text:
Why can potassium hydroxide not be used to remove atmospheric carbon dioxide?
As we know that $\ce{KOH}$ absorbs $\ce{CO2}$, then why can't it be used to remove the excessive $\ce{CO_2}$ present in the atmosphere?

$\ce{KOH}$ is produced by electrolysis of aqueous $\ce{KCl}$ solutions. Taking into account that the electrical energy still mostly comes from power plants that burn fossile fuels, it doesn't look like the most clever idea to release $\ce{CO2}$, just in order to capture it from the atmosphere again.
While it doesn't make sense to tackle the global $\ce{CO2}$ problem with $\ce{KOH}$ or other hydroxides, they do nevertheless bind $\ce{CO2}$ just as you said. Consequently, these hydroxides were and are still used for the individual protection against $\ce{CO2}$ (self)poisoning in emergency cases and/or in confined spaces. Rebreathers used by fire brigades, divers or miners are typical examples. Initially, $\ce{KOH}$ was indeed used, but nowadays, these $\ce{CO2}$ filters mostly contain mixtures of sodium hydroxide and calcium hydroxide.

KOH can affect us when inhaled and by passing throught the skin.It is a highly corrosive chemical and contact can severely irritate and burn the skin and eyes leading to eye damage.Inhaling it can irritate the lungs. Highly exposures may cause a build-up of fluid in the lungs  (pulmonary oedema) a medical emergency.In addition to these adverse effects, it can also cause headache, dizziness, nausea and vomiting and also may cause a skin allergy.
So by observing these we can surely say that KOH can never be an option to absorb carbondioxide of the atmosphere.If we use it then we will just be gambling our life to death.

You are absolutely right on the health risks of these hydroxides! Nevertheless, when properly embedded in filtering devices, they can safe lifes because they can prevent self-poisoning by carbon dioxide.

This is completely irrelevant to the question. Almost every industrial process in chemistry involves the use of dangerous chemicals but the systems are designed to avoid exposing people to them.

"Potassium hydroxide is usually sold as translucent pellets, which will become tacky in air because KOH is hygroscopic." from Wikipedia. This clears that if KOH is used to absorb the excess carbon dioxide from atmosphere ,there is greater chance of absorption of water molecules from the atmosphere which in turn can change the atmospheric contents resulting in some imbalance atmospheric conditions if KOH is used in excess.

Wet $\ce{KOH}$ doesn't stop to react with $\ce{CO2}$.

Its not about reaction with carbondioxide because there is large amount of carbondioxide in our atmosphere nowadays that to control that, large amount of KOH should be used which will surely absorb large amount of moisture as well changing the atmospheric condition which might cause a lot of problems.

Taking into account that 70% of the surface of the earth is covered by water, which is in equilibrium with the vapour in the atmosphere, I'd like to see a rough estimation on how much $\ce{KOH}$ is needed to cause an ecological problem due to "drying the atmosphere".

Oh i'm so sorry that i didn't take account on the fact but i have come up with one of the best answer for the question please have a glance of it.


text:
Spectrophotometric assay by pNPP
Here is a series of related questions that I want to ask.
Background
The activity of acid phosphatase is measured by an enzymatic reaction that converts para-nitrophenyl phosphate (pNPP) to para-nitrophenol (pNP), liberating phosphate. The product, pNP, absorbs light whose wavelength is $\pu{400 nm}$ with an absorption coefficient ($\pu{400 nm}$) of $\pu{19000 M-1 cm-1}$ at extremely alkaline pH. Reaction mixture for an acid phosphatase is slightly acidic. Thus, it must be alkalinized for quantification of pNP.
Two enzyme concentrations are to be examined - They are $\ce{1X}$ and $\ce{0.1X}$ (The $\ce{0.1X}$ enzyme is made by mixing $\pu{1 ml}$ of $\ce{1X}$ enzyme in $\pu{9 ml}$ of $\ce{NaCl}$.) Reaction times for each enzyme is $1$, $10$ and $20$ minutes.
Procedure

Protocol for measurement of acid phosphatase activity:

Mix $\pu{0.12 ml}$ of $\pu{0.5 M}$ $\ce{Na}$ acetate buffer (pH $5.6$) and $\pu{0.24 ml}$ of $\pu{5 mM}$ pNPP in a test tube. Start the reaction by adding $\pu{0.24 ml}$ of an enzyme solution.
After the reaction times of $1$, $10$, and $\pu{20 min}$, respectively, stop the reaction by adding $\pu{0.6 ml}$ of $\pu{0.5 M}$ $\ce{NaOH}$. $\ce{NaOH}$ stops the reaction and converts the pNP produced into a yellow-colored (A400-absorbing) form.
After all reactions are stopped, measure A400 of the samples.

Assay of potato acid phosphatase:
\begin{array}{ll}
\pu{0.5 M} \text{Na acetate buffer (pH 5.6)} & \pu{0.12 ml}\\
\pu{5 mM} \ce{pNPP} & \pu{0.24 ml}\\
\text{Enzyme} & \pu{0.24 ml}\\
\pu{0.m M} \ce{NaOH} & \pu{0.5 ml}\\\hline
\text{Sum} & \pu{1.2 ml}
\end{array}

There are some things I am supposed to calculate after I obtain the results.
First I am going to plot a graph of absorbance versus time and find the slope for each of the lines ($\ce{1X}$ and $\ce{0.1X}$).

I have been asked to find the absorbance change/min/1ml of $\ce{1X}$ enzyme.

My answer: Divide the slope obtained by $0.24$ (the amount of enzyme).

I have to convert the absorbance change to concentration change when
$L$ is $\pu{1 cm}$ and e400 of pNP is $\pu{19000 M-1 cm-1}$.

My answer: This can be found out by $A = eCL$.

The next question is to convert the concentration change to a change in the amount of substance of pNP.

I have no clue how to do this.

Finally, I have to calculate total activity (in moles per minute) in $\pu{4 ml}$ of $\ce{1X}$ enzyme solution.

My answer: Multiply the answer I obtain in the above question by $4$.

For the first subquestion:
Using unit analysis, the answer is very quickly found: The unit of the slope $m$, assuming dimensionless values for the absorbance, is
$$[m] = \frac{[\Delta y]}{[\Delta x]} = \frac{[\Delta A]}{[\Delta t]} = \frac{1}{\text{min}}$$
Now you need a volume in the denominator as well, so dividing by a volume seems like a good idea. Since you need the change in absorption per $\pu{mL}$ of $\ce{1X}$ solution, you divide the slope of the $\ce{1X}$ reaction by $\pu{0.24 mL}$ for the correct value.

For the subquestion with which you seem to have the most trouble: simply apply $c=n/V$.
You will find the amount of substance changed via
$$ \Delta n = \Delta c\times V = \Delta c \times \pu{1.2 mL}$$

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text:
Why is ionization energy measured in gaseous phase?
When I looked up the definition of ionization energy, it says:-

it is the energy required to remove one electron from the shell to
  infinity in an atom in gaseous state.

Why is it measured in gaseous phase and not in any other phase?

You can define the energy necessary to remove an electron from a solid instead of a free atom/molecule, but it takes on another name, the work function. As expected the extra intermolecular interactions stabilize the material after losing an electron, and so it's much easier to remove an electron from a solid than a single atom/molecule.

You want to determine the energy for a single molecule and avoid stabilization effects through intermolecular interactions.

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text:
Why is fluorine the most electronegative atom?
It seems related to the atomic size but hydrogen has a smaller atomic size than fluorine. Why is fluorine the most electronegative atom?

One reason is hydrogen has one shell and one proton, as compared to fluorine with 2 shells and 9 protons.

It's interesting to note that if you define electronegativity as the "strength" with which the valence electrons are attracted to the nucleus (for a certain definition of strength), then the noble gasses also have very high electronegativities. In particular, neon becomes the element with the highest electronegativity, since it has the same number of shells as fluorine while having an extra proton! It's kind of odd that neon should be more electronegative than fluorine, because we often associate extreme electronegativity with high reactivity, but it makes a lot of sense.

Electronegativity is the tendency of an atom to attract a shared pair of electrons (not the valence electrons ) towards itself

Fluorine is the most electronegative element because the definition of electronegativity makes it so. The electronengativity scales are defined based on experimentally determined properties of the elements. Fluorine has appropriate values for all of the common scales to ensure it has the highest electronegativity.
The Pauling scale, which is the first electronegativity scale proposed, relates the difference in electronegativity $\Delta\chi_{A,B}$ between two atoms $(A\ \& \ B)$ is related to the bond dissociation energies $(E_d)$ of the diatomic species $AA,\ BB,\ \& \ AB$:
$$\Delta \chi_{A,B}=|\chi_A - \chi_B| = (\text{eV})^{-1/2}\sqrt{E_d(AB)-[E_d(AA)+E_d(BB)]/2}$$
Since only differences can be determined, one element (hydrogen) was chosen to have an arbitrarily set value. On the Pauling Scale Hydrogen has a value of $2.20$. Since the $\ce{H-F}$ bond is much stronger $(E_d=586\ \text{kJ/mol})$ than the average of the  $\ce{H-H}$ bond $(E_d=436\ \text{kJ/mol})$ and the $\ce{F-F}$ bond $(E_d=157\ \text{kJ/mol})$. Data from here. In general, since fluorine forms stronger bonds to all other elements than it does to itself, fluorine will have the highest electronegativity.
The Mulliken scale relates electronegativity to two other properties: electron affinity $E_{ea}$ and ionization energy $E_i$:
$$\chi_A = \frac{E_{ea}(A)+E_i(A)}{2}$$
Fluorine has a high electron affinity (easy to gain an electron) and a high ionization energy (hard to lose an electron).
There are other scales, including the Allred-Rochow scale, based on effective nuclear charge and covalent radius, the Sanderson scale based on atomic size, and the Allen scale based on the average energy of the valence electrons.

Fluorine has a higher electronegatevity if compared with other than elements in the periodic tables due to:

atomic size is decreased.
nuclear charge increase due to gain two electrons.
attraction force between nucleus and adding electron increase so electronegatevity increase too.

It seems that (although I am no chemist) that the valence electrons in the 2nd shell have a higher priority than any other shell, if the 2nd shell is the valence shell. The closer it gets to being full, we know that electronegativity increases. With Fluorine only needing one more electron to be complete. The big battle for second place is interesting. Oxygen only needs 2 electrons in its valence shell. Chlorine only needs one,  but in its 3rd shell. But we know Oxygen wins out by a slim margin.

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text:
How many moles of the substance are in 2.5 g of lead?

How many moles of the substance are in 2.50 g of lead?

What I did:
Atomic mass of Pb: 207.2 amu (or g?)
Applying unitary method, 1 mol of Pb -> 207.2 g; so, x -> 2.5 g
$x = \frac{2.5}{207.2}= \pu{0.0121 mol}~\text{of}~\ce{Pb}$
Shouldn't it be like this?:
1 mol of Pb -> $(\pu{207.2 amu} \times1.66\times(10^{-24}))~\pu{g}$ so x -> $\pu{2.5 g}$

The atomic mass unit (unit symbol: amu) is obsolete since 1961. The accepted units are dalton (unit symbol: Da) and the unified atomic mass unit (unit symbol: u).

