{dP}{dt}=\frac{-dX}{dt} =kX^2$
2-  what would the rate laws look like for an "elementary" reaction 2X -> 2P?
$\frac{dP}{dt}=\frac{-dX}{dt} =2kX^2$

@porphyrin Thanks. Correct my answer:
**  what would the rate laws look like for an "elementary" reaction A + A-> 2P?**
$\frac{1}{2}\frac{dP}{dt}=\frac{-dA}{dt} =kA^2$


text:
Can 100% covalent bonds exist?
Every covalent bond has some ionic character and every ionic bond some covalent character. I can understand why a completely ionic bond is an ideal situation. But completely covalent bonds can exist(?).
Take the case of $\ce{H_2}$ or any other diatomic molecule between identical atoms. How can these have any ionic character?

It is common to see diatomic Bromine act ionically when being used as an electrophile.

As a combination of $\ce{H+}$ and $\ce{H-}$. The problem with this question is that ‘covalent bond’ and ‘ionic bonds’ — heck, actually ‘bonds’ in general — is a lot more theoretical of a construct than it might look. It does a good job of explaining a lot of things, but it remains a concept to explain reality; it is not reality. The closest to reality we can get is the calculation of molecular orbitals. Through mathematical tricks, we can derive ‘bonds’; that’s all.

@Alizter: But Bromine behaves ionically in a very specific chemical environment. The real question is if isolated bromine atoms have any ionic tendencies?

@Jan: I understand what you mean. Bonds are a theoretical construct. Yet theories are self-consistent (reasonably), and the fact that covalent bonds have ionic character is a statement the theory makes. I want to know if this statement is correct, why it is correct, and the reason behind stating it like this. The fact that bonds are theoretical should not arise in this discussion because it is a discussion purely about the theory itself.

If you have symmetric molecule you should think about about metallic not ionic character:.https://en.wikipedia.org/wiki/Metallic_hydrogen

http://chemistry.stackexchange.com/questions/31055/is-there-a-characteristic-that-characterizes-all-and-only-metals, http://chemistry.stackexchange.com/questions/17064/why-is-kf-the-most-ionic-compound

According to Pauling's famous The Nature of the Chemical Bond , 3rd edition, at page 73:

In the hydrogen molecule a quantum-mechanical treatment has shown that the two ionic structures $\ce{H+H-}$ and $\ce{H- H+}$ enter into resonance with the extreme covalent structure $\ce{H-H}$ only to a small extent, each ionic structure contributing only about 2 percent to the normal state of the molecule.

He goes on to state that other diatomics such as chlorine should have even less ionic character.
So all molecules, even homonuclear diatoms, have some ionic character.
For further information on diatomic hydrogen's ionic character see section 1.5 of the book and The Normal State of the Hydrogen Molecule Journal of Chemical Physics vol. 1, pages 593-596.
See also Overlaps of Trial Functions for the Hydrogen Molecule. II. Covalent and Ionic Character of H2 J. Chem. Phys. 23, 176

In other words, the bond is not something static - no stick gluing together two balls. In the end, the bonding is a quantum phenomenon, and has to be treated as such to get a precise picture of what's actually happening. The older bonding theory is just an approximation of the actual interaction.

@DavePhD when you apply hannay smith equation,the % ionic character will be zero(by applying pauling equation too),doesn't that mean homonuclear diatomic molecules are perfectly covalent??(since % ionic character is 0)

@VidyanshuMishra See the two definitions of "ionic character" discussed in the following reference:  https://digital.library.unt.edu/ark:/67531/metadc228342/m1/34/ By the first definition, homonuclear diatomic molecules have ionic character, while by the second definition they do not.

Check your first link.  It sends me to wrong pages of the book.


text:
What is GMEC (Global Minimum Energy Conformation)?
In a paper I have read the term GMEC, which was not further explained.
GMEC stands for Global Minimum Energy Conformation. But what is it?
Is it just the conformation with the lowest energy?

Yes. It is the conformer with the lowest energy out of all possible conformers.
There are also local minima which have a lower energy than conformations either side of the them on the energy diagram but do not have the lowest possible energy.

In this diagram showing the conformers of butane we can see that all three stable conformers; the two gauche conformers and the anti conformer are at local energy minima but the anti conformer is the global minimum.

GMEC is typically used within the GMEC-model, a model within the protein design field that entails the following. The backbone of the protein is not allowed to move, the interaction energies between atoms are pairwise additive, the amino acid side chains are allowed to change only between discrete states, c.q. rotamers, and the goal is to compute the GMEC and its corresponding amino acid sequence.
So depending on the context where you came across this ambiguous term, it may carry more assumptions than would be clear at first sight.


text:
In spectroscopy, is it possible for the sample to be excited multiple times?
The context of my question is rotational spectroscopy (using microwave radiation), where there is the $J = \pm 1$ selection rule. The way I understand spectroscopy is that the sample is irradiated with a range of frequencies. Is it possible, during the irradiation of the sample (say CO), for the molecule to be excited by 2 levels by 2 successive photons of the appropriate frequencies?
Now this does not affect the locations of the peaks per se, because the relaxation of course has to happen one level at a time. I am specifically wondering about the distribution of peaks, which is supposed to follow the Boltzmann distribution. Let us say, for example, we have a sample at absolute zero, where all the molecules are at the ground state $J = 0$. Is it possible for a molecule to be excited twice (or more), leading to the rotational spectrum showing more than one peak?
Thank you in advance.

Very technically? Yes.
Realistically? The probability is small enough that even if it does happen, the peaks for the multiple transitions are going to be small enough that we cannot really observe them on the spectrum. The lifetime of a given excited state is so small compared to the analogous time in the ground state that it can basically be considered zero for the situation you're talking about. I'm sure it has happened, but compared to the one at a time transitions, your spectrum isn't really going to show it.

I see. I was thinking that the pulse width would be short enough such that this possibility would be very small, so I was wondering more about any theoretical possibilities that would rule this out (e.g. some rule of QM that I don't know about). Thanks for the explanation!

