text:
Standard Hydrogen Electrode
Could someone, please, explain to me how the standard
Hydrogen Electrode works and how it is used to measure electrode potentials? Also, why does it have a potential value of zero?

The SHE (Standard Hydrogen Electrode) is a standard equipment for measuring the relative electrode potential, as the process of measuring the actual potential is quite difficult. It consists of a platinum foil dipped in 1.0M $\ce{HCl}$ in the presence of $\ce{H2}$ gas at 1 bar of pressure.
Now when the SHE is connected to another half-cell, either oxidation or reduction may occur in the SHE.

Suppose we have to find the electrode potential of a half cell represented by "A" Now, on setting up the cell and connecting a voltmeter we get the standard potential(Estandard) of the cell i.e.Estandard = EA - EH2
Now as our our aim is to just find EA, we take EH2 to be equal to 0
Therefore,
Estandard = EA - 0
or,
Estandard = EA

The value of Estandard is taken from the voltmeter.
It should be noted that the electrode potential calculated using this method is relative(to the electrode potential of hydrogen). It isn't the absolute value.
Also, the potential for the SHE is taken to be 0 for the simplification of the process.

The SHE also has hydrogen gas in it so that either half reaction can occur: $$\ce{2H+ +2 e- -> H2}$$ $$\ce{H2 -> 2H+ + 2e-}$$


text:
Coloration of substances because of F-Centers
In case of metal deficient defects in the crystal lattice of a metal, the positive voids are filled by electrons.

How does this result in the coloration of a substance?
Does this condition arise only in the case of metals?

For a simple explanation, we could explain the coloration of the substance by the absorption of the energy of the electrons in the void, which leads to the color of the compliment color (as is suggested by http://rspa.royalsocietypublishing.org/content/204/1078/406)
However, a more rigorous explanation could be found here: http://ptp.oxfordjournals.org/content/4/2/181.full.pdf
For the second question, it is written in Ceramic Materials: Science and Engineering that "the term color center is now applied to any defect, including an impurity, that produces color in an insulator". It also notes that the original observation was made by the production of F centers, which suggests that the modern interpretation of the phenomenon applies to the defects which product color in an insulator.
I'm not sure if I answered you question exactly as you hoped, but if you look into the resources, there is a wealth of knowledge there.

source: Stack Exchange

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text:
Synthesis of ethyl methyl ketone from 2-(4-hydroxyphenyl)ethyl methyl ketone
Is it possible to reduce raspberry ketone (4-(4-Hydroxyphenyl)butan-2-one) to ethyl methyl ketone? How would I do this in practice?

In practice you should simply buy butanone ;)

Well, that should probably be the right answer. I just need a small amount though, I don't want to buy a whole litre, and I have 4-(4-Hydroxyphenyl)butan-2-one already. I thought I would ask to see how difficult it is, since it may be easy (I don't know a lot of chemistry)

There are a number of ways to accomplish your goal, but they all involve several synthetic steps. As mentioned in the comments, it's always easier to buy a compound then run a reaction, but if you had a small amount of some exotic derivative, you could convert it to methyl ethyl ketone.
One possible route would be to convert the raspberry ketone to an ester using the Baeyer-Villager reaction.  In this reaction the secondary (ethyl) branch migrates in preference to the primary (methyl) side.  The resulting ester can then be converted to the desired ketone by any number of methods such as the Weinreb route.

Wow epic. Chemistry blows my mind. So after the baeyer-villager reaction, which particular ester would be left?

After the Baeyer-Villiger you would have the ester pictured in the top-right corner of my drawing.  It is just your starting Raspberry ketone with an oxygen inserted between the carbonyl and the p-hydroxyphenethyl group.  If this answer was helpful please consider accepting and\or up-voting it, thanks.

Also, if I did a Catalytic Baeyer-Villiger oxidation using hydrogen peroxide, and hydrochloric acid as a catalyst, would that give me Methyl ethyl ketone peroxide, an explosive?

No, peroxides are thought to be intermediates in the reaction, but they are not isolated.  Rather the presumed peroxide intermediate rearranges further during the reaction to yield the final product.

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text:
What is an effective extraction solvent for removing hydrocarbons from soil (using a soxhlet extractor)?
I am interested in improving the extraction efficiency of hydrocarbons from soil using a soxhlet extractor, and I would like to know what solvents might improve this extraction efficiency, and why. Currently, I am using 1:1 hexane:acetone.