Do not mix amu with mass of mol of substance! In your last formula you've forgot about Avagadro number.
One Pb atom weighs $\pu{207.2amu}$ or $\pu{207.2amu} \cdot \pu{1.66\times10^{-24}g}$.
But one $\text{mol}$ of Pb weighs $=\pu{207.2amu}\cdot \pu{1.66\times10^{-24}g}\cdot6.02\cdot10^{23}=\pu{207.2g}$

A gram-mole is the numeric value of atomic weight of the formula unit times grams.  Given an amount of said stuff in grams, divide by the numeric value of atomic weight of the formula unit to output the number of gram-moles of stuff.  One gram-mole contains Avogadro's number of formula units.  Embroidering the answer atom-by-atom undoes the installed shortcut to arrive at the same answer the hard way.  Avogadro's number is subject to small changes depending on what defines a kilogram.
A mole is the archetypal 5 pound bag.  A mole of sulfur atoms is about 32 grams.  Rhombic sulfur is eight-membered rings.  A mole of those is (8)(32 grams).  The crystallographic unit cell contains 16 rings.  A mole of unit cells is (16)(8)(32 grams).  If you are reacting zinc and sulfur, all you care about is moles of atoms.  If you want to know how many unit cells you have for x-ray diffraction of a rhombic sulfur microcrystal, you want to take those atoms 128 at a time.


text:
Why is this organic compound trans, and not cis?
I'm reading a high school chemistry course online, and I'm now reading about geometric isomers.
They provide this example:

And in terms of naming it, they say this:

Begin by identifying the parent chain - it is the continuous chain that contains the carbon-carbon double bond. The two groups that belong to the parent chain are on opposite sides.

Then supply this illustration:
And with that explains that it's trans-3-ethyl-3-hexene for that reason. I don't quite get their explanation though, becuase to me it seems like this could also be cis.
Why can't it be this one:
As far as I understand, that's also a valid way of expanding it, because it has the same amount of carbons in the chain and the numbers are still as low as they get. The only difference is that this one would be cis-3-ethyl-3-hexene instead of trans-3-ethyl-3-hexene.
So, why exactly is it that only one of them is right?

You are right, there is no difference between the two molecules that result if you rotate 180 degrees about the $\ce{C=C}$ double bond - they are the same exact molecule.  So for this molecule, the terms "cis" and "trans" would not be used, they have no meaning here.  The name of the molecule is simply 3-ethyl-3-hexene.  If the on-line course says otherwise, it is wrong.

Well, I didn't draw these. I photoshopped the third one to illustrate the problem and I saw, but otherwise it's taken straight out of the course. I'll e-mail them about it, then.


text:
Why does silicon tetrafluoride have a higher melting point than sulfur tetrafluoride?
So looking at the Wikipedia pages of sulfur tetrafluoride and silicon tetrafluoride, the melting points are −121 °C and −90 °C respectively, and so $\ce{SiF4}$ has the higher melting point. However, their boiling points are −38 °C and −86 °C, respectively, giving $\ce{SF4}$ the higher boiling point.
I can justify that $\ce{SF4}$ has the higher boiling point because it is more polar since its Lewis structure has a lone pair on the sulfur atom, and so it experiences greater dipole-dipole forces. However, why does $\ce{SF4}$ have a lower melting point?

High symmetry molecules fit into crystal lattices especially well (higher m.p.), but are volatile for having fewer van der Waals interactions (lower b.p.).
$$\begin{array}{lrr}
\hline
\text{Compound} & \text{m.p.}/\pu{°C} & \text{b.p.}/\pu{°C} \\
\hline
\text{pentane} & −130 & 36.1 \\
\text{isopentane} & −160 & 27.2 \\
\text{neopentane} & −18 & 9.5 \\
\hline
\text{octane} & −57 & 125 \\
\text{2-methylheptane} & −110 & 117 \\
\text{2,2,4-trimethylpentane} & −107 & 99 \\
\text{2,2,3,3-tetramethylbutane} & 101 & 107 \\
\text{cubane} & 131 & 133 \\
  & & \text{(sub. at r.t.)*} \\
\hline
\text{decane} & −30  & 174 \\
\text{isodecane} & −75 & 167 \\
\text{adamantane} & 270 & \text{sub.} \\
\hline
\end{array}
$$
* sub. — sublimes; r.t. — room temperature.

Linear alkanes crystallize especially well.  Look at the volume expansion (density decrease) of paraffin wax upon melting.  This is not true for iso-paraffins, substantially lowering the mp.

here's another one which you can try with a regular fridge. mp / bp: Cyclohexane 6.5C / 81C, n-Hexane -95C / 69C,  isohexane -154C / 60C. I fondly remember the inconvenience of frozen 200L drums of cyclohexane in the winter in a previous job: dropping a handpump in and feeling the inlet pipe hit solid wax.  It rarely if ever happened with the diesel. Per your data decane has mp -30C despite having nearly twice the mw of cyclohexane. (cetane/hexadecane is +18C but real diesel is of course a mixture which depresses the freezing temp.) Benzene has mp 5.5C, toluene -95C, another striking example.

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text:
Xenon and the human body
Reading this article on Wikipedia: Xenon Medical applications I see that Xenon can be used as an anesthetic, neuroprotectant and doping agent.
If it is a noble gas, and thus, chemically stable, how can it have those properties?
If Xenon does not react with any other substance, how does this happen, from a chemical point of view?

Very similar to this question put on hold. It would have been better to edit the previous question to improve it than to ask a new question.

I sent myself that question, was under an old account and can not edit. Rewritten anyway.

@BenNorris Thanks for point that out.  I have deleted that one, since it had the account issue that the OP mentioned.

From what I have briefly gathered, the concept behind the usage as a doping agent is basically letting the subjects breath less oxygen thus achieving effects similar to altitude training. This basically exploits the physical properties of Xenon (its heavier than the main components of air and thus stays in the lung) and its inertness (it does not have any harmful effects).

Somehow, this reminds me of the sign I saw at MIT many years ago: "DANGER: ARGON STORAGE AREA". I always wondered whether that was a chemistry joke or if they were concerned about a possible smothering hazard...

Inhalation of an inert gas could drive the nitrogen out of your bloodstream to bends symptoms.

Xenon was also tested as a marker of brain perfusion in acute stroke using xenon-ct scanning. It's not used anymore.

Chemist Neil Bartlett in 1962 discovered that although Xenon was a noble gas, it is able to form compounds with other substances even though it is chemically "inert."
Neil Bartlett, at the time in 1961, produced an unidentified red solid and discovered that the red solid was a reaction between gaseous fluoride, platinum hexafluoride ($\ce{PtF6}$) and oxygen that was oxidized. Finally, determining the chemical structure, they found that the red solid was $\ce{O^{2+}PtF6}$.
Oxygen is a powerful oxidizing agent, being consumed in many combustion reactions. However, Bartlett believed that the platinum hexafluoride ($\ce{PtF6}$) was a stronger oxidizing agent than oxygen, and oxygen is instead reduced. Bartlett then believed if the oxidation of oxygen was possible, he could apply this xenon.

Because my co-workers at that time (March 23, 1962) were still not sufficiently experienced to help me with the glassblowing and the preparation and purification of $\ce{PtF6}$ [platinum hexafluoride] necessary for the experiment, I was not ready to carry it out until about 7 p.m. on that Friday. When I broke the seal between the red PtF6 gas and the colorless xenon gas, there was an immediate interaction, causing an orange-yellow solid to precipitate. At once, I tried to find someone with whom to share the exciting finding, but it appeared that everyone had left for dinner!

The discovered new compound was xenon hexafluoroplatinate ($\ce{XePtF6}$). Bartlett's important contribution to chemistry is a series of noble gas compounds with specific and useful properties.
In regards to your question on how xenon can be so diverse, with over "100 noble gas compounds known today..."

Noble gas compounds have already made an impact on our daily lives. $\ce{XeF2}$ has been used to convert uracil to 5-fluorouracil, one of the first anti-tumor agents. The reactivity of radon means that it can be chemically scrubbed from the air in uranium mines and other mines. Excimer lasers use compounds of argon, krypton or xenon to produce precise beams of ultraviolet light (when electrically stimulated) that are used to perform eye surgery for vision repair.

Material derived from a highly interesting article from the American Chemical Society.
http://www.acs.org/content/acs/en/education/whatischemistry/landmarks/bartlettnoblegases.html
Looking at the literature:

Neuroprotection

For the use of Xeon during mechanical ventilation, an article examining the neuroprotective effects of xenon,
Neuroprotection against traumatic brain injury by xenon, but not argon, is mediated by inhibition at the N-methyl-D-aspartate receptor glycine site.

Anti-Tumor Agent

Noble gas compounds have already made an impact on our daily lives. $\ce{XeF2}$ has been used to convert uracil to 5-fluorouracil, one of the first anti-tumor agents.
The American Cancer states that

Fluorouracil belongs to the class of chemotherapy drugs known as anti-metabolites. It interferes with cells making DNA and RNA, which stops the growth of cancer cells.

Anesthesia

I'm still looking for literature to add here. Some of these cover very complex physiological processes, and this is not my area of expertise. I will try to find more pertinent articles from peer-reviewed journals with high enough impact and relevance. It will be up to the poster of this question to explore in-depth as much as they need by starting at some of the literature I provide.

A nice introduction to "real" xenon chemistry. But you will certainly agree that the conditions for their synthesis aren't really physiological ;) So, it is not very likely that xenon is oxidized in the binding site of a protein. There must be another sort of interaction between the noble gas and the protein.

@KlausWarzecha Ah, you're right! Thanks for letting me know. Surely there must be some literature about xenon compounds and physiological response. There are many such compounds and I will try to find some good examples tomorrow.

If you have some additional insights on http://chemistry.stackexchange.com/questions/235/what-is-the-geometric-configuration-of-the-four-fluorine-atoms-during-the-synthe, I'd enjoy hearing them.  I first learned about Neil Bartlett in high school and I have always been fascinated by his work.

All that being said, as @KlausWarzecha has pointed out in his comment, this doesn't directly answer the question.  Since you have agreed to rework it, and it has the beginnings of a great answer, I will let it stand for now.

If Xeon does not react with any other chemical, how does this happen, from a chemical point of view?

If molecules of phospholipids do not react with each other, how do they form biomembranes?
If Xenon doesn't react, how can it form host-guest complexes with cryptophanes?

It's not necessary to form covalent or ionic bonds to hold the pieces together.
Weak dispersion forces, recently described by Ben Norris here on this site are apparently enough in the examples above. These forces are also responsible for the binding of xenon to proteins.
For further information, you might want to have a look at the work of Seth M. Rubin, who did a lot on $\ce{^{129}Xe}$ NMR spectroscopy in proteines.


text:
Are entries ever removed from the Merck Index?
I was re-sorting books the other day, and thought about how many years it has been since I bought my Merck Index. I'm sure there are new entries in the more recent additions, but I do not have such a need to stay that up-to-date.
At first thought, it seemed to me that a book of this nature only adds entries, and does not remove any entries. But is this the case? Are compounds de-listed because they are bad science? Other reasons?
Iit would be great to know other details about the editorial process, as well, if relevant.

Interesting question, but I have no idea. You might want to ask the guys at the RSC.

Btw, did they ever have an entry on polywater? ;)

Well, at least you've done half of your research

What do do you mean?