It's certainly a different story for rotational spectroscopy, but excited state absorption can occur quite readily in some materials as one of the phenomena leading to photon upconversion.

see: https://en.wikipedia.org/wiki/Hot_band
or: https://en.wikipedia.org/wiki/Two-photon_absorption

From memory something like this has been used as a basis for isotope separation using intense $\ce{CO2}$ lasers to fragment molecules. A process of 'ladder climbing' takes place aided by the fact that the electric field of the laser is so intense that it can bring levels into resonance that would otherwise not be so. Hence ladder climbing is possible. Multi-photon effects should also occur.

Multiple excitations are unlikely to occur incidentally (at least to the extent where you would get an appreciable signal), but I know of research groups that look at multiple excitations as a way of triggering reactions or as a means of studying excited states via pump-probe spectroscopy


text:
What happens if a polymer's length is equal to its persistence length?
So, I've been looking into polymer chemistry to model breaking polymers for a biological model, and I've been a bit stumped by the concept of persistence length.
I know that if a polymer is significantly longer than its persistence length, it needs to be treated as though the ends are completely independent of each other thanks to thermal fluctuations bending it all over (the analogy I have is treating the polymer like a piece of cooked spaghetti). If I model a polymer like this, I need to use the words "3 dimensional random walk" to describe its shape mathematically.
I also know that if a polymer is significantly shorter than its persistence length, it can be treated like a rigid rod that might be a bit bendy, like a steel rod. This is significantly easier to model: I can just treat it as a line/arc that moves through space as a single particle.
But what happens if a polymer's length is about equal to its persistence length? How can the polymer be modeled then? What precisely does it mean for the polymer's movement?
EDIT: I've been debating the merits of just treating it like a flexible rod (like the significantly shorter than persistence length example) and calling it good just because of model simplification. If I do that, how off would I be from reality?
EDIT2: I've been doing further research and found something called a "worm like chain" model. I'm pretty fuzzy on what it means for the fiber, and I'm having trouble grasping how the fiber would move from an intuitive perspective; the visual for how I'd need to treat it isn't working, and the visual would help immensely with the actual modeling for the rest of the equations I need.

Can you give us more information about the specifics of the model you want to develop? Is this a computational model (like a simulation), or a mathematical model, for example? The persistence length is a statistically defined property, and so it is usually only important and meaningful in statistical models. When the chain length is much longer or shorter than the persistence length, the effect is usually that one or the other terms drop out of the equation. If they are about equal, then neither term would drop out.

The model is statistical, and the issue I have is that I know the length of my polymer is about equal to its persistence length. I'm trying to get a more thorough understanding of what that means so that I can derive a more detailed model of the polymer fragmentation.

You should read https://en.wikipedia.org/wiki/Persistence_length - spaghetti is nowhere near its persistence length.

Wikipedia refers to uncooked spaghetti.

If my answer doesn't give you enough information, you might also try Polymer Physics by Rubinstein and Colby for an introductory look at polymer statistics, or The Theory of Polymer Dynamics by Doi and Edwards for a more detailed and complete overview.

The persistence length is a statistically defined property - in other words, it is defined by looking at many lengths of many polymer molecules over a long period of time. It is the length (L) at which a vector drawn tangent to the polymer is no longer correlated in space with another tangent vector at distance (L). In other words, it tells you how well the position of one section of a polymer is correlated to another section. Beyond the persistence length, you would not expect to find any correlation. Below it, you would expect correlations.
In this case "correlations" are measured by the cosine of the angle between the two vectors. When the angle is 90 degrees on average, the cosine is zero, and when it is zero degrees, the cosine is 1. So, for lengths beyond the persistence length, the amount of correlation is zero, and for lengths much less than the persistence length the amount of correlation is one. For lengths close to the persistence length, the correlation is between 0 and 1.
To give you an idea of how this works, take a look at the following illustrations:

The red lines represent tangent vectors along the backbone of a polymer chain, spaced at regular intervals. We can translate all of these vectors and put their origin at the same point, and then find the angle between them:

If we find another polymer chain, or watch this one for a little while and take another "snapshot" and do the same thing, we might end up with something that looks like this instead:

We can even take the same very long chain and find all of the vectors that are a distance 1 apart, 2 apart, and so on. If we do this for all possible distances over hundreds of polymer molecules, and average over many snapshots for a very long time (nano- to micro- seconds for a melt or something in solution), what we will eventually find is something like this:

The correlations drop off quickly with length, then more slowly, until finally they are completely uncorrelated.
It is important to realize here that at any given time, for any given polymer, the tangent vectors could be almost anything. You wouldn't expect two adjacent bonds to have a very small angle between them, and in general you would expect more flexibility over longer lengths of chain compared to shorter lengths, but you would not be able to predict the shape or angles with any degree of certainty, other than to say that on average, beyond the persistence length the tangent vectors would not be correlated, and that the amount of correlation would decay exponentially with length.
The examples you gave of a stiff rod vs. cooked spaghetti are meant to demonstrate this intuitively, but your question is about what happens when the persistence length and the polymer itself are the same length. The answer is that the polymer will be somewhere between a very flexible piece of cooked spaghetti and a stiff rod. Imagine taking a piece of spaghetti and bending it into a loop, but not enough that it breaks. The circumference of 1/4 of the loop would be something like the persistence length, and the response of the spaghetti in the loop would be similar to a polymer that is about the same length as its persistence length. However, keep in mind that polymers are constantly in motion, and so "flexible rods" and "cooked spaghetti" are not great analogies in the first place. It's more realistic to think of them as constantly wriggling worms, or something like that. Statistical descriptions of correlation lengths and times are the best we can get, unfortunately.
Another concept that might help you visualize this is that of the Kuhn length - this is twice the persistence length, and is the length at which sections of polymer are uncorrelated with each other in space. Practically, it means that the motion of the polymer within the Kuhn length is irrelevant to the motion of the overall chain. On a large scale, the motion of the Kuhn "blobs" is what matters. In your case, the polymer is about as long as the persistence length, which means it is one-half of the Kuhn length. This means motions along the polymer are correlated, and so you can't ignore them (if they are important to your model.)
Practically, what this all means is that your model is probably going to be fairly complicated, because you can't average out or ignore terms that can be ignored in either of the extreme cases.
If you post some more details about your model, I could probably give you more specific information.

The main specific thing I'm trying to find is a way to determine how the polymer will arc: in detail, whether parts of it will do so semi independently of other parts, and statistically, the distribution how many degrees each arc will be.