The answer depends on what type of hydrocarbon (alkanes, polycyclic aromatic, etc.) you are interested in removing.  If you google "solvent to extract hydrocarbons from soil", you'll find many useful links.  It seems that methylene chloride:acetone (1:1) is commonly used.

Unfortunately I cannot narrow the focus down to a class of hydrocarbons. Alkanes and polycyclic aromatics could both easily be in my samples. Thanks for the advice, and the suggestion of solvent choice!

possible duplicate of What types of chemicals/ionic liquid could perform this type of separation?


text:
What is the pH of a 5M solution of hydrochloric acid?

What is the $\mathrm{pH}$ of a $\pu{5M}$ solution of hydrochloric acid?

So in order to solve this, apparently all you need to do is plug it into the equation:
$$\mathrm{pH} = -\log[\ce{H3O+}]$$
where the concentration of $\ce{H3O+}$ is $\pu{5M}$. But why can we assume this? Don’t we need to create an ice table from $\ce{HCl + H2O-> H3O+ + Cl-}$ and use the $K_\text{a}$ value of $\ce{HCl}$ to solve this?

It's because HCl is so strong in aqueous solution. In certain other solvents, you may need to actually calculate it. Also, note that in especially high concentrations you may need to consider the activity rather than just plugging in concentration

@Shadock if the dissociation constant were really only 1, at 5M, the HCl would only be about 36% dissociated, but because the dissociation constant is more like 1000000, HCl is almost 100% dissociated in 5M HCl

@DavePhD yeah you right  i will erase my comment. But I'm not sure like you said that in HCl 5M, HCl is almost 100% dissociated. Because we are speaking about dissolution of HCl in a mixture of HCl and water which can't be approximate at just water here. My knowledge in dissolution is not really important but I have some doubts. If you know some good lectures please tell me. Friendly :)

@Shadock right, it can't just be approximated as water, so you need to use activity instead of concentration. See the references in my answer and also http://pubs.acs.org/doi/abs/10.1021/ac60302a030 for more information

@DavePhD Yes we use activities but I'm not sure we can speak about dissociation as easely as you said. And than you for the document. :)

As explained in pH Paradoxes: Demonstrating That It Is Not True
That pH ≡ -log[H+] Journal of Chemical Education Vol. 83 pages 752-757.
pH = -log (hydrogen ion activity).
and "pH of 7.6 M HCl is about -1.85 (not -0.88)"
So how far off is -log (5) = -0.69 from the real answer?
From this table of activity coefficients 5m HCl has an activity coefficient of 2.38, so activity is 11.9, which yields pH = -1.1.
Unfortunately the table is in terms of molality rather than molarity.  5M HCl is about 5.5m, and activity coefficient is dramaticly increasing with concentration (see table I of the pH paradox reference) so the true pH deviates somewhat further from -0.69, but pH = -1.1 is a reasonable estimate.

Method to calculate the pH of a solution with a strong monoacid
$$
\begin{array}{|c|c|c|c|c|}\hline
&\text{Before}&&\text{After}&\\\hline
&\ce{AH}&\ce{H2O}&\ce{A-}&\ce{H3O+}\\\hline
\text{Initial}&n0&\text{excess}&0&0\\\hline
\text{Equilbrium}&\epsilon&\text{excess}&n0&n0\\\hline
\end{array}
$$
Then we have $\mathrm{pH}=-\log\left[\ce{H3O+}\right]$ Now we have to verify if our result is true.
We consider that $\ce{H3O+}$ in the solution has exactly the same concentration as that of the acid. However they can react with $\ce{OH-}$ because of the water autoprotolysis! $K_\text{a}=10^{14}$ total reaction.
$$\ce{H3O^+ + HO^- -> 2H2O}$$
So the concentration of $\ce{H3O+}$ is approximatively constant if $\left[\ce{H3O+}\right]\gg\left[\ce{HO-}\right]$
For example if $\frac{1}{10}\left[\ce{H3O+}\right]>\left[\ce{HO-}\right]$ then we can find the limit for which our reasonning is false :
$$
\left[\ce{HO-}\right]\times \left[\ce{H3O+}\right] < \frac{1}{10} \times \left[\ce{H3O+}\right] \times \left[\ce{H3O+}\right] \iff K_\text{e}<\frac{1}{10}\left[\ce{H3O+}\right]^2
$$
Then we have $\mathrm{pH}<6.5$
So if with a strong monoacid you find $\mathrm{pH}=6.8$ with this formula the result is not correct and you have to make an other approximation or reasonning!
Here we have $\mathrm{pH}=-0.69<6.5$ so you find a reasonable answer. And in water $\mathrm{pH}$ is between $0$ and $14$ so your solution is at $\mathrm{pH} \approx 0$

There is no non-metrological reason for pH to be restricted to $[0,14]$ in water. Perhaps adding a point about activity vs concentration would also be a good idea.