Without several print editions side-by-side, I restrain my answer to the electronic version. This public link or here regarding the online version based on the 14th edition may be read like it may be the case.  To quote:
 "The 14th Edition also contains "retired" records from earlier editions of The Merck Index which still contain valuable information but are no longer being updated with new information."
Of course, beside RSC you may contact them (Merck in Rahway, NJ), too.


text:
How do I overlay spectra in an ACD/NMR Processor, version 12.01?
I am having great difficulty getting spectra to overlay (the option doesn't even appear) after setting to new window mode and selecting each spectrum.

Did you know that you can submit a real answer to your own qeuestion?

I have moved your answer out of your question, since as the above comment suggests, it should be in the form of an answer.  You can add your own answer if you would like and I will delete the temporary one.

If anyone else is running into the same issue, after selecting which spectra you would like to overlay go to series -> collect spectra and then select "manual offset" if you want to adjust the height.


text:
Do catalysts undergo physical changes during a reaction?
I know that catalysts don't undergo chemical changes during a reaction, but do they undergo physical changes? If they do, are there any examples?

I'll just give an example of a catalysis to show you that catalysts do really temporarily react with the starting product and eventually will be regenerated after the catalytic reaction.
An example is the catalytic decomposition of formic acid to carbon monoxide and water by using sulfuric acid as a catalyst. The sulfuric acid donates one proton ($\ce{H+}$ ion) which is then bonded to the central carbon atom in the formic acid molecule. Then next another hydrogen atom from the sulfuric acid reacts with the hydroxyl group present on the formic acid molecule to yield water. In the end the sulphate ion snatches the last two present hydrogen atoms from the formic acid molecule and yields carbon monoxide. The sulfuric acid is regenerated.
$\ce{HCOOH + H2SO4 -> H2COOH + HSO4- -> H2CO + SO4^{2-} + H2O -> CO + H2SO4}$
So catalysts do undergo chemical changes in the catalytic reaction but are regenerated after the catalytic reaction is finished.
This means that if you are using a catalyst, you can normally reuse it after your catalytic reaction because it has the same chemical (and physical) properties as before the catalytic reaction.
Also : when you use manganese dioxide as a catalyst to decompose hydrogen peroxide, you will see that the black solid compound is the same after the catalytic reaction.
There are also so called 'sacrificial catalysts' (which aren't regenerated and do physically/chemically change, but aren't real catalysts) -> see http://en.wikipedia.org/wiki/Catalytic_cycle

I know that catalysts don't undergo chemical changes during a reaction [...]

They don't? How then, other by forming a complex with the starting material, is the activation energy of that reaction lowered?
I do however agree that - in the ideal case - this step is reversible and leaves the catalyst unchanged after the catalytic cycle ;)

You didn't explain pretty much anything. I actually came here to find out how they decrease activation energy.


text:
What's Standardization, and why do it?
Surprisingly, there are very few resources online (outside of obscure lab pages) that explain exactly what standardization is (not even Wikipedia!). The context is the phrase standardize the titrant for a titration. Does this mean you titrate it with a known solution?

Imagine that you need to titrate the sample S with a solution of the reagent R.
Unfortunately, R isn't available as a stock solution and maybe not even in solid form. All you have is it's precursor P and a recipe for the reaction $\ce{P + X + Y -> R}$. You never know how good that works out.
At this point, titration of S with R doesn't make any sense - unless you can determine the concentration of R!
Here, a stock solution with an exact concentration of the helper H, maybe in a sealed vial from a commerical source comes into play.
The titration of R with H to give the exact concentration of the solution of R is the standardization of the titrant.

Standardization is doing a titration to work out the exact
  concentration of the solution you want to use to determine the
  concentration of an unknown solution.

Where the solution you want to use is the titrant
See : http://answers.yahoo.com/question/index?qid=20090903183139AAXfQGm

Standardization is the process of determining the accurate concentration of a standard solution by titrating it against a solution of accurate concentration with high degree of purity


text:
Is this analogy explaining the difference between enantiomers and diastereomers correct?
I tried writing an analogy to explain the difference between enantiomers and diastereomers. However, I'm not entirely sure if it's actually a correct representation of the differences. It might also be more confusing than it has to be:

To use an everyday object as an illustration of the difference, imagine two forks with four tines each. If fork 1 has its first and third tine longer than the other ones, and fork 2 has its second and fourth tine longer than the other ones, then the two forks would not match completely if you tried to stack them on top of each other. However, if you held fork 2 up to a mirror, then you would see it now has both its elongated tines in the same place as fork 1. These forks would be enantiomers of each other because they are the same thing, but mirror images of each other
If you changed fork 2 to have its first and second tine longer than the other ones, instead of the second and the fourth, then the two forks would instead be diastereomers. This is because now they both cannot match completely when stacked on top of each other, and holding either one up to a mirror will not make them look any more similar. This is exactly how diastereomers work too, because it only partially mirrors its counterpart. As you can see, fork 2 now has its first elongated tine actually match the first elongated tine of fork 1, while also retaining the effect that holding it up to a mirror will have its second elongated tine match that of fork 1. That is, in the mirror fork 2 will have its third tine elongated, just like fork 1 does when not viewed through a mirror. In other words, fork 2 has its first elongated tine match fork 1’s, but its second elongated tine is the mirror image of fork 1’s. This makes them diastereomers because they are the same thing, but neither structurally identical nor complete mirror images of each other.

So, is there anything off about my analogy, or does it correctly and comprehensibly explain the difference between the two?

Enantiomers are non-superposable mirror images (all the signs of one spatial coordinate are reversed).  Diastereomers are  formed by inverting some but not all stereocenters.
http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf
 pp. 11-13.  Excellent overall.
Have some fun with your prof!  An acyclic, aliphatic, unstrained, tetrahedral carbon atom is $\ce{sp^3}$-hybridized.  It is an uncharged, valence-saturated singlet.  It bears four freely rotating, rigorously identical substituents, e.g., $\ce{CR4}$, with identical bond lengths to that central carbon.  Said carbon is a chiral center.  It has no ${S_n}$ improper axes of symmetry.
1) Show how this can be true.
2) Use any system of nomenclature to name left-handed and right-handed stereoisomers.
Acyclic, unstrained, etc., trigonal planar $\ce{sp^2}$-hybridized carbons ${R2>C=C<R2}$ are each identical stereocenters.  Each carbon and the olefin overall have no ${S_n}$ improper axes of symmetry.  The olefin is resolved chiral.  (1) and (2) as above.

C'mon, folks.  Show me how isolated CR_4 can be a chiral center.  The answer is trivial.  End of the week I'll show you how you already know the answer simply by reading the question.

The enantiomer of CR_4 is CS_4 for four idenntical substituents, Cahn-Ingold-Prelog notation.  There is no way to name chirality.  If you suggest by default, consider CR_aR_bR_cR_d in which the "R" priorities are arranged in S order.  The olefin uses the  same loophole.  A mirror plane may not pass through a chiral center.  The mirror image of R is S, and the reverse.

Go into HyperChem or whatever, make dodecahedrane.  Put a tetrahedral sp^3 carbon at its center, make four internal bonds to the shell (remove those four external hydrogens).  Minimize.  Make the inversion.  Superpose the two central carbons, and rotate to exactly align all four bonds.  Enantiomers! Name the central carbons' chiralities.  Good luck on that, for all four "groups" in each molecule are identical, point group T (not T_d or T_h).


text:
The energy gap between a pi-conjugated system with (2 bonding and 1 anti-bonding orbital) and (1 bonding and 2 anti-bonding) orbitals
I asked a question previously about "why" it is the case the expanding the size of pi-conjugated systems decreases the required energy to excite an electron from a HOMO to a LUMO band: Why does the energy gap for π - π* transitions shrink with the size of the pi-conjugated system?.  The user "Philipp" (and others) very patiently and kindly explained the reason for this by pointing out the expansion in the number of possible bonding and anti-bonding configurations for a polymethine-like pi-conjugated system.
However, there's something I'm missing that has to do with what is probably my lack of knowledge about chemistry.  If we look at this image: https://i.stack.imgur.com/Ti3cx.png (provided by "Philipp") why is that that we have an energy gap between $\Psi_2$ and $\Psi_3$ smaller than the energy gap for a simpler, smaller pi-conjugated system where the HOMO corresponds to a single pair of p-orbitals in a bonding conformation and the LUMO corresponds to the same single pair of p-orbitals in an anti-bonding conformation (i.e. this system: https://i.stack.imgur.com/9cscv.png)?
My guess that this would be because the bonding interactions that exist in the larger pi-conjugated system mitigate the penalty for having the one anti-bonding interaction?
Update --- Could we justify the above guess using something like an overlap integral (if we know the shape of the two bonded p-orbitals forming a pi-bond)?  If so, this would be quite intuitive.

Luckily enough, you have chosen one of very first studied problems in theoretical chemistry. Have a look here:
http://en.wikipedia.org/wiki/Hückel_method

I can only second ssavec. HMO is the way to go. Does your library carry a copy of The HMO model and its application by Edgar Heilbronner and Hans Bock?

@KlausWarzecha I wish I had easy access to that text.  Is it good enough to be worth purchasing?

Just the 1st of three volumes is between USD 70 and 250 for a used copy on amazon! Volumes 2 and 3 are exercises and tables, resp. Are you sure that you'll be using it in 2-5 month? Might be better to save the money and use free online resources from universities. Pesonally, I wouldn't buy it.

If you want to really understand why energy levels of conjugated $\pi$ systems are the way they are I suggest you take a good look at the Hückel theory as @ssavec suggested - this method is only an approximation but because of its simplicity it is really good for helping you to understand the underlying concepts.
Nevertheless, doing as you suggested and looking at the overall overlap between the $\text{p}$ orbitals in a given molecular orbital does indeed seem to work although I don't have a strict justification for that.
From Hückel theory it is known that the coefficients of the $\text{p}$ orbitals in a given molecular orbital can be calculated as
\begin{equation}
  c_{j,r} = \sqrt{\frac{2}{N + 1}} \sin\Bigl(\frac{r j \pi}{N + 1}\Bigr)
\end{equation}
giving the coefficient $c_{jr}$ for atom $j$ in molecular orbital $r$ of a conjugated system of $N$ atoms (so that $j$ and $r = 1, 2, 3, \dots , n$).
Let's apply this to butadiene.