The persistence length isn't really related to an arc angle in the way I think you mean. I am not sure that there is a well defined concept of arc angle for polymers at all - polymer dynamics are very random and the conformation of a single polymer at any given instant could be anywhere from linear to tightly coiled. How do you want to define the arc angle?

The arc angle would be the central angle of the theoretical circle the arc would be a part of: eg a quarter circle would be a 90* arc.

How would you map the polymer backbone to a circle? What I mean is, over what length scale do you expect to find two dimensional arcs? I would be very surprised if you could define an arc in two dimensions over more than three backbone bonds. In that case, the distribution of arc angles would be the same as the distribution of the dihedral angle.
It might help if you could explain what physics your model is intending to capture - I have a feeling that you have a conceptual picture of polymer dynamics that is much more orderly than how they behave in reality.

I'm hoping I can at least simplify the model a little by saying that all the bends in the polymer are a linear combination of circular arcs. What I'm trying to find is basically a rough distribution of the angles of these arc elements. If there's a distribution (not just a mean, but an actual distribution) for these angles, then that would be absolutely perfect for what I am looking for.

The dihedral angle will tell you the angle between bond vectors leading into and out of adjacent monomers. Energies of dihedrals are published for most common polymers. It has been a while, so I am fuzzy on the details, but I believe you can generate a probability distribution for the dihedral angles based on their energies and the temperature of the system you are studying, similarly to a Boltzmann distribution. Keep in mind that these are 3-D and there are lots of other considerations - so this still won't capture much reality, and might be more difficult than other options.

Thank you very much, thomij. I know I must seem to be just a little bit abrasive, asking four billion questions on top of the original, but you've been an enormous help.


text:
Hypervalency in elements in the second period
In my experience, most texts that address hypervalency say that it only occurs from elements in the 3rd period and onwards. This explains the occurrence of $\ce{Cl2O7}$ or chlorine heptoxide. However, some 2nd period nonmetals like $\ce{C}$ and $\ce{O}$ show hypervalency.
Examples:

$\ce{CH5}$ - This is unlikely to occur but it does sometimes happen that carbon bonds to 5 atoms instead of 4.
$\ce{H3O+}$ - Here oxygen is hypervalent.

How is it possible for carbon and oxygen to each have 9 electrons if each orbital only holds 2 electrons? Do they switch between electrons or something?

I have improved the formatting of your question using $\LaTeX$. For more information on how to do this yourself please see here.

Do you mean $\ce{CH_5^{+}}$?  Also, why do you call $\ce{H3O^{+}}$ hypervalent?  There are just 8 electrons around the central oxygen; 6 (3x2) in the 3 $\ce{O-H}$ bonds and 2 residing in a lone pair.

But it has 3 bonds which oxygen doesn't normally have since it hates being positive and would rather be negative if charged at all. Wouldn't that extra bond cause oxygen to have 9 electrons and thus be hypervalent?

And yes I do mean the methyl cation.

Normally when we talk about a single covalent bond, we are referring to a 2-centre 2-electron bond, which means that there are two electrons holding two atoms together.
Carbon never forms 5 bonds. The only exception that I know of is the $\ce{CH5+}$ methanium cation, the bonding in which can be explained by a 3-centre-2-electron bond. The same kind of bond appears in diborane ($\ce{B2H6}$). In both cases, the octet rule (or duplet rule in the case of the bridging hydrogens in diborane) is not violated. It is just that those 2 electrons are shared amongst 3 different atoms, so each "bond" is effectively half a bond (in MO theory parlance we say that the bond order is 0.5). You could think of it as three of the C-H bonds being normal 2-electron bonds, and two of the C-H bonds being half-bonds (having one electron each). The total number of electrons around carbon is therefore $3 \times 2 + 1 + 1 = 8$.
The neutral species $\ce{CH5}$ does not exist, because it has one more electron than the $\ce{CH5+}$ cation. That would mean that you either have to put 9 electrons around carbon, or put 3 electrons around hydrogen, both of which are of course not allowed.
The hydronium ion $\ce{H3O+}$ is not actually hypervalent. It is similar to the ammonium ion $\ce{NH4+}$ in that a dative bond is formed from the lone pair on O to a $\ce{H+}$ ion.

Nice answer!  BTW, $\ce{CH5^{+}}$ is not the only example of "5 coordinate" carbon.  All non-classical carbocations fall into this category of "5 bonds to carbon" through the use of the 3-center 2-electron bond.

Whether the carbon in $\ce{CH5^+}$ has five bonding linkages is ambiguous. There could be four bonds to carbon with one having side-on H2 at the other end. See the description of the ions in this answer.


text:
Does a catalyst increase atom economy?
Somebody I know insists that the use of a catalyst increases the atom economy. They did chemistry at school and were told that a catalyst increases the atom economy. He pointed me to several past exams where this was written as a valid answer.
However, I am convinced that this is false. After all, atom economy is:
$$\text{atom economy} = \frac{\text{molecular mass of desired product}}{\text{molecular mass of all reactants}} \times 100\%$$
I feel this has nothing to do with a catalyst.
Am I missing something?

There are a couple of ways to think about this. First, considering your atom economy equation. Typically a catalyst is used in place of a stoichiometric reagent. A catalyst is (strictly speaking) not a reagent, so its effect on the atom economy is neutral. The alternative stoichiometric reagent is a reactant, so it will reduce the atom economy. This leads to the conclusion that a catalytic method offers a better atom economy than an alternative, non-catalytic method.
From a slightly different point of view, atom economy is one aspect of measuring the amount of waste that comes out of a reaction. A more relevant quantity is really the ratio of the mass of the desired product(s) to the total mass of all of the reaction inputs (including reactants, catalysts, solvents, purification materials, etc). Since catalysts can be used in smaller amounts than stoichiometric reagents, there will physically be less mass put into the process, which increases that ratio.

It's a bit of a blanket assumption as processes exist where the catalyst is over-stoichiometric. This would knock the atom economy argument on the head.  Also one could argue that the catalytic reaction is no different as without a catalyst, the reaction still occurs albeit more slowly, all other things being equal.  It's a completely reaction specific consideration. Technically the catalyst mass shouldn't be included in the calculation.  Atom economy is a bit of a misleading measure, mass intensity is better.

using catalysts mean lower temperatures and pressures can be used. so energy saved, meaning less co2 is released, and fossil fuel reserves are preserved. catalysts can also reduce waste by allowing a different reaction to be used with a better atom economy.