@LinearChristmas I can't reach the server of doi.org from where I am; when I'll have some more time I'll try to add a point on that it should be great yes.

It seems the link is also missing a slash /, which is why it won't open (doi should be fine). That's my mistake.

There exists seven strong acids that completely disassociate with hydrogen ( $\ce{HCl}$, $\ce{HBr}$, $\ce{H2SO4}$, $\ce{HClO3}$, $\ce{HClO4}$, $\ce{HI}$, and $\ce{HNO3}$), so the only calculation you need to do is the negative log of the concentration (in molarity).
Therefore
$\mathrm{pH} = -\log5.0=-0.69$. Yes, $\mathrm{pH} = −0.69$ even though the $\mathrm{pH}$ scale "normally" ends at $\ce{pH} = 0$.
(in case of $\ce{HCl}~\mathrm{pH}$ will be negative for any solution more then $\pu{2M}$)

Oh and here is proof : http://en.wikipedia.org/wiki/Hydrochloric_acid#Physical_properties

"disassociate with hydrogen" - I think you meant Bronsted acids that completely dissociate? Also, there's uncountable number of strong acids.

7 Acids that I know of, completely disscoiates according to my textbook

That your textbook didn't mention others doesn't mean they don't exist: http://chemistry.stackexchange.com/questions/6450/why-is-fluoroantimonic-acid-so-strong/6454?s=15|0.1391#6454, http://chemistry.stackexchange.com/questions/17124/which-is-the-strongest-acid?s=1|2.5878, http://chemistry.stackexchange.com/questions/20283/whats-the-strongest-known-organic-acid?s=2|1.1192

Just to give a quick and logical answer, if you look at the following acids:

$\ce{HCl}$
$\ce{HBr}$
$\ce{HI}$
$\ce{H2SO4}$ - first dissociation only
$\ce{HClO4}$
$\ce{HClO3}$
$\ce{HNO3}$

These acids above are Strong Acids, meaning that they completely dissociate, so for that purpose there is no need of a certain K value, since none of this is an equilibrium reaction.

Just a forward reaction. $\ce{->}$ not $\ce{<->}$

NOTE: This is usually for general purposes. Not to be super accurate.

This really isn't true in the concentration range of the question.  For example quantitatively for nitric acid the degree of dissociation is 4M: 90%, 6M: 75%, 8M: 55%.  See Ionization of Strong Electrolytes. III. Proton Magnetic Resonance in Nitric, Perchloric, and Hydrochloric Acids J. Chem. Phys. 22, 2067 (1954) http://scitation.aip.org/content/aip/journal/jcp/22/12/10.1063/1.1739994


text:
I need to find the Ka of a weak acid in titration with a strong base
I have been working on this lab forever and have not been able to get a $K_\text{a}$ anywhere close to the accepted value. I’m sure I’m missing something really simple.
We titrated $\ce{CH3COOH}$ with $\ce{NaOH}$.
I used $10.00\ \mathrm{mL}$ of $\ce{CH3COOH}$. Volume of $\ce{NaOH}$ at the equivalence point was $10.14\ \mathrm{mL}$, and the concentration of $\ce{NaOH}$ was $0.1530\ \mathrm{mol \cdot L^{-1}}$.
First I found the concentration of $\ce{CH_3COOH}$ by using $c_\text{a} \cdot V_\text{a} = c_\text{b} \cdot V_\text{b}$, so $c_\text{a} \cdot 10.00\ \mathrm{mL} = 0.1530\ \mathrm{mol \cdot L^{-1}} \cdot 10.14\ \mathrm{mL}$
$c_\text{a} = 0.1551\ \mathrm{mol \cdot L^{-1}}$
Then I used the equation: $\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$
and the pH (at the midpoint of $5.07\ \mathrm{mL}\ \ce{NaOH}$) which was $4.87$
$$K_\text{a}= \frac{[\ce{H3O+}] [\ce{CH3COO-}]}{[\ce{CH3COOH}]
[\ce{H3O+}]}=[\ce{CH3COO-}]= 10^{-4.87}= 1.36 \times 10^5
$$
$$
K_\text{a}= \frac{(1.36 \times 10^5)^2}{ 0.1551} = 1.19 \times 10^{-9}
$$
Any help on what I am doing wrong would be greatly appreciated!