The $\pi$ molecular orbitals of butadiene are $\Psi_{1}$, $\Psi_{2}$, $\Psi_{3}$, and $\Psi_{4}$.
They are built from a linear combination of 4 atomic $\text{p}$ orbitals ($\phi_{1}$, $\phi_{2}$, $\phi_{3}$, and $\phi_{4}$), so
\begin{equation}
  \Psi_{r} = c_{1,r} \phi_{1} + c_{2,r} \phi_{2} + c_{3,r} \phi_{3} + c_{4,r} \phi_{4}
\end{equation}
When you calculate the atomic orbital coefficients for each molecular orbital $r$ then you get to the following picture:

where the numbers above the different $\text{p}$ orbitals are the coefficients.
Doing the same for ethene you get the following:

Now, you need to compare the overall overlap in the HOMOs of ethene and butadiene.
Since it is already assumed in the Hückel theory only the overlap between neighboring $\text{p}$ orbitals is considered.
Furthermore, it is useful to set the overlap integral of two "normal" $\text{p}$ orbitals (both with coefficient $c = 1$) to 1 (its exact value doesn't matter because only the relative amounts of overlap in different MOs are important for this comparison), so that you can calculate the overall overlap in a molecular orbital simply by multiplying the coefficients of neighboring $\text{p}$ orbitals and summing over these products.
So, for the overlap in the ethene and butadiene HOMOs you get:
\begin{align}
  S_{\text{HOMO, ethene}} &= 0.707 \cdot 0.707 \\
  &= 0.500 \\
  \\
  S_{\text{HOMO, butadiene}} &= 0.600 \cdot 0.371 + 0.371 \cdot (-0.371) + (-0.371) \cdot (-0.600) \\
  &= 0.307
\end{align}
There is less overlap and thus less bonding in the HOMO of butadiene than in the HOMO of ethene which would imply that the HOMO of butadiene is higher in energy than that of ethene.
Now, what about the LUMOs:
\begin{align}
  S_{\text{LUMO, ethene}} &= 0.707 \cdot (-0.707) \\
  &= -0.500 \\
  \\
  S_{\text{LUMO, butadiene}} &= 0.600 \cdot (-0.371) + (-0.371) \cdot (-0.371) + (-0.371) \cdot 0.600 \\
  &= -0.307
\end{align}
There is more overlap and thus more bonding (or better to say less antibonding) in the LUMO of butadiene than in the LUMO of ethene which would imply that the LUMO of butadiene is lower in energy than that of ethene.
So, the HOMO of butadiene is higher in energy than that of ethene and the LUMO of butadiene is lower in energy than that of ethene.
Thus, the HOMO-LUMO gap in butadiene is smaller than that in ethene.

I was hoping I'd see a more mathematical description like this in the other question. Great work!

@NicolauSakerNeto Thanks. As for the other question: I had the feeling that the OP wanted something intuitive rather than mathematical that's why I gave a very pictorial description in terms of in-phase and out-of-phase combinations in the first place. But I think about adding another answer with a full mathematical treatment of the Hückel model if I find the time (and motivation :)).

Very sweet graphics!

@tschoppi ChemDraw magic :)

@Philipp Actually I'm fairly certain this is more a case of Philipp magic.

Amazing answer!

@Philipp Indeed a great answer! Would you mind add the general mathematical expresion (in Hückel's theory) that shows that increasing the conjugation decreases the HOMO-LUMO gap?

Here's an alternate approach to answering this question using the "Particle in a Box" approach.  The $N^{th}$ energy level for a particle in a one-dimensional box is given by $$E_n=\frac{n^2h^2}{8mL^2}$$ where h is Planck's constant, m the mass of the particle and L the length of the box.  The energy of the N+1 energy level is equal to $$E_{n+1}=\frac{(n+1)^2h^2}{8mL^2}$$.  Subtracting these two terms gives us the energy separation between these two energy levels as $$\Delta E=\frac{(2n+1)h^2}{8mL^2}=K\frac{2n+1}{L^2}$$.  Pictured below is the beginning of our polyene series, ethylene, butadiene, hexatriene...

For simplicity, let's assume all bond lengths are equal.  Then the "length" of our box increases by 2L each time we move from ethylene to butadiene to hexatriene, etc.  If we now compare the difference of the $\Delta E's$ (the $\Delta\Delta E's$) for the $N^{th}$ and $(N+1)^{st}$ energy levels between ethylene and butadiene, we find that the difference in energy level separation is given by $$\Delta\Delta E=K(2n+1)(\frac{1}{L^2}-\frac{1}{9L^2})= K'(\frac{8}{9L^2})$$
If we perform the same comparison for butadiene and hexatriene we see that $$\Delta\Delta E=K(2n+1)(\frac{1}{9L^2}-\frac{1}{25L^2})= K'(\frac{16}{225L^2})$$ and for hexatriene compared to octatetraene it is $$\Delta\Delta E=K(2n+1)(\frac{1}{25L^2}-\frac{1}{49L^2})= K'(\frac{24}{1225L^2})$$  In other words as we increase the length of our box (increase the length of our polyene), the energetic separation between any corresponding pair of energy levels (HOMO-LUMO being one case) will shrink.  Hope this helps!

+1, retreating to those simple model systems is very useful for getting an idea of the general behavior.

@Philipp  Thank you.  Over the years I've learned that some people will see the answer more clearly with one approach, other folks will see the answer more easily with another approach.  So presenting different ways to arrive at an answer is always helpful.

@ron This is really getting close to an explanation that makes complete sense to me.  However, how do we understand the above "Particle in a Box" system at the limit of an ultra-large pi-conjugated system?  I.e. for $(D = 1)$ a $k$-unit polyene chain as $k \to \infty$, for $(D = 2)$ graphene, and for $(D = 3)$ a metal like Cu?  Regarding the $D = (2,3)$ scenarios I just outlined, I understand the expression above is for the $(D = 1)$ case only, but nevertheless, does this question make sense?

An example of an ultra-large pi-conjugated system would be polyacetylene and other similar molecules.  In such a case the "Particle in a Box", as well as other MO approaches, tells us that our discrete energy levels bur into a continuum (what's called a "band") with a small band gap between the occupied band (valence band) and unoccupied band (conduction band).  So in effect our system has become a quasi-metal, and begins to behave like a metal.  It becomes more like a conductor and less like an insulator.  If you "Google" "particle in a 2 (or 3) dimensional box", you'll find that

the solution is analogous to our 1-dimensional solution.  Instead of a $1/L^2$ term, we now have $1/L^2,  1/W^2 and  1/H^2$ terms

@ron That makes good sense to me, however, we should still have a reasonably large band-gap in a metal right (consider the photoelectric effect)?  The above expression would appear to indicate that there should be a band gap of size zero as $L \to \infty$?  Or am I not thinking correctly about the nature of the band gap in a metal / pseudo-metal?

If I recall correctly, the PE effect knocks an electron out of the metal.  That's not what we're doing, we're just promoting an electron, not effecting ionization.  Thermal energy should be enough to allow an electron to be promoted from the valence band to the conduction band in a metal.  But you are right about the band gap shrinking as the box size increases.  However crystal defects, impurities, etc. often serve to limit the size of our box.  That's why semiconductors are prepared under such stringent conditions - trying to make the box bigger, trying to make conduction easier.

@ron Ok, that makes sense.  So in a metal, there is no HOMO / LUMO gap, and we need to think about the "band gap" as corresponding to some exciton binding energy (i.e. the energy to split an electron and a hole).

@user4717 Yes, that's my understanding

However, the relation $\Delta E = K\frac{2n+1}{L^2}$ only shows the energy gap for transition with same quantum numbers. But the HOMO for a compound A (might be for n=2) whereas for a compound B (might be for n=3). It isn't guaranteed just by the increasing length.


text:
What is a product of water ionization?
I've read that radioactive compounds are dangerous because they emit electromagnetic waves (gamma-radiation etc.) that are dangerous for living organisms because they can ionize atoms in molecules.
So if one water molecule is ionized by using radiation, would you then get:
$\ce{H2O -> H^+ + OH^-}$
or would you get free radicals : $\ce{H2O -> H^{.} + OH^{.}}$?

Water, when hit by radioactive particles, will most likely suffer ionization (or radiolysis). The ejected electron very quickly leaves the vicinity of the affected molecules (in the case of exposure to gamma or beta radiation) or a very chemically stable neutral helium atom is formed (in the case of exposure to alpha radiation), so you obtain short-lived species which will tend to react with most other substances nearby, even if neutral. The electron, after being slowed down considerably, becomes hydrated and can participate in reactions itself. This article has several examples of reactions triggered by radiolysis, and also separates them by their relevant timescales.
Some further examples of reactions involving radiolysis of water via beta particles and gamma rays are (from the very well-cited article "Critical review of rate constants for reactions of hydrated electrons, hydrogen atoms and hydroxyl radicals (⋅OH/⋅O−) in aqueous solution"):

Of course, not only are the decay particles damaging, but so are the entities produced by their interaction with matter in the vicinity. If the nuclear decay particle doesn't directly ionize organic matter, reactive species such as $\ce{OH^{.}}$ or $\ce{H^{.}}$ could then react with biomolecules, degrading cells and tissues.
This Wikipedia link has some details on the radiolysis of water. Interestingly, it seems that dissolved hydrogen in water can efficiently suppress the formation of unusually reactive species, presumably as the side reactions eventually cause the formation of hydroxyl anions (if electrons are present in excess) or hydronium cations (if positive charges are present in excess). This image (source) shows a collection of possible reactions in sequence. If you search for articles on water radiolysis, there are many other results, such as these two. It's interesting to note that the effect of radiation on water still doesn't seem to be entirely understood, in part due to our incomplete knowledge of the behaviour of hydrated electrons.

Here's an extra bit of background on radiation and its biological effects (at least for the most common types. There are many ways nuclei can decay).
Depending how far the radiation source is from you, gamma radiation is either the least or the most dangerous type of nuclear decay. If the radiation is coming from well outside your body, then the danger scale goes alpha < beta < gamma. If you happen to have a radioactive nuclide in your pocket, the scale may well become alpha < gamma < beta. However, if the radiation is coming from inside your body (avoid ingesting glowing substances), then the danger scale goes gamma < beta < alpha. Why does radiation behave like that? There are two main effects in play: penetration capacity and ionization capacity.
In collision theory, different types of projectile and target can be characterized by something called an collision cross-section (or absorption cross-section, in nuclear physics), which basically gives you a probability of how likely it is for the projectile to hit a certain target. If the combination of projectile and target have a high collision cross-section, they they are very likely to interact, and so almost any projectile will be stopped by even a thin target. If the collision cross-section is low, then the projectile has a low chance of hitting the target, and therefore will tend to pass through it entirely. To compensate, you need to make a very thick target so there are more chances for the projectile and target to interact.
The different types of radioactive decay have different penetrating capacities. All of matter as we know it is composed of atoms, which contain well-defined oppositely charged regions (the nuclei and the electron shells), and so we can expect them to interact strongly with other charged particles. Alpha and beta(-minus) decays emit particles with a charge of $+2e$ and $-e$, respectively ($+e$ for beta-plus decay). We then expect these particles to interact strongly with matter, being stopped easily. Indeed, most alpha particles are efficiently stopped even by air. Beta particles have half the charge of an alpha particle and are going quite a bit faster (having approximately 1/7000th the mass), so they manage to go further before interacting. A few centimetres of of metal shielding or a concrete wall are usually enough to stop them. However, gamma rays have no charge, so they interact comparatively weakly with matter. These highly energetic photons will go though dozens of centimetres of solid matter. To maximize the odds that they will hit something and be stopped, you want to use a shield made of a very dense substance (which would pack a lot of possible targets in a small space). This is why lead is the substance of choice for radioactive shielding; its high density combined with low cost allows the fabrication of relatively compact shields which are capable of stopping even gamma radiation, when around 10 cm thick.
Now, for the mechanism of how particles from radioactive decay are stopped. For uncharged particles, the absorption cross-section is low because interactions only happen when the uncharged particle comes very close to a target particle. For gamma rays (very high energy photons), when there is an interaction, it is usually due to Compton scattering, where an energetic photon interacts with an electron, giving it a large amount of kinetic energy, while the photon is re-emitted with a smaller amount of energy. The target electron (not necessarily a valence electron) often gains enough energy to be completely removed from the atom, causing ionization.
For collisions between charged particles (both same and opposite charges), interactions happen much more easily, as charges interact strongly even at comparatively much larger distances. Beta-minus particles (very fast electrons) can more efficiently repel electrons and shove away from atoms, creating cations. Alpha particles (bare helium-4 nuclei) are interesting because their positive charge means they can tear electrons from any atom or ion. In fact, it's defensible that alpha particles are the strongest Lewis acid one can commonly find; they are effectively free protons with twice the charge.
Notice that penetration capacity and ionization capacity are inversely correlated; if a radioactive decay particle is strongly ionizing, it interacts well with matter and is stopped in short order. If it is weakly ionizing, one of the main interaction mechanisms is suppressed, and the decay particle can go further before it is stopped. In order to not damage you, you want the nuclear decay products to not dump their energy in your body, i.e. you want them to pass right through you. Here, it is clear that alpha particles are the most dangerous, followed by beta and gamma radiation. However, alpha particles interact so strongly with matter that they literally need to be released closer than your skin to dump energy into living tissue, so you need to ingest, breathe or inject alpha emitters. When this happens, they are extremely deadly. It is a very rare occurrence, however, since radiation usually comes from far outside. In that case, gamma radiation has a larger chance of reaching you, so it is said to be the most dangerous.