Huh, this is a bit going far.

Most of this is irrelevant to the question.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. At the moment this reads more like a comment than an actual answer - could you elaborate a little more. With a bit more rep, you will be able to post comments on any question/answer.

Catalysts do have an effect on the atom economy, since catalysts have act sides, they eliminate  side products( waste), and they are not considered in the equation for atom economy, thus they will increase the atom economy.

That's wrong on so many levels. A catalyst might reduce side reactions, but he doesn't have to. It might even give you the possibility to run a reaction with worse atom economy (but cheaper or faster than alternatives). I don't know what "act sides" should be and google doesn't find anything either. And strictly speaking if you don't recover the catalyst it is lost and decreases atom economy.

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


text:
Role of Acetonitrile and Ammonium Acetate buffer in Leucomalachite Green stock solution?
I am trying to repeat an experiment conducted by FDA on Malachite Green (MG).
According to this paper,
Prepare a series of LC calibrants by aliquoting into individual 15 mL graduated
glass or disposable polypropylene centrifuge tubes, 50 µL, 100 µL, and 200 µL
of the 0.1 µg/mL MG2 solution and 50 µL and 100 µL of the 1.0 µg/mL MG1 solution.
Dilute each to 5.0mL with (1:1 vol) acetonitrile/ammonium acetate buffer
(mobile phase A), vortex mix or stopper and invert to mix thoroughly.
Based on a 5 g sample weight, these solutions will generate standards with
concentrations of 1, 2, 4, 10 and 20 ng/g of MG.

If I am not wrong, Malachite Green used in this paper is from the salt, Malachite Green Oxalate.

Is Acetonitrile used here as a pH buffer?
What is the role of Ammonium acetate?

If I plan on using Raman Spectroscopy to detect MG in my samples, shouldn't I be using the pure forms of MG as my reference signal? If so, wouldn't the addition of these two chemicals affect my readings?
References
The pdf document originally linked in this post (see edits) is no longer available.
The most similar reference identified (available here via the Internet Wayback Machine) is:

Wendy C. Andersen, José E. Roybal, Sherri B. Turnipseed. Determination of Malachite Green and Leucomalachite Green in Salmon with In-Situ Oxidation and Liquid Chromatography with Visible Detection. Laboratory Information Bulletin (LIB) 4334: Malachite Green and Leucomalachite Green in Salmon. Volume 20, No. 11, November 2004.

Since they're using ACN/ammonium acetate to elute from solid phase extraction cartridges and that's what they're injecting into their LC analysis, they probably want to use the same mixture to avoid problems with unmatched matrices with their standards. The ammonium acetate is the pH buffer. Malachite green has different protonation states at different pH, so it's important to set the pH properly for elution from the SPE and LC analysis. MG is fairly soluble in water, but still has a lot of organic functionality, so adding an organic solvent would help elute from the SPE column and keep the analyte from being too strongly retained by the LC stationary phase.
As for raman analysis, water has rather weak raman scattering, so it's not usually a problem. Acetonitrile has one very strong peak at 2942 $\mathrm{cm^{-1}}$ and a few moderate peaks, so it might be an issue. Since you're not doing LC, you can likely prepare standards in buffer only. Unfortunately, ammonium acetate is also raman-active, but if you're not doing LC, you might be able to do without as well. Fortunately, one of the great powers of IR and raman spectroscopies is that there is often enough bandwidth to find one or two non-overlapping peaks in a complex mixture and use that for quantitation, so it may still work as written. They're both also very amenable to chemometric techniques like partial least squares when measuring complex samples.
The big problem is going to be sensitivity if you're interested in the concentrations they use in the paper.. Normal raman is not that sensitive and the smallest standard is only ~3 nM, so it's probably going to be hopeless. These guys do surface-enhaced resonance raman and are able to get to about 1 nM, and the raman setup's not actually that hard since MG has a huge absorbance at 620 nm, right where HeNe lasers emit. Unfortunately, MG doesn't fluoresce much in solution or it would be very easy to measure.

Thanks for the detailed answer. It is very informative. Can I just prepare my standards in methanol without the buffer? And further dilute it into various ppb solutions also in methanol? Since methanol evaporates fast, this allows it to dry quick on the SERS substrate. What do you think?

They use acetonitrile in that paper, but I don't see why you couldn't use methanol, you'll just need to be more careful with your standards as it evaporates very quickly. You may also want to use the chloride salt of MG rather than oxalate, to avoid complicating your spectrum.


text:
What's the difference between a tetrahedron and a trigonal pyramid?
The shapes respectively denoted by 'tetrahedron' and 'trigonal pyramid' seem to be the same. Is there a difference between the two? If not, why are the two presented as different concepts in introductory chemistry courses?

trigonal pyramid has lone pair instead of substituent

colloquially, a tetrahedron is regular and trigonal pyramid is not. Methane (with 4 identical bonds) is a tetrahedron (4 identical bonds enclosed in a regular tetrahedron). But ammonia is less symmetric with three bonds and a lone pair enclosed in, effectively, a squashed tetrahedron usually called a trigonal pyramid.

In a rigorous geometrical sense, there is no difference between tetrahedron and trigonal pyramid--the terms both mean the same thing. In colloquial and chemical use, however, 'tetrahedron' typically implies the 'regular tetrahedron', where all four faces are equilateral triangles.
Chemically speaking, when referring to these two shapes as descriptors of molecular geometries, there is (usually) a central atom in addition to the four atoms at the vertices.  In the symmetrically-proper regular-tetrahedral geometry, the central atom is inside the solid volume of the tetrahedron and equidistant from all of the four vertex atoms.  In the trigonal pyramidal geometry, the central atom can be located inside the solid volume, contained within a plane of the trigonal pyramid/tetrahedron, or I believe even outside the tetrahedral volume.

Can you please explain this line: "the central atom can be located inside the solid volume, contained within a plane of the trigonal pyramid/tetrahedron, or I believe even outside the tetrahedral volume."