I’m not sure if I’m missing your question, but I thought that perhaps you could use the relation $\mathrm{pH}=\mathrm{p}K_\text{a}$ at the half equivalence point? The answer would be still incorrect but closer?
Otherwise, you could probably use stoichiometry to solve for $K_\text{a}$ at the beginning or the end of the titration by constructing an ICE table or otherwise. By considering the deprotonation at the start, so the concentration of protons would be equivalent to concentration of conjugate base. Then, instead of inserting the pH at the half equivalence point, insert the pH before the titration.
If you need help doing the algebra, I could try...
But I think what you did wrong was inserting the concentration at the half equivalence point, rather than before the titration.

Thank you for the help, I believe using the relation of pH=pKa at the half equivalence point was the way to go. I was just trying to make it more difficult than it needed to be.


text:
Mixing borax and shampoo
I use borax to clean everything. the other day I thought it would be a good idea to mix a little borax with my shampoo while showering.  After lathering it on my scalp about 30 seconds it began to give off a really bad chemical smell, almost like a perm solution. What was the borax reacting with the product in my hair or the shampoo?

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

Every type of shampoo have slightly something different


text:
Using enthalpies of formation and combustion to find formation enthalpy of Acetylene (Hess' Law)
I understand this is a simple question (I’m in high school), but for some reason the numbers just are not working for me.
Given:
$$\Delta H^\circ_\text{c} = -1301\ \mathrm{kJ}\ \text{for}\ \ce{C2H2(g)}$$
$$\Delta H^\circ_\text{f} = -394\ \mathrm{kJ}\ \text{for}\ \ce{CO2(g)}$$
$$\Delta H^\circ_\text{f} = -242\ \mathrm{kJ}\ \text{for}\ \ce{H2O(g)}$$
Calculate $\Delta H_\text{f}$ for $\ce{C2H2(g)}$.
My process:
$$(1) : \ce{C2H2(g) + 5/2 O2(g) -> 2CO2(g) + H2O(g) }\quad \Delta H^\circ_\text{r} = -1301\ \mathrm{kJ}$$
$$(2) : \ce{C(s) + O2(g) -> CO2(g)} \quad \Delta H^\circ_\text{r} = -394\ \mathrm{kJ}$$
$$(3) : \ce{H2(g) + 1/2O2(g) -> H2O(g)} \quad \Delta H^\circ_\text{r} = -242\ \mathrm{kJ}$$
$$\text{Target:}\ \ce{2C(s) + H2(g) -> C2H2(g)}$$
After some experimentation I arrive at:
$$-1\times(1) + 2\times(2) + (3) = \text{Target}$$
As shown through:
$$-1\times(1){:}\ \ce{2CO2(g) + H2O(g) -> C2H2(g) + 5/2 O2(g)} \quad \Delta H^\circ_\text{r} = 1301\ \mathrm{kJ}$$
$$2\times(2){:}\ \ce{2C(s) + 2O2(g) -> 2CO2(g)} \quad \Delta H^\circ_\text{r} = -788\ \mathrm{kJ}$$
$$(3){:}\ \ce{H2(g) + 1/2O2(g) -> H2O(g)} \quad \Delta H^\circ_\text{r} = -242\ \mathrm{kJ}$$
$$\text{Sum:}\ \ce{2C(s) + H2(g) -> C2H2(g)} \quad \Delta H^\circ_\text{f} = 271\ \mathrm{kJ}$$
Now as far as I know, all the givens are accurate empirical values. But the empirical value for $\Delta H^\circ_\text{f}$ for $\ce{C2H2(g)}$ is $227\ \mathrm{kJ}$.
What’s even weirder is that this disparity $(271-227=44)$ is equal to the difference between standard enthalpies of formation for water in gas and liquid states $(-242 - -286 = 44)$.
Have I made an error somewhere, or is water formed in the liquid state during the combustion of acetylene?

The standard enthalpy of reaction involves the species in their standard states. So what is the standard state of water?

Oh. The reason water is formed as a gas is a result of the enthalpy change, but if we're looking just at the product in standard conditions it's just a liquid? And if we were to wait long enough... I guess the water would return to liquid state as the heat returned to 25°C?

Welcome to chemistry.SE BTW!

Could I get confirmation on my above conclusion? I think it makes sense, but I want to be sure.
I guess my teacher has been using $\ce{H2O(g)}$ for simplicity's sake.