You can get the free radicals upon direct UV irradiation of water at wavelength shorter than $\lambda$ = 190 nm.


text:
How do you calculate the total heat capacity of a reaction
I need help with this one homework question:
I am combining $100\,\text{mL}$ of $0.5~\text{M}~\ce{HCl}$ and $100~\text{mL}$ of $0.5~\text{M}~\ce{NaOH}$ to obtain $200~\text{mL}$ of $0.25~\text{M}~\ce{NaCl}$. The change in temperature of this reaction was $\Delta \vartheta  =3.38~\text{°C}$. I have to calculate the heat of reaction for 1 mole of the reactant. I also need to calculate the $c_\text{total}$ of this reaction. How do I go about doing this?
So the equation is $$\ce{HCl + NaOH -> H2O + NaCl}$$
Suppose the heats of formation of $\ce{HCl}$, $\ce{NaOH}$, $\ce{H2O}$, and $\ce{NaCl}$ are $-166.19$, $-469.07$, $-285.85$, and $-406.77$, respectively, all in $\text{kJ/mol}$.
I went about doing this calculating the heat $$q=\Delta H=[-(-166.19-469.07)+(-285.85-406.77)]~\text{kJ/mol} = -57.36~\text{kJ/mol}$$
I then found the mass of the liquids used for the reaction:
$$\begin{aligned}m &= 0.1~\text{L}\times 0.500 ~\text{M}\times 36.4611~\mathrm{g\,mol^{-1}} \\
&\;\;+0.1~\text{L} \times 0.500~\text{M} \times 39.99715 ~\mathrm{g\,mol^{-1}}\\
&=3.822~ \mathrm{g} \end{aligned}$$
I then solved for $c$ using $q=mc\Delta T$.
$0.05\times 57360~\mathrm{J} = 3.822\ \mathrm{g}\times c_\text{total}\times 3.38~\mathrm{K} \Rightarrow c_\text{total}=222.01 ~\mathrm{J \, g^{-1}\, K^{-1}}$
By the way, I multiplied the $q$ by $.05$ because that's how many moles of reactant I have (since we have the molarity and volume of the reactant).
I am pretty sure I am doing something wrong, but I am not sure what.

This question is nicely showing your work, but could you elaborate why you think something is wrong? What conceptual problem you have with the question? As it is, this is a 'check my calculation'-type of question that is off-topic on this site.

You correctly calculated the heat per mole, $-57.36kJ/mol$...Consider that you do not have a mole of acid and base and then you're done !

Two things are wrong in your calculation. Well, it's really just one thing, but the other one bugs me more.

The total mass of the solution.
Units.

1. Mass of the Solution
To calculate the mass of the total solution, I think it is fair to make the following assumption:
$$ \rho_\ce{HCl} \approx \rho_\ce{NaOH} \approx \rho_\ce{NaCl} \approx \rho_\mathrm{H_2O} \approx 1000~\mathrm{kg\, m^{-3}}$$
$$ V_\text{tot} = V_1 + V_2$$
It says that we assume the densities of all the solutions to be the same, and that the total volume is the same as the sum of the two initial volumes.
Using the above assumptions, we can calculate the total mass of the reaction mixture:
$$ m_\text{tot} = \rho V_\text{tot} = \rho (V_1 + V_2) = 1~\mathrm{g\, cm^{-3}} \times (200 ~\mathrm{cm^{3}}) = 200~\mathrm{g}$$
You have calculated the reaction enthalpy to be $\Delta H = -57.36~\mathrm{kJ\,mol^{-1}}$.
Now we have to calculate the total heat that was released during this reaction, which leads me to
2. Units
Using unit analysis (we have $\mathrm{kJ\,mol^{-1}}$ and need $\mathrm{J}$) we know that we need to multiply the reaction enthalpy with some amount of substance (unit: $\text{mol}$). Now that we know what we have to plug in, we need to know how much of it. Since the reaction enthalpy that you calculated is per $\text{mol}$ of reactant, and we have exactly $0.25~\mathrm{M} \times 0.2~\mathrm{L} = 0.05~\mathrm{mol}$ of reactants, we calculate the total heat as follows:
$$ q_\text{tot} = n_\text{tot} \times -\Delta H_\text{reaction} = 0.05~\mathrm{mol} \times -57.36~\mathrm{kJ\,mol^{-1}} = 2.868~\mathrm{kJ}$$
Why the $-\Delta H$? Well, the heat that goes out from the reaction goes into the mixture. That is why the sign has to change here.
We also need the temperature difference, which was measured to be $\Delta T = 3.38~\mathrm{K}$.
The final result (the heat capacity) is calculated via
$$ q_\text{tot} = m_\text{tot}c_\text{tot}\Delta T \Rightarrow c_\text{tot} = \frac{q_\text{tot}}{m_\text{tot} \Delta T} = \frac{2.868~\mathrm{kJ}}{200~\mathrm{g} \times 3.38~\mathrm{K}} = 4.243~\mathrm{J\, g^{-1}\, K^{-1}
}$$
This is a little bit more than the specific heat capacity of water, so we're good!
I hope my walk-through has demonstrated all crucial points of where you went wrong, and that it clarified things for you.


text:
Reaction Order Greater than 1
What is the chemistry behind why reactions of order >1 exist? And what role does the number of activated complexes formed have in determining reaction rate?

I believe you should breakdown your question into different questions. All of them are good questions, but together, the compound question becomes too broad to answer.

I agree with @SatwikPasani, breaking these down will help the answers stay organized.

The following answers might be vague or overtly simplified or basic, but the questions by nature are fairly open-ended, aren't they? feel free to inquire further if something isn’t clear enough.

Take $$\ce{2NO2 \to 2NO + O2}$$ You can (extremely basically) think of this reaction as two $\ce{NO2}$ particles colliding, (forming a complex), then coming out as $3$ molecules, $2$ being $\ce{NO}$ and $1$ being $\ce{O2}$. But you needed two $\ce{NO2}$ molecules to collide. Collisions are unlikely, first the two molecules have to find each other, then they have to have enough energy, and be in the right orientation.
I’m not sure about multiple activated complexes, but I would only imagine that the reaction becomes even less likely if more than one activated complex is necessary for it; that basically means you need a second reaction from the first activated complex, which means a second collision. (likely slower rates? Someone else feel free to correct if I’m wrong)
Three molecules colliding at once ($3$ rd order reaction): way less likely, and that’s why it’s not studied so much, it’s much harder getting $3$ molecules to collide (especially in the gas phase without) in the right orientation with enough energy.
By general rate law do you mean finding the constant $k$? Or deriving its form entirely? Finding the orders for each reactant and the constant can only be done experimentally by using different concentrations and seeing what comes out.
I think there is more than one way to derive the general rate law’s form. One way is to take the steady state approximation. It has a useful problem that might explain more clearly. The steady state approximation is a commonly made assumption in various situations. Essentially, it assumes that the intermediate is consumed as soon as it’s produced (so there’s no net accumulation of intermediate – which makes sense for many situations). Using the info gained from this approximation you can use the intermediates and the relations that result to form the general rate law. Of course, it turns out the $k$ is composed of multiple constants.
It’s true because they told you it was. This gives you details that you wouldn’t have necessarily been able to deduce just by having the reaction equation there. Importantly, it’s there to tell you straight up that the equation you have can’t be broken down to further reactions... Again you can think of it like this: $\ce{a A}$ molecules have to collide with $\ce{b B}$ molecules to produce a single activated complex. That’s one step. If they didn’t specify that, maybe there could have been intermediate steps that could change the rate law.
Maybe it'll be easier to see if you consider extremes. Say Reaction $R$ can be divided into $4$ reactions, $r1, r2, r3, r4$. Now, say $r1, r2$, and $r4$ happen in microseconds, while $r3$ takes $100$ seconds. Already you're at the point where you can basically ignore the times of $r1, r2$, and $r4,$ because they're so small compared to the time $r3$ took (comparing $1\mathrm E-6$ to $1\mathrm E+2$ here). So as you can see in this example, $r3$ was literally the rate determining step.

http://www.science.uwaterloo.ca/~cchieh/cact/c123/steadyst.html

I have closed this as too broad, but when the OP breaks down the questions, you can rearrange your responses for each one and paste them as answers to the new questions.

Yes, for question 2 I do mean entirely. Why is each component of the rate law correct and significant in determining rate? I get the idea that higher concentration means higher rate, but why do we even have a constant?

Also, with regard to question 3, the answer was not helpful because it doesn't address why mole ratios in a single-step reaction equate to corresponding reaction orders in A and B. Please elaborate.

Hopefully someone else can answer #2 in depth.
For 3: The problem is specifying the conditions for you. You need experimental data to determine rate law for a chemical reaction. Just looking at the mole ratios isn't a valid way of determining the rate law. Hope that clarifies both a bit of 2 and 3.


text:
How do I calculate the temperature of hot air in kelvin?
The question simply says “calculate the temperature of hot air in kelvin” on a data calculation section for this lab. We recorded:

volume change on cooling = 170 mL
original volume of hot air = 20 mL
volume of the cooled air = 150 mL
temperature of heated water = 100 °C
temperature of cool water = 20.4 °C

and that is all.
The lab was focused on Charles’ law, namely to predict how much air should contract when cooled from 100 °C to room temperature. We took a flask and inserted a rubber stopper with glass tubing through the center. We then heated the flask in a water bath until the water boiled, then keeping a finger over the top of the glass tubing we inverted the flask in a “cool” water bath and held it there for 10 minutes. Then we equalized the pressure within the flask with the atmospheric pressure by raising the flask until the water level inside equaled the outside.

Should the volume change not be 20 mL and the hot volume 170 mL. Otherwise it doesn't make any sense?