A handy guide to visualise and explain the chemical differences between tetrahedron and trigonal pyramid geometries comes from the The Nobel Prize in Chemistry 1971: Gerhard Herzberg page:

There is a handy explanation of the geometries in a YouTube clip referenced to Grand Valley State University: Molecular Geometry and Polarity: Tetrahedral (trigonal pyramidal)

This certainly is one paradigm for illustrating the difference. However, I'm pretty sure it's still possible to have trigonal pyramidal geometry even with four substituents, as long as one of the four differs from the other three.


text:
What is the explanation of the odd even effect in liquid crystals?
What is the most accepted theory for explaining the odd even effect in liquid crystals? Can anyone explain and refer the corresponding paper?

Do you have access to a literature search tool like e.g. SciFinder, ISI Web of Science, or even Google Scholar? Have you already done some searching using any of those methods?

I hope you mean liquid crystals with LC. I also added the reference request to the question, if I introduced any mistakes, please [edit] yourself.

I have been busy in going through more and more about main chain LC and did not search rigorously. But I did search about odd even effect but did'nt get any good results. Searched in Google. I will try to look whether my Institute has SciFinder access or not


text:
Perchlorate salts: (explosive) reactivity with methane?
I read everywhere about the explosive reactivity of perchlorate salts (e.g., $\ce{Na}$, $\ce{Ca}$, $\ce{NH4}$ [especially]) when mixed with organic materials or solvents, yet I find no information about its reactivity with methane.
Can anyone comment please?

In general, explosion of alkali metal perchlorates (at least in small quatities) is deflagration, with the perchlorate providing oxygen for combustion of another component of a mixture, rather than detonation (due to decomposition of the perchlorate). However, perchloric acid itself and other inorganic and organic perchlorates such as cis-bis-(5-nitrotetrazolato)tetraminecobalt(III) perchlorate can transition from deflagration to brissant detonation.
So, to answer your specific question, gaseous methane would burn more readily in the presence of Group I perchlorates, but the flame-front would be limited by the small contact area and low concentration of methane at the solid/gaseous interface. On the other hand, if one were to synthesize methyl perchlorate or  ethyl perchlorate, one might not be able to perform subsequent syntheses.
So, to avoid maiming or perhaps a terminal experiment, eschew perchlorates.

Thank you for your detailed response and valid safety concerns. However, mine is a natural (i.e., geochemical) laboratory, so the 'career-closing' risk to my syntheses should not be an issue.

The perchlorate in question is Calcium perchlorate hydrate and the solution has methane bubbling through it. The temperature is low (~ 5°C), but the solution is under sustained pressure. While deflagration may be implied, observation (at distance!) has confirmed detonation.
I presume that reaction remains thermodynamically favourable (but with a high kinetic barrier?), so is my reactant list incomplete somehow or does the pressure contribute catalytically?

There is also reason to believe that someone has subjected the site to impact-testing (but without leaving any body-parts behind!)

Though methane is pretty stable, is it possible that methyl perchlorate formed? If the pressure bomb(!)  is unlined metal, it could be that the container helped catalyze the reaction, or even reacted with the calcium perchlorate solution. Sorry, these are just suppositions.

Thank you again Dr. P -- the possibility of methyl perchlorate synthesis is tantalising, so I shall ask a new question on this very subject


text:
Computing initial reaction speed given reaction constant
Problem: $\ce{N2O}$ falls apart in oxygen gas and nitrogen gas according to a first order reaction rate. If you know that the rate constant is $0.38\ \mathrm{s^{-1}}$ at $760\ \mathrm{^\circ C}$, compute then the initial rate for the decomposition of $3.00\ \mathrm{g}$ $\ce{N2O}$ in a barrel of $5.0\ \mathrm{L}$. When the initial concentration doubles, what happens with the rate?
Attempt at solution: I set up the reaction as $$\ce{2 N2O -> O2 + 2 N2}$$ The molar mass of $\ce{N2O}$ is $44.019\ \mathrm{mol/g}$, and since we got $3.00\ \mathrm{g}$, this is $132.057\ \mathrm{mol}$. Then I computed $[\ce{N2O}] = 27.01\ \mathrm{M}$ since we had to divide by $5$ because the barrel has a volume of $5.0\ \mathrm{L}$. Then we have $$v = k [\ce{N2O}] = 10.26\ \mathrm{M/s}$$ But I’m not sure if this is the correct approach. Any help please?

Your calculations are correct except for the step where you calculated the amount of N2O (i.e. the number of moles). What are the units of molar mass?

Oh thanks, my bad. They are g/mol. So I have to do $1/44.019$, then multiply this with $3.00$ g to get the amount of mol. Then I divide by 5 to get the correct molarity, and substitute that into the equation for $v$?

yeah, that would work.


text:
Relative acidities of alkanes, alkenes, and alkynes
How does one explain the trend in bond lengths and acidity of the following hydrocarbons?
$$\begin{array}{ccc}
\hline
\text{Species} & \ce{C-H}\text{ bond length / Å} & \mathrm{p}K_\mathrm{a} \\
\hline
\text{Ethane, }\ce{C2H6} & 1.102 & 50 \\
\text{Ethene, }\ce{C2H4} & 1.085 & 44 \\
\text{Ethyne, }\ce{C2H2} & 1.061 & 25 \\
\hline
\end{array}$$
Shouldn't the shorter bonds between $\ce{C-H}$ be easier to break than longer ones, making ethane the most acidic?

You're right in that bond length, and therefore bond strength does affect acidity (see: $\ce{H2S}$, $\mathrm{p}K_\mathrm{a} = 7$ and $\ce{H2O}$, $\mathrm{p}K_\mathrm{a} = 15.7$). If we defined acidity with the following equation
$$\ce{HX -> H + X}$$
then the bond strength would indeed be the only deciding factor in the acidity of $\ce{HX}$, since the enthalpy change of that process is literally the bond dissociation energy.
However, acidity actually corresponds to the heterolytic dissociation of the $\ce{H-X}$ bond, with both electrons in the bond going to $\ce{X}$:
$$\ce{HX -> H+ + X-}$$
so bond strength isn't the only factor. In the case of ethane, ethene, and ethyne, the most important factor is the type of orbital that the electrons from the $\ce{C-H}$ bond end up in after deprotonation. For ethane, the two electrons end up in a carbon $\mathrm{sp^3}$ orbital; for ethene, an $\mathrm{sp^2}$ orbital; and for ethyne, an $\mathrm{sp}$ orbital.