You already know the temperature of the hot air in Celsius: 100 degrees
Using the equation Kelvin = Celsius + (273.15 degrees), you would get 100 degrees + 273.15 degrees, or 373.15 degrees.
I am assuming that this is a preliminary step that is required in order to find the temperature to plug into Charles' law (which must be in Kelvin), in which case you can use the formula I mentioned above for your other calculations as well.


text:
What are pharmaceutical uses for Vanillin?
What are some pharmaceutics that use vanillin? Is vanillin just an aroma and taste agent to pharmaceutics?

Vanillin is a pharmaceutical intermediate and flavoring, but not a pharmaceutical itself.  One can easily convert vanillin into dihydrocapsaicin (15 million Scovilles versus 16 million for capsaicin):  $\ce{NaBH3CN/NH3/NH4Cl}$, then isodecanoic amide of that benzylamine.  Surprise the downwind neighborhood.

To my knowledge, vanillin nowadays does not have any pharmaceutical uses.
But according to Elisabeth Vaupel in Pharmazeutische Zeitung online, 2002, 38 (SRC), Michael Bernhard Valentini wrote 1714 in his book Museum Museorum, oder vollständige Schau-Bühne aller Materialien und Specereyen, nebst deren natürlichen Beschreibung:

Sehr erwärmende und zerteilende, anbei aber auch stärkende Kraft, womit sie den Magen stärken, die Winde zerteilen und dem Gehirn, der Mutter und anderen nervösen Gliedern sehr gut tun. Sie treiben den Harn, befördern die monatliche Reinigung, natürliche Geburt und Schwierungen: Treiben auch die Nachgeburt und tote Kinder fort und kommen also dem weiblichen Geschlecht in ihren meisten Krankheiten wohl zu pass. Ingleichen werden sie gegen die erstarrend-machenden giftigen Bisse and andere dergleichen giftige Sachen gebraucht [...] Am meisten aber werden die Vanillen zu Verfertigung der Chocolaten gebraucht, welche sie anmutiger und kräftiger machen. Die Tabaksbrüder brauchen sie auch, den Tabak wohlriechend zu machen.

As a starting material in synthesis, vanillin can be converted to MMDA in 5 steps.


text:
Why is it possible to separate NaOH from certain compounds using acetone?
Is it due to the aldol reaction that happens? But the product of that reaction is still soluble in acetone.
Apparently if you have a solid mixture of 6-ethoxycarbonyl-3,5-diphenyl-2-cyclohexenone with NaOH and Na2CO3 impurites, and you dissolve them in acetone, after which you centrifuge, the solid impurites will settle out.
Why is this?

Are you sure that the name of the organic compound is correct?

The question is different from the procedure in the body. What you are actually doing is separating your organic compound (that I think is soluble in acetone) from $Na_2CO_3$ and $NaOH$ that are not soluble in acetone!

Ask yourself what kind of reaction happens?
What glassware have you used to remove impurities?
By controlling pH and polarities you are able to separate compounds because of different affinities between them and solvents. You change molecular solvatation of your analyte and impurities. Then by controlling boiling point, using a rotavap, you may be able to separate two phases that are miscible. Or by making impurities insoluble you can remove by mechanical operation such as centrifugation. Such post-treatments are very common in chemistry and they scale up very well.
So, there is no chemical reaction (except acid/base reaction if any) there are physical reactions (solvatation, phase extraction, phase change: evaporation, condensation, crystallisation and sedimentation).

In this particular case, the separation utilizes the low solubility of $\ce{NaOH}$ and $\ce{Na2CO3}$ in acetone.


text:
Why does HNO3 not give off H2 when reacting with Cu?

A dilute solution of which acid is most likely to produce a  reduction
product other than $\ce{H_{2}}$ when it reacts with a  metal?
(A) $\ce{HF}$
(B) $\ce{HCl}$
(C) $\ce{HNO3} $
(D) $\ce{H2SO4}$
$\tiny{\text{Question from 2012 local chemistry olympiad}}$

The given answer is (C), $\ce{HNO3} $ but I don't understand why.
First off, what does it mean by reduction product -- is it the product that is formed when the acid is reduced in the redox reaction between the metal and the acid?
Second, why is the answer $\ce{HNO3} $?

Whenever any metal is added to an acid, the metal is oxidised to one of its positive ions, at the cost of either the acid which is reduced or water which is reduced to hydrogen. The product formed by reduction is what is referred to as the reduction product.
The reduction potential of $\ce{NO3-}$ is $0.8$ which makes the reduction of nitrate instead of hydrogen ion $\ce{H+}$ ($E^\circ=0.0$) more favourable. Thus it is the nitrate in nitric acid which is reduced instead of $\ce{H+}$ or $\ce{H2O}$. But at very low concentration of nitrate ion and low temperature, as explained by Nernst equation, the reduction potential decreases significantly and combined with overpotential effects, it is $\ce{H2}$ which is formed and not the reduction product of nitrate which is $\ce{NO2}$.
In the case of halic acids, the halogens cannot be further reduced. In the case of the sulfate anion, although it has a reduction potential of $0.17$, owing to overpotential effects, sulfate is not reduced, instead, water is reduced yielding $\ce{H2}$. At high concentration, the reduction potential rises sufficiently to overcome the overpotential effects, allowing the reduction of sulfate to form $\ce{SO2}$ at higher concentrations of sulfuric acid. Hence under normal conditions, only nitric acid gives a product other than hydrogen gas.
Reduction potential data
$$\ce{SO4^{2-} + 4H+ + 2e^- -> SO2 + 2H2O}(E^\circ=0.17)$$
$$\ce{NO3- + 2H+ + e- -> NO2 + H2O} (E^\circ=0.8)$$
$$\text{Nernst equation } E=E^\circ- \frac{RT}{nF}\ln Q$$
$Q$= reaction quotient, $E$=redox potential, $E^\circ$=Standard redox potential
EDIT
The oxidation of metal at the cost of reduction of another substance occurs whenever it is thermodynamically favourable. Usually, (except silver, copper) metals have negative reduction potentials, which makes their oxidation at the cost of hydrogen thermodynamically favourable as the $\Delta G<0$. This leads to a spontaneous redox reaction. This is not true if there is a kinetic or physical barrier to reaction. For example, Iron is rendered passive by nitric acid due to the formation of an oxide layer at the surface which prevents any further contact and hence stops the reaction.

I found this answer to a similar question on Yahoo Answers:
"If the metal is below hydrogen in the reactivity series, then the oxidizing agent is the nitrate ion, not hydrogen." I don't understand how the metal being below H in the reactivity series determines the oxidizing agent. Could you please explain how this is?

You should really mention standard conditions. Dilute Nitric acid is  close to 1M which is one standard condition.

@user2617804 The $\circ$ in the superscript of the redox potential symbol $E$, already indicates the standard state as per the IUPAC recommendation.

@user2612743 Activity series is another way to state the redox potentials. The elements above hydrogen have negative reduction potential and can hence liberate hydrogen by reducing it in water. The elements below have positive electrode potentials and hence cannot reduce hydrogen, therefore they must reduce the nitrate anion which has a more positive reduction potential.

@1110101001 Please don't rely on Yahoo Answers. Please.


text:
How to separate HCl and NaOH from salt and water solution?
I mean, how can we reverse the neutralization reaction of $\ce{HCl}$ and $\ce{NaOH}$ to get back $\ce{HCl}$ and $\ce{NaOH}$.
I want to extract $\ce{HCl}$ and $\ce{NaOH}$ from a mixture of common salt and distilled water in equal proportions at home.
Basically, how can I make this reaction happen?
$$\ce{H2O + NaCl -> HCl + NaOH}$$
What temperature or pressure conditions are required?
Also, would the extracted components be pure? (I know common salt contains iodine and even distilled water is not pure, that's why I am asking this.)

Unless time travel is an option, you could

Electrolyze the $\ce{NaCl}$ solution to obtain a solution of $\ce{NaOH}$, and $\ce{H2}$ and $\ce{Cl2}$ as gases.
Collect the gases and photolyze them. The dissociation energy of $\ce{Cl2}$ is $243\, \mathrm{kJ \cdot mol^{-1}}$, irradiation  at $\lambda$ < 490 nm will cleave $\ce{Cl2}$ to chlorine radicals and initiate the chain reaction to yield $\ce{HCl}$ gas.

$$\ce{Cl2 ->[h\nu] 2 Cl*}$$
$$\ce{Cl* +\  H2 -> HCl + H*}$$
$$\ce{H* +\  Cl2 -> HCl + Cl*}$$

EDIT 1
Unless you know exactly about the risk of both processes, don't do it!
This video gives a nice impression of how violently hydrogen and chlorine react upon irradiation! Note that it has nothing to do with the laser involved. A flashlight would give the same dramatic effect!

EDIT 2
The answer above, giving practical advices on how to get $\ce{HCl}$ and $\ce{NaOH}$ back apparently suggests that you can not simply revert the neutralisation reaction.
The neutralisation is irreversible.
In order to find out why this is the case, you might want to have a look at thermodynamics of this exothermic reaction, particularly at the entropy changes.

That vid seems really cool. It would be fun trying that out! Just kidding. Just wanna know, after the burst in the video, was HCl finally formed or was the reaction stopped?

Yes, HCl is definitely formed here. I updated the answer and added the mechanism. The video is really impressive and it is fun to carry out - in a well-vented lecture hall or in a fumehood, but not in the living room ;)

Is it really entropy that dominates the huge free energy release on neutralization? I figured most of it would come from the enthalpy change on formation of the additional, strong O-H bond.

Strictly speaking, it is theoretically possible to reverse neutralization using super-critical steam on crystalline $\ce{NaCl}$. However, it requires extreme conditions.

The video link doesn't work.

The question is whether you want to get pure $\ce{HCl}$ and pure $\ce{NaOH}$ or whether you want a hydrochloric acid solution and a sodium hydroxide solution. If the former, I hereby direct you to Klaus’ answer. If the latter, read on.
The neutralisation of hydrochloric acid and sodium hydroxide is often simplistically given as in equation $(1)$.
$$\ce{HCl + NaOH -> NaCl + H2O}\tag{1}$$
However, that reaction is only true for gasous $\ce{HCl}$ reacting with solid $\ce{NaOH}$. When one neutralises the acidic, aquaeous solution and the basic, aquaeous solution, you need to first consider equation $(2)$ (for $\ce{HCl}$) and $(3)$ (for $\ce{NaOH}$).
$$\ce{HCl (g) + H2O -> HCl (aq) + H2O <=>> Cl- (aq) + H3O+ (aq)}\tag{2}$$
$$\ce{NaOH (s) + H2O -> Na+ (aq) + OH- (aq)}\tag{3}$$
In both solutions, the compounds can be considered completely dissociated. Once you mix these two together, a standard acid-base reaction happens in which only half of the ions react at all — see equation $(4)$.
$$\ce{Cl- (aq) + H3O+ (aq) + Na+ (aq) + OH- (aq) -> Cl- (aq) + Na+ (aq) + 2 H2O}\tag{4}$$
Removing the spectator ions gives us equation $(4')$
$$\ce{H3O+ (aq) + OH- (aq) -> 2 H2O}\tag{4'}$$
And this, when viewed closely, is the autoprotolysis of water (equation $(5)$) reversed.
$$\ce{2 H2O <<=> H3O+ + OH-}\\ K_w = [\ce{H3O+}][\ce{OH-}] = 10^{-14}\tag{5}$$
The ion product $K_w$ I added there shows the concentration of $\ce{H3O+}$ and $\ce{OH-}$ ions in a neutral solution. To be able to reverse that reaction you would somehow need to generate substantial quantities of $\ce{H3O+}$ and $\ce{OH-}$ — but they just aren’t present at equilibrium in the concentrations you would need, since their formation is strongly endothermic (and thus their destruction strongly exothermic).
And that is the true reason why an acid-base neutralisation is not reversible in many cases.


text:
Proton transfer equilibrium in bisulphite adducts
In the addition of $\ce{NaHSO3}$ to aldehydes and methyl ketones (higher ketones do not respond well to this reaction), crystalline addition products are formed. An $\ce{-SO3H}$ and an $\ce{-OH}$ group are attached to the same carbon.
$$\ce{R2C=O + NaHSO3 -> R2C(OH)(SO3^{-}Na^+) <=> R2C(O^{-}Na^+)(SO3H)}$$
The line I read online, bothering me is:-

"The proton transfer equilibrium lies to the right for most ketones
  and to the left for most aldehydes."

What is the reason for this (assuming that it is correct in the first place)? Since the sulphonic acid group is more acidic than the hydroxyl group, should not the proton tranfer equilibrium lie towards the left for both the cases?

Could you provide a reference to the source where you read that claim?

@Philipp Can't seem to find my original source. But found another one here pg 359. This appears to be comparatively more credible than the original source (some sort of notes), but still am not sure whether this is true.

Hmm, your linked book says it is so "for steric reasons". That is rather vague especially because they don't explain in detail which steric interactions get relieved. Another problem is that they don't say whether their statement is true only for the crystalline compound or for the compound in solution too. If I had to guess, I'd say that maybe in the crystalline structure it might be true (although I don't think anyone could have predicted that without knowing the crystal structure) but not for the compound in solution.

A little bit of searching gives no indication that the structure on the right is formed. See here, here, here and here

@NicolauSakerNeto I even searched a few books. At best this equilibrium is shown without commenting on which products predominate.

I tried to calculate some simple model, apparently (in gasphase) $\ce{MeHC(OH)(SO3^{-}Na+)}$ may exist, while $\ce{MeHC(O^{-}Na+)(SO3H)}$ is blown into pieces. I have not yet tried a ketone. (Solvent might also have to be considered. While I have some time for that I might be coming back on that one.) I honestly cannot think of a plausible reason why the right structure should exist (esp. in solution). In what solvent would this reaction be carried out?

@Martin probably in aqueous acidic medium or a protic solvent. But i guess the proposed stabilities refer to the stabilities in solid state for the crystalline adduct.

I think your confusion arose because of an error in the translation of the text you linked (the discussion on p. 359) and that this text was unclear about which equilibrium it was referring to. In that figure, the equilibrium that the authors are presumably referring to is the (planar) carbonyl- (tetrahedral) hemiacetal equilibrium (as described on page 358). So the equilibrium that the text refers to isn't the "proton transfer equilibrium," but the planar-tetrahedral equilibrium.

In contrast, the scheme you posted with your question assumed that the formation of a hemiacetal (or hemiketal) is irreversible, as indicated by a unidirectional arrow. This reaction should be treated with an bidirectional equilibrium arrow as shown in p. 359 of the document that you linked.
In this context, the steric argument is reasonable and the reaction as a whole analogous to the more well-known relative hydration equilibria of aldehydes and ketones (which is related the general notion of aldehydes being more electrophilic than ketones).


text:
Will monosodium dihydrogen phosphate react as base or acid in water?
Will $\ce{NaH2PO4}$ react as as an acid or a base if you make it react with water?
I got stuck here. I can figure out the reaction:
$$\ce{NaH2PO4 + H2O -> Na^+ + H2PO4^- + H2O}$$
But when I look $\ce{H2PO4^{-}}$ up, it says that it can both be an acid and a base.
Any ideas?

Monosodium dihydrogen phosphate, $\ce{H2PO4^-}$, is an amphoteric species and it will act as both an acid and a base.
These are the related chemical equations:
\begin{align}
  \ce{H2PO4- + H2O &<=> HPO4^2- + H3O+}  & \mathrm{p}K_\mathrm{a} &= 7.21\\
  \ce{H2PO4- + H2O &<=> H3PO4 + OH-}     & \mathrm{p}K_\mathrm{b} &= 11.88
\end{align}
As an example, $\pu{0.1 M}$ solution of such a salt will have a pH of about $4.7$.
In other words, the acidic behaviour dominates over the alkaline one. This is due to the value of $\mathrm{p}K_\mathrm{a}=7.21$ (related to the acidic behaviour) compared to the $\mathrm{p}K_\mathrm{b}=11.88$ (related to the alkaline behaviour).

Wouldn't it be acidic on every concentration in water?

Acidic yes,  but pH would vary with concentration although slightly.

if acidic behavior dominates, why would it be amphoteric ?


text:
How to increase the velocity of a reaction?
I'm doing relatively simple Chemistry but I've been given a formula for the velocity of a reaction.
$$v= k \cdot c_\mathrm A^a$$
and
$$k= A \cdot \mathrm e^{E_\mathrm a/RT}$$
As a result, from that formula, surely decreasing the temperature $T$ would increase the velocity of the reaction? As that would increase the value of $k$ and therefore drive up the reaction speed? Also, same thinking, increasing the activation energy $E_\mathrm a$.
I know that's wrong, because common sense. But the formula is lying to my face.

It should be $-E_a/RT$
Therefor, increasing the temperature, lowering the activation energy (catalysis), or increasing the concentration of reactant(s) would increase the reaction velocity.


text:
Could this be a feasible fuel system?
First off, please forgive me, I am a chemistry novice!
Overall Process

Fuel for the system is distilled water
Split water into Hydrogen and Oxygen (electrolysis or similar process)
Exhaust Oxygen
Burn Hydrogen to power a generator/engine
Generator/engine charges battery
Battery powers Step 2 as well as whatever else we want to power

Problems

Energy gained needs to be greater than the sum of lost energy in all steps, as well as the energy required to perform electrolysis
We could compress the exhaust oxygen and use that as an energy source?
Is there a danger of running out of hydrogen? (Forgive my naivety, but I've heard about a helium shortage...)

"Energy gained needs to be greater than the sum of lost energy in all steps"
http://www.lhup.edu/~dsimanek/museum/unwork.htm
 Perpetual motion machines
http://news.sciencemag.org/chemistry/2014/01/rhubarb-battery-could-store-energy-future
 Flow battery, which works.
1) You cannot win (First Law of thermodynamics from Noether's theorems and the isotropy of time)
2) You can only break even on a very cold day (Second Law of thermodynamics from Carnot cycle and the law of large numbers).
3) It never gets that cold (Third Law of thermodynamics.   If there were an entropy difference at absolute zero, T = 0 K could be reached in a finite number of steps. However, at T = 0 K there is no entropy difference, so an infinite number of steps are needed).

Understood that I cannot create energy.  Perhaps I should specify that energy gained should be greater than the sum of lost energy in Steps 2 through 6.

Water, being the fuel in the system, would have potential energy that I am not attempting to have replenished by this system.

Energy storage is simplicity or value regain from discards. Magnets grab ferrous alloys, linear induction windings repel conductors. PET is valuable. Recycle! Burn other plastic and fiber as fuel in secondary regeneration. Landfill anaerobic ferment everything else.


text:
Specific Case of the Rate law
My textbook says that for a chemical reaction with two reactants, $\ce{A}$ and $\ce{B}$ with equation $\ce{aA + bB -> Products}$, where $a$ and $b$ are coefficients, the general rate law is:
$\text{Rate} = k[\ce{A}]^m[\ce{B}]^n$
where $m$ and $n$ are the reaction orders for $\ce{A}$ and $\ce{B}$ respectively. Then it says that $m=a$ and $n=b$ if  the reaction between $\ce{A}$ and $\ce{B}$ occurs in a single step and with a single activated complex. Why is this true?

According to collision theory, for a chemical reaction to occur, the reacting molecules must collide with each other.
For a bimolecular reaction (as given by you), collision occurs in the single step and thus becomes the rate determining step, as there are no other steps involved.
If you consider reactions like bromide ions reacting with bromate ions in the presence of acid, it involves 12 reacting particles. If all the 12 reacting particles collide simultaneously you can write rate law expression considering it to be elementary, but those events are extremely rare. In reality such reactions are found to be complex i.e reactions which take place through a sequence of two or more consecutive steps. Here, there are many steps involved, and one should consider slowest step to write rate law expression. But, why? That will become another SE question. In case of reactions taking place in single step, as there are no other steps left to compare with, whether it is slower or not, so you need to consider  the theoritical equation.
One more thing to be noted is that elementary reactions are associated with single transition state and not single activated complex. For more info read this page: Activated complex

This does not tell why the relationship between a and m is that one is equal to the other. Why is it that if you have 2 particles of A and 1 particle of B reacting (mole ratio) then the rate law has the reaction order being 2 in A and 1 in B?

In elementary reactions, 2 particles of A and 1 particle of B collide simultaneously to form products. So, rate of disappearance of A must be twice as that of B. Suggesting that order of reaction being 2 in A and 1 in B.

Twice is not quadratic, which is what order 2 means, though.

Your question is not clear.

Basically I'm saying, why is it what you said 2 comments ago implies that order of reaction is quadratic. How does a directly proportional relationship in concentration lead to a quadratically proportional relationship in order?

I didn't say total order of reaction to be 2. I said order in A as 2.

That's what I meant, sorry.

First some definitions:-
Elementary reaction= The reaction that takes place in a single step i.e one transition/activated complex according to the transition theory or one single collision event according to the Collision theory of kinetics.
Molecularity= Number of independent species involve in the collision event in case of an elementary reaction. These all collide together with sufficient energy (threshold) and proper orientation to convert themselves into products.
Now, let us assume the collision model for gaseous reaction where there are two species $A$ and $B$ which react together in a reaction according to :-
$$\ce{aA + bB ->}\text{ products}$$
This reaction will occur whenever $a$ molecules of $A$ collide with $b$ molecules of $B$ in proper orientation. Suppose out of all collisions of $a$ molecules of $A$ with $b$ molecules of $B$, a fraction $P$ occur in proper orientation. The probability of collision of these specified number of reactants is proportional to the $a^{th}$ power of $[A]$ and $b^{th}$ power of $[B]$. This is because the concentration of the reactant gives crudely the probability of finding that reactant in a unit volume. For collision probability, you multiply the probability of finding one reactant to the probability of finding another reactant. Since there are $a$ molecules of $A$, therefore the probability of finding $A$ will be multiplied $a$ times and similarly for $B$. Thus the resultant rate based on collision probabilities will be :-
$$k=PZ[A]^a[B]^b$$
Where Z is an appropriate constant (since the probablities are only proportional to the concentrations and not equal to them, there is some appropriate constant for each proportionality which finally multiplies to give $Z$). This is true only for elementary reactions adhering to the Collision theory-Mechanism of reaction.

I understand this, except for the notion that probability of collision is proportional to the ath power of A. Could you please elaborate?


text:
Reaction Rate Limiting Reaction
Why exactly does the slowest reaction limit the instantaneous rate of a reaction even if the faster processes come after? Theoretically, if the products of the slow reaction are required for the fast reaction, the fast reaction limits the rate because even though it is fast, it will not finish in exact synchronization with the end of the slow reaction. There will be a lag between when the net reaction is done and when the slow reaction is done.