Now, of these three, the $\mathrm{sp}$ lone pair is the most stable, since the $\mathrm{sp}$ hybrid orbital contains the most s character ($50\%$) and therefore has the lowest energy. So, of the three carbanions, the lone pair in the acetylide ion $\ce{HC2-}$ is most stabilised, making ethyne the most acidic molecule.
The acidity of terminal alkynes is in fact pretty useful from a synthetic viewpoint, since you can deprotonate them with a strong base like $\ce{NaNH2}$, forming a carbon-based nucleophile which can then be used to make new $\ce{C-C}$ bonds.

As for why the electronic stabilisation outweighs the bond strength, it's worth looking at some quantitative data. The $\ce{H-O}$ and $\ce{H-S}$ bonds have mean bond enthalpies of $463$ and $338~\mathrm{kJ~mol^{-1}}$ respectively (source: Physical Chemistry 9th ed., Atkins & de Paula, p 932), which is a difference of $\mathbf{125~kJ~mol^{-1}}$. This can, to some extent, explain the difference in acidities of $\ce{H2O}$ and $\ce{H2S}$.
On the other hand, the dissociation energies of the $\ce{H-C\mathrm{(sp)}}$ and $\ce{H-C\mathrm{(sp^2)}}$ bonds are given in J. Phys. Chem. 1987, 91, 17-19 as $132.6$ and $116~\mathrm{kcal~mol^{-1}}$ respectively. The difference is roughly $17~\mathrm{kcal~mol^{-1}}$, which is equivalent to $\mathbf{69.5~kJ~mol^{-1}}$ - a somewhat smaller value than before. You can see why bond length variations might therefore play a smaller role in the hydrocarbons.

An alternative
Another equivalent way of looking at it is that an $\mathrm{sp}$-hybridised carbon is more electronegative than an $\mathrm{sp^2}$-hybridised carbon, which is in turn more electronegative than an $\mathrm{sp^3}$-hybridised carbon.
As mentioned earlier, an $\mathrm{sp}$ orbital has the greatest s-character, and consequently electrons in a carbon $\mathrm{sp}$ orbital experience a greater effective nuclear charge and are more tightly bound to the nucleus than electrons in $\mathrm{sp^2}$ or $\mathrm{sp^3}$ orbitals.
Of course, the $\ce{C-H}$ bond also involves a contribution from the hydrogen 1s orbital. However, the above means that the electrons in $\ce{C(sp)-H}$ bonds are more strongly attracted to the carbon nucleus than electrons in $\ce{C(sp^2)-H}$ or $\ce{C(sp^3)-H}$ bonds. This means that the $\ce{C(sp)-H}$ bond is more polarised towards carbon, leaving less electron density on hydrogen, and a greater acidity.
The amount of electron density on the hydrogen can in fact be probed using $\ce{^1H}$ NMR spectroscopy. A lower electron density translates into a higher chemical shift. Even though there is a second, unrelated, factor - anisotropic shielding - that serves to decrease the chemical shift of acetylenic hydrogens, they still tend to show up at higher chemical shifts than hydrogens on saturated hydrocarbons (~2 ppm, compared to ~1 ppm). This is a clear indication that the local electron density on a proton attached to $\ce{C(sp)}$ is less than that of a proton attached to $\ce{(sp^3)}$.

How does less electron density on hydrogen imply greater acidity?


text:
Energy of photon
What is a photon and how is its energy directly proportional to frequency of light (please don't tell that it is because $E=h\nu$ relation)?

Please explain in detail .

I'm voting to close this question as off-topic because it is primarily about physics.

A photon is a quantum of light.
To explain more thoroughly, everything in the universe has elements of both particles and waves. So while light is usually thought of as a wave in classical mechanics, it also consists of particles. One photon is the smallest amount of light you can have while still having light at all.
The relation between the energy of a photon and its frequency (as well as how its particle nature was realized) was determined by Albert Einstein in a treatise on something called the photoelectric effect. The basic experimental set up was fairly simple. People had observed that, when shining light on a metal's surface, sometimes electrons would start popping out. The hypothesis proposed by classical mechanics was that this was proportional to the intensity of the light shining on the metal.
However, when the experiments were carried out, the results were a bit more surprising. The light had to be a minimum frequency before electrons were ejected at all. If the intensity of the light were increased at that frequency, more electrons would come flying out, but each electron would still individually have the same kinetic energy. If the frequency of the light was below the threshold, then no electrons would come out regardless of how intense the light was. And the higher the frequency of the light (once you got above the threshold frequency), the faster the electrons flew out.
So, Einstein realized what this meant: light had to exist as a particle as well as a wave. Each light particle (which he called a "photon") could interact with one electron at a time. Higher intensity light just meant more emitted photons, so the individual photon had to have enough energy to dislodge the electron from the metal surface. According to the experiments on the photoelectric effect, this energy was proportional to the frequency of the light, as noted before. When Einstein did the calculations, the constant of proportionality was $6.626\times10^{-34}~\mathrm{J~s}$, better known as Planck's constant and represented by the symbol "h."
The fact that this exact constant appeared in something seemingly completely different (the law for how blackbodies emit light) was a major basis for more investigation and eventually yielded quantum mechanics as we know it today.

I get two points clear from your answer that no. of electrons to be ejected depends on intensity and ejection of electron from surface depends upon the frequency but I didn't get that how energy of photon depends upon frequency of light

The observation was made that the energy of the photon depends on frequency by measuring the kinetic energies of ejected electrons, and higher frequency light made electrons travel faster. The reason why this is the case is tautological: higher frequencies of light give more energetic photons because higher frequencies of light give more energetic photons. It's simply the way the universe is structured, just like how gravity exists or opposite charged particles attract each other.

Why did you write " It's simply the way the universe is structured .......... each other " Please explain.

Basically, the fact that the energy of a photon scales with its frequency doesn't really have a reason. It's just a fact of existence.


text:
Computing the concentration of a reactant given formation of products
At high temperatures nitrogen dioxide decomposes into nitrogen oxide and oxygen gas according to $$\ce{2 NO_2 -> 2 NO + O_2}$$
I have a question now. Let's say at a certain temperature I'm given data for the formation of $\ce{O_2}$. For example:
At $t = 0$ we have $[\ce{O_2}] = 0$ mol/L, at $t = 50$, $[\ce{O_2}] = 7.10 \cdot 10^{-4}$ mol/L, at $t = 100, [\ce{O_2}] = 1.205 \cdot 10^{-3}$ and so on until a certain time.
If I'm given that the initial concentration of $\ce{NO_2}$ is $0.006M$, how can I know what the concentration of $\ce{NO_2}$ is at certain times? I'm asked to determine the order of the reaction on the basis of the given data, and also to present graphically the rate constant.
To do that I need to know if $\ln([\ce{NO_2}])$ versus time $t$ is a straight line. But how to find $[\ce{NO_2}]$ in the first place?

If you had a reaction A -> B, with [A] = 1 M and no B present at the start, and you are told that after 60 seconds [B] = 0.5 M, how would you calculate [A] at t = 60 s?

I would do (initial concentration of A) - (amount of formation of B). So 1M - 0.5M = 0.5M of A at 60 s. But can we apply this? This assumes theres a linear relationship between reactants and products. Is this the case also in my example? Since the stoichiometric coefficient is 2 there.

Stoichiometry is always linear. If one molecule of $\ce{O2}$ is produced, two molecules of $\ce{NO2}$ have decomposed, so you just have a factor of 2: $[\ce{NO2}]_t = [\ce{NO2}]_i - 2[\ce{O2}]_t$.

What about the presence of the other product, NO. Won't that influence the amount that is taken away?

That would be the case if you have two competing reactions: A -> B and A -> C. But here the production of NO, the production of $\ce{O2}$, and the decomposition of $\ce{NO2}$ are all linearly dependent on each other because they are part of the same reaction. Essentially, for every molecule of $\ce{O2}$ you produce, you also have to produce two molecules of NO, and two molecules of $\ce{NO2}$ have to be decomposed. So their concentrations are all linked: $[\ce{O2}]_t = 0.5 [\ce{NO}]_t = 0.5([\ce{NO2}]_i - [\ce{NO2}]_t)$.

In this reaction, the production of $\ce{O2}$ is tied to the decomposition of $\ce{NO2}$; for each molecule of $\ce{O2}$ that is produced, there have to be 2 molecules of $\ce{NO2}$ that have decomposed. So:
$$2 \Delta n_{\ce{O2}} = -\Delta n_{\ce{NO2}}$$
where $\Delta n$ represents a change in the number of moles. The question does not specify, but I will assume that the volume of the reaction is kept constant; since concentration is amount divided by volume, this means that the same relation holds for concentrations:
$$2\Delta [\ce{O2}] = -\Delta [\ce{NO2}]$$
Since the initial concentration of $\ce{O2}$ is zero, $\Delta [\ce{O2}] = [\ce{O2}]$, and therefore
$$2[\ce{O2}]_t = [\ce{NO2}]_i - [\ce{NO2}]_t $$
$$[\ce{NO2}]_t = [\ce{NO2}]_i - 2[\ce{O2}]_t$$
To use the numbers in the question, at $t = 50$ s we have $[\ce{NO_2}] = (0.006 - 2(7.10\cdot 10^{-4}))$ M = $4.58\cdot 10^{-3}$ M.


text:
Evaporation of acetic acid
I performed aan acid catalysed haloform reaction in acetic acid to an acid. I need to evaporate acetic acid which I used as solvent and It is around 150 ml. What are the alternatives , I have a rotavap in lab but 150ml AA is too much for a rotavap I guess.

Are you evapotating to dryness? Don't worry about alternatives - use the rotavap with a not too small wide-neck (NS29) flask.

Alternative: destillation followed by neutralisation/extraction. I used that to remove propionic acid (whose boiling point is a lot worse for rotavaps than acetic acid’s) from my porphyrin condensation.

Rotavap is fine even until > 500 mL volume.


text:
solution concentrations
I think I'm making this more complicated than it needs to be, but here it is:
I have 7 drug solutions, all 1 mg/mL which need to have the following concentration in urine:
A: 25,000 ng/ml
B: 25,000 ng/ml
C: 100,000 ng/ml
D: 100,000 ng/ml
E: 10,000 ng/ml
F: 1000 ng/ml
G: 1000 ng/ml
How much of each 1 mg/ml solution do I need to add to urine to get a stock with the above concentrations? My ultimate goal is to take several 75 ul aliquots of urine for testing. Any suggestions?

To answer your question, you also need to know how much of your final solution you want to make.

Since you did not tell us how much of this final solution you would like to make, let's imagine that you want to make a generic $V_f\ \ce{mL}$ of it.
Take $\text{A}$ as an example:
$$
n_A \ \text{in final solution} =  25000\ \ce{ng/mL} \times V_f\ \ce{mL}
$$
$$
V_{A}\ \text{stock to take} = \frac{n_A}{c_{A\ stock}} = \frac{25000\ \ce{ng/mL} \times V_f\ \ce{mL}}{1\ \ce{mg/mL} \times 10^6\ \ce{ng/mg}} \frac{1000\ce{\ \mu L}}{\ce{mL}}= \frac{25000}{10^3}\ V_f\ \ce{\mu L}
$$
So let's suppose that you want to make $10\ \text{mL}$ of such a solution containing $\text{A}$ at $25,000\ \text{ng/mL}$, then you need to measure out $25000 \times 10\ /\ 1000\ \ce{\mu L} = 250\ \ce{\mu L}$ of the $\text{A}$ stock and bring it up to the final volume with urine.
You will want to carry out these calculations for all your compounds, then add all the corresponding quantities to a calibrated vessel, e.g. a volumetric flask, then bring this mixture up to the final volume with your solvent, i.e. urine.


text:
Can the Bohr atomic model be applied to sodium and other single valence electron species?
Is $\ce{Na}$ a single electron species and can I apply Bohr's Model on it?
Please explain why not as it also contains one electron in its outermost shell.