It is the measured overall reaction rate that counts and not which reaction finishes the sequence.
If you measure the rate of a reaction that consists of two elemental reactions with rate constants $k_1$ and $k_2$ where $k_1 \gg k_2$ you will get an overall rate constant $k_{\mathrm{tot}} \approx k_2$.
This is so, because if the faster reaction takes place after the slower reaction, the faster reaction uses up its reactants (which are the products of the slower reaction) faster than the slower reaction produces its products, so that the faster reaction will consume every new product molecule produced by the slower reaction nearly instantly compared to the time it takes the slower reaction to produce it.
The time it takes the faster reaction to consume the newly formed molecule is insignificant.


text:
Rate and mechanism of the reaction of Mg and HCl
In the reaction between hydrochloric acid and Magnesium the overall reaction is:
$$\ce{Mg + 2HCl -> MgCl2 + H2}$$
My proposed mechanism of reaction is that $\ce{HCl}$ dissociates in $\ce{H2O}$ like so:
$$\ce{H2O + HCl <=> H3O+ + Cl- }$$
And then the $\ce{H3O+}$ oxidises the magnesium like so:
$$\ce{2H3O+ + Mg -> Mg^{2+} + H2O + H2}$$
This, in an experiment to determine the rate of the reaction, measuring the $\ce{H2}$ produced would measure this mechanism, because it is the step in which hydrogen gas is evolved. Presumably these Magnesium ions are then in solution. I would therefore deduce that the step:
$$\ce{2H3O+ + Mg -> Mg^{2+} + H2O + H2}$$
Is the rate-determining step, because it is the step in which magnesium ions are formed and go into solution.
My rate law, deduced from these mechanisms, would look like this:
$$\ce{rate  = k * [H3O+] }$$
Because it is the concentration of the $\ce{H3O+}$ which determines the formation of the Mg ions. From there, I would assume, the $\ce{Mg^{2+}}$ and the $\ce{Cl-}$ would remain in solution, as the reaction is carried out in the acid, obviously consisting of water.
From this, can I say that the rate law for the reaction
$$\ce{Mg + 2HCl -> MgCl2 + H2}$$
is merely:
$$\ce{rate  = k * [H3O+] }$$
But $\ce{[H3O+]}$ is:
$$\ce{[H3O+] = (K*[H2O][HCl])/[Cl^{-}] }$$
Where K is the equilibrium constant of the acid equilibrium.
So can we write our rate law like:
$$\ce{rate = k*(K[H2O][HCl])/[Cl^{-}] }$$
Where K is the equilibrium constant and k is the rate constant and might look like this (following the Arrhenius equation):
$$\ce{k = A e^{-E_a/(R T)}}$$
To make our rate law like this:
$$\ce{ rate = A e^{-E_a/(R T)} *((K*[H2O][HCl])/[Cl^{-}]) }$$
My question is: Is my rate law correct for this reaction? If not, where have I gone wrong?
Additionally: This was one of our practicals. We ended up concluding that rate goes up quadratically or exponentially with concentration. This rate law would seem to indicate a linear increase, not exponential. I assume in my rate law there is no contribution from the solid Mg. Obviously in a powder, more surface area is exposed, and a high rate is experienced. Is there a way to factor this into my rate law?
Thanks for the help! :)

Your rate law should be
$$ r = k_\text{obs} \times [\ce{H3O+}]^2 = A_\text{obs} \exp\left(\frac{E_\text{a, obs}}{RT}\right) \times [\ce{H3O+}]^2\; ,$$
which nicely fits your experimental data of a quadratic dependence on the concentration for the rate.
The exponent is necessary because the reaction is seemingly second-order in the concentration of $\ce{H+}$.

Why? What steps/mechanisms would be in the reaction if this was the rate law? :) Thanks

Is H3O order 2 because there are 2 molecules needed to oxidise every atom of Mg?? Thanks :)

Well, according to the mechanism you propose: yes. If it really happens concertedly or via an intermediary step ($\ce{H+ + Mg -> Mg+ + H^{.}}$) I don't know.

OK. Thanks for clarifying. I would think that it would happen with an intermediate, so Mg is first oxidized to Mg+ then to Mg+2 a second time. But I don;t know. :)

Yeah, it seems more reasonable. But hey, if it fits your data (and it's not valuable, to-be-published research), go ahead and propose the concerted mechanism ;)

For the reaction $\ce{2H3O+ + Mg -> H2 + Mg^{-2}}$; the actual rate law is
$r=k \times [\ce{H3O+}]^2 \times (\ce{Mg})^1.$
The most obvious error is that HCl is a strong acid, thus not in equilibrium. Cl- is merely a spectator ion.
A more accurate reaction step method is:
$\ce{H3O+ + Mg -> H2O + MgH}$
$\ce{H3O+ + MgH -> H2 + Mg^{-2}}$
The surface area of magnesium directly affects the k value, but magnesium is still a valid part of the reaction. The magnesium term may appear zeroth order if the magnesium is in far excess, so significantly lowering the initial input of magnesium will probably result in the correct rate law.

I hope while editing I haven't messed up anything!! Feel free to rollback if i have messed up stuff :-)

I didn't think solids could be considered in a rate law - [] indicates concentration, which can only happen in liquids/gases, not solid. Clearly the surface area of the Mg has an effect - typically taken into account in the A term of the Arrhenius equation


text:
Why is HF considered a covalently bonded compound?
Why is the bond between H and F in HF considered polar covalent whereas HCl, HI, and so on are all ionic? The electronegativity difference between them suggests that it too should be ionic, yet all textbooks say that HF is covalent. Is there a good reason why?

possible duplicate of Do hydrogen halides (HX) form strong or weak acids?

Not necessarily a duplicate. This question is about the covalent character of the H-X bond. I would argue that all of the H-X bonds are covalent (until water becomes involved).

I don't think I've ever seen a reputable source claiming the bonds in HCl, HBr and HI are mainly ionic. The fact that they are all gasses at room temperature clearly suggests otherwise. HF has the most ionic character of all the hydrogen halides, but it too has a rather low boiling point (just below room temperature), which is uncharacteristic of ionic compounds.
It is best to think of all hydrogen halides as covalent polar molecules, with the polarity increasing in the order HI < HBr < HCl < HF, as suggested by the electronegativity differences.

From the Wikipedia article on hydrogen chloride: "Hydrogen chloride is a diatomic molecule, consisting of a hydrogen atom H and a chlorine atom Cl connected by a covalent single bond." You should find similar sentences in the articles for the other compounds.

A few words of support of Nicolau's answer: charge separation can be a bad thing. Noting that H has a proton, there is good reason to leave on the order of one electron in its vicinity. This would imply a covalent-like bonding scenario.

@EricBrown Indeed, that's an interesting fundamental reason why an ionic salt containing discrete non-solvated $\ce{H^+}$ ions is all but impossible; the free hydrogen cation has an immense ability to polarize electron clouds (which I have discussed elsewhere), demanding a high degree of covalent bonding with even the most unwilling species. However, I guess that argument does not strictly preclude an ionic $\ce{HX}$ compound with, for example, discrete $\ce{HX2^{-}}$ and $\ce{H2X^{+}}$ ions, at least qualitatively.

@EricBrown See, e.g., bifluoride and fluoronium, for examples of $\ce{HX2-}$ and $\ce{H2X+}$ species (more links to related content therein).

As already stated in the comments, this question includes the FALSE assertion that HCl, HI, and HBr are ionic. No other reply is appropriate. https://en.wikipedia.org/wiki/Hydrogen_halide. Ionic bonds are those which can be nearly completely characterized as electrostatic. That is, by the assumption that one ion is positively electrically charged and the other is negatively charged - or by combinations of various integer charged ions resulting in a net attraction between the ions. The key point is that the electron(s) responsible for bonding are concentrated around the anion(s) and the cation(s) have lost one (or more) of them so can be considered to have a positive charge. This is to be contrasted with covalent bonds in which the electrons are shared between atoms in the bond. This simple definition ignores the amount of "sharing" and the amount of "concentration" which occurs. An ionic bond is "mostly" unshared electrons (substantial charge separation) and a covalent bond is "mostly" shared electrons. In fact, most chemical bonds between dissimilar atoms are somewhere between pure covalent and pure ionic. There is no requirement that the compound be in the solid state. I know of no ionic gasses (in their ground state) near room temperatures (how could you explain the separation of charge?). There are many ionic liquids, and of course ionic solids are common.

According to dipole moment we can get the information about "percentage ionic character of a bond", so for H-F bond the %age of ionic character comes out to be about 43% ionic and 57% covalent. Therefore this bond is a covalent bond.

General definition of ionic says that there should be an electronegativity difference of more than 2.  H is 2.2, F is 3.98, so not quite a big enough difference, which is true of all hydrogen halides.  These are polar covalent bonds.
What I find interesting is that they are all normally named by ionic convention, when they should be termed Hydrogen monoflouride, hydrogen monochloride, etc. by binary covalent naming conventions.


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Oxygen in water: How is it single if it's always double?
This is simple question, I think. At college (I'm an Intro Chem. student), the professor explained that oxygen is always found in molecules of 2 oxygen atoms combined together - $\ce{O2}$. On the other hand, molecule of the water is $\ce{H2O}$. So I don't get, how is it that O is single in this molecule, if it's always found coupled?

Free single oxygen atoms are unstable and will bind to another free single oxygen atom that is present in the air because this is more stable.(So you get $\ce{O2}$) When you look at electron configurations then you see that both oxygen atoms share their electrons with each other and so have the same electron configuration as Neon, which is very stable: $\ce{1s^2 2s^2 2p^6}$
This is also the same for other non metals such as nitrogen gas : $\ce{N2}$ and also chlorine gas ($\ce{Cl2}$) and fluorine gas ($\ce{F2}$)
Now, the oxygen atom in a water molecule is bonded to not another oxygen molecule, but to two hydrogen atoms.($\ce{H2O}$) The oxygen atom has a partial negative charge of -2 and both hydrogen atoms have a positive partial charge of +1. This has as a result that all atoms once again have stable electron configurations : oxygen once again has a stable electron configuration, equal to Neon.
$\ce{O2}$ is the pure element, water is a molecule. But they both achieve stable electron configurations. Just with different atoms.

I think you are confusing something here: Molecular oxygen is most commonly found as the gas we know, $\ce{O2}$ and ozone, $\ce{O3}$. However, this doesn't limit it to only that. Oxygen can appear in a wide variety of (bio-) chemicals and there, it only needs two bonds to other atoms (or a double bond to one atom).
Good examples for this are:

The aforementioned water $\ce{H2O}$ where the oxygen is bonded to two hydrogen atoms.
One of my most loved molecules: Ethanol. Here, the oxygen is bonded to a hydrogen and a carbon atom.
In an ester, which usually are quite fragrant compounds, there actually are two oxygen atoms. But they are not connected to each other. Instead, one has a double bond and the other two bonds, each of them to carbon respectively.

I hope this has made it clear that oxygen doesn't have to bond to itself (but it can, see peroxides), and that oxygen doesn't have to come in pair of twos for the molecule to not fall apart.