One of the great successes of the Bohr model was the ability to explain the emission spectrum of hydrogen (and hydrogen-like monoelectronic ions), by allowing a simple theoretical derivation of the experimentally determined and remarkably accurate Rydberg formula:
$$\frac{1}{\lambda}=RZ^2 \bigg(\frac{1}{n_f^2}-\frac{1}{n_i^2}\bigg)$$
$\lambda$ is the wavelength of light emitted, $R$ is a proportionality constant known as the Rydberg constant (which actually varies slightly with changes in atomic mass), $Z$ is the amount of positive charge attracting the electron (the nuclear charge in multiples of $q=1.6022\times 10^{-19}\ \mathrm{C}$), and $n_f$ and $n_i$ are positive integers (called quantum numbers) which previously had no experimental interpretation, but which now arise from the postulates of the Bohr model, being attributed to the concept of electron shells.
After this great new insight into atomic physics, naturally one would like to investigate how generally applicable the model is. Given the success in describing the emission spectra of hydrogen, one of the first places to look would be the spectra for other elements. So here they are:

(Source - isn't it a beautiful image? Kudos to the designer! Open in a tab for gigantic resolution)
As you can see, most elements have rather elaborate spectra. After undoubtedly many frustrated attempts, it was found that the Bohr model, as accurate as it was for hydrogen, applied quite poorly to the rest of the elements, and no spectra could be reliably predicted.
However, the alkali metals stood out in that a modified version of the Bohr model had some success, producing a Rydberg-like formula, though salvaging it came at the cost of additional experimental parameter fitting, again with no theoretical background.
The formula for the energy of a electron in a hydrogen atom according to the Bohr model is:
$$E_n=-\frac{hcR}{n^2}=-\frac{13.6\ \mathrm{eV}}{n^2},\ \ \ \ n \in \mathbb{N} \ \ (n=1,2,3,4...)  $$
For the alkali metals, the formula had to be modified to:
$$E_n=-\frac{hcR}{n_{eff}^2}=-\frac{hcR}{(n-\delta_l)^2},\ \ \ \ n_{eff} \in \mathbb{R_+}  $$
To keep the same structure, the mismatch between theory and experiment is brushed into the quantum numbers $n$, which are replaced by effective quantum numbers $n_{eff}$ that are no longer positive integers. To recover integer quantum numbers, a value called the quantum defect ($\delta_l$) had to be introduced. This value is different for each value of $n$, and also varies with another parameter $l$. This may seem like an arbitrary mathematical attempt to salvage the equation, but while the values of $\delta_l$ had to be tabulated, it was found that they were relatively constant with changes in $n$, and that there were a limited number of different cases ($l$ was a parameter limited to only a few values). The resulting Rydberg-like equation is:
$$\frac{1}{\lambda}=R \bigg(\frac{1}{(n_f-\delta_{l_f})^2}-\frac{1}{(n_i-\delta_{l_i})^2}\bigg)$$
While not as tidy as the result from the Bohr model, at least the equation can be used to approximately explain the atomic emission spectra for the alkali metals. Details of an experimental procedure for chemistry undergrads to determine the emission spectrum of sodium and apply the modified Rydberg formula can be found here, along with theoretical background.
Nowadays, we know why the alkali metals seemed to be a better fit to the Bohr model than the rest of the elements. With finer knowledge about the details of the electronic structure of atoms, borne out of the full quantum mechanical model of the atom, we know that alkali metals share an important property with hydrogen; they all possess a single electron in their outermost shell, i.e. have a valence shell configuration of $(ns)^1$. This is the only electron which is important in electronic transitions in the near-ultraviolet, visible and infrared regions of the electromagnetic spectrum, as the core electrons of the alkali metals are held much more tightly to the nucleus and require high energy photons to excite. This creates a much simpler spectrum.
Furthermore, atoms with multiple electrons suffer from electron-electron repulsions, which are not taken into account at all in the Bohr model, and thus are a significant source of error. However, in the alkali metals the consequences of these repulsions are simplified, as all electrons are in core shells, except the single valence electron. For all other multielectronic elements, electron-electron repulsions in the valence shell preclude a simple application of the Bohr model.

That's a wonderful writeup. I was trying to provide an answer that would suit the needs of OP (no way I would have thought of discussing quantum defects!) but I'm sure yours is a much more helpful answer in general for others who might chance upon this question.

@orthocresol Thank you! Simpler answers certainly have their merit, and I'd upvoted you even before commenting. Lack of brevity is one of my most common shortcomings, but hopefully it means there's useful/interesting information for everyone.


text:
Industrial synthesis of hydrazine
Not a duplicate of Hydrazine synthesis - I know what hydrazine is and am not going near it voluntarily!
Looked at this resource on the Internet which cites three major commercial-scale processes:

$\ce{H_2O_2}$ process (in use at Arkema and Mitsubishi Gas Chemical, developed by Produits Chimiques Ugine Kuhlmann and patented in 1976)
Bayer process (at the eponymous company)
Urea process

What processes do other major large-scale producers use and what are the driving forces behind adoption of a particular hydrazine synthesis? I can guess that Chinese producers may use the urea process, but I'm not sure.
Background: am a Space Exploration.SE regular and am interested in learning more about worldwide production of hypergolic propellants.

I think that Chemistry of the Elements by Greenwood and Earnshaw might have something to say about this topic. If nobody else has checked it within the next few hours I'll pop out to my library and report back to you.

The Kirk-Othmer Encyclopedia of Chemical Technology reports that: "About 60% is made by the [Bayer] hypochlorite–ketazine process, 25% by the peroxide–ketazine route, and the remainder by the Raschig and urea processes." Here is a link to the original article, if you can access it (e.g. at a university library): Hydrazine and Its Derivatives, by Eugene F. Rothgery. The data cited dates back to 2004, so it may have changed somewhat, but probably not too much.
Consider this also: the above makes hydrazine hydrate. Propellant-grade hydrazine must be anhydrous, so there is an extra step in which water is removed from the hydrate, typically by azeotropic distillation with aniline.

Thanks for the Kirk-Othmer tip, have looked the data up. Could you please explain why these processes were chosen?

@DeerHunter I'm afraid that I don't really have any good insight as to the choice of these methods.

I would hazard a guess that the economies of scale of where the process is located comes into play. Arkema may make bulk $\ce{H2O2}$ at industrial concentration so can just feed it in without transport and hence dilution isn't required . Bayer (or a neighbour) may operate a chloralkali process on site, so use that. Everything will come down to the input costs.