Yes that is correct

Like many past chemistry.se questions, the answer to this one hinges on the presence of the $^\circ$.
$\Delta H^\circ_\text{c}$ is the $\Delta H$ value for a "standard" reaction where acetylene at 1 atm of pressure is combusted isothermally at 298 K with 1 atm of oxygen gas, forming 1 atm of $\ce{CO2}$ and liquid water (which is the most thermodynamically stable form of water at 1 atm and 298 K).
I think it's safe to say that no combusion of acetylene, ever in the history of the world, has occured under those conditions.  But those "standard" conditions are the only conditions where the  $\Delta H^{\circ}_f$ is guaranteed to be -227 kJ/mol.
By the way, in practical and engineering literatures, $\Delta H_c^{\circ}$ of a fuel is commonly called the higher heating value of the fuel.  It corresponds to $\Delta H_{rxn}^{\circ}$ of a reaction like
$\ce{C2H2(g) + 5/2 O2(g) -> 2CO2(g) + H2O(l)}$
The lower heating value, conversely, corresponds to $\Delta H_{rxn}^{\circ}$ of reactions like
$\ce{C2H2(g) + 5/2 O2(g) -> 2CO2(g) + H2O(g)}$


text:
Why should 2,6-Dimethyl-4-nitrophenol be more acidic than 3,5-Dimethyl-4-nitrophenol
I was trying to compare acidity of some organic compounds but can't understand some:
Why should 2,6-dimethyl-4-nitrophenol be more acidic than 3,5-dimethyl-4-nitrophenol?
I have been trying to figure it out using various factors like hyperconjugation and induction effect of Me group but its not working. Suppose i take H.C effect 2,6-dimethyl-4-nitrophenol be "less" acidic than 3,5-dimethyl-4-nitrophenol since it destabilizes the negative charge on conjugate base. But this logic is not working.

In the 3,5-dimethyl isomer, steric effects come into play and influence the acidity.
Resonance structures like B in the figure below, that stabilize the phenoxide anion explain why 4-nitrophenol ($\ce{R=H}$) is more acidic than phenol.  Resonance structure B requires that the nitro group be planar with the aromatic ring.  If methyl groups are placed adjacent to the nitro group ($\ce{R=CH3}$), the steric interactions between the two methyl groups and the nitro group becomes large and forces the nitro group to rotate out of the aromatic plane,  Hence resonance structure B can no longer contribute and the compound is less acidic.

So that's basically ortho effect? Thanks i got it now :-)

Yes exactly, in this case it's a double ortho effect.

In3,5 dimethyl 4 nitro phenol , the nitro group is between two ortho methyl groups . Therefore due to this steric inhibition of resonance occurs and this nitro group is pushed out of plane by two methyl groups. Such that nitro group does not take part in the resonance. And there have no extra stable structure. That is why it is less acidic than2,6 dimethyl 4 nitro phenol.

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text:
After mixing, which species has the dominant affect on the pH of the solution that remains?
A solution is created by mixing $250\,\mathrm{mL}$ of $1\,\mathrm{M}$ $\ce{HCl}$, $250\,\mathrm{mL}$ of $1\,\mathrm{M}$ $\ce{CH3COOH}$ and $500\,\mathrm{mL}$ of $1.5\,\mathrm{M}$ $\ce{CsOH}$.
The $K_\mathrm{a}(\ce{CH3COOH}) = 1.8 \times 10^{-5}$
What is the $\mathrm{pH}$ of this solution?
I am stuck between $\ce{HCl}$ and $\ce{CsOH}$ since they are both a strong acid and base (respectively) meaning that they should both have a substantial affect on the $\mathrm{pH}$.
Also when determining what the $\mathrm{pH}$ is, which species would we use in our ICE table?

CsOH is a strong base.
0.5 L X 1.5 M = 0.75 moles of hydroxide.
Similarly, there are 0.5 moles of monoprotic acid.
After mixing there will be 0.25 moles of hydroxide in 1 liter of solution.
In other words [OH-] = 0.25M
pH is approximately 14 + log (0.25) = 13.4

You need to provide the concentration of acetic acid. However, if I were to guess, I would say that the hydrochloric acid completely reacts with the cesium hydroxide, and the remaining cesium hydroxide reacts with the acetic acid. Finding the quantity of acetic acid, which probably was more than the concentration of cesium hydroxide, you could use hendersen hassebach.


text:
How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?
For example, which is the right way to name this compound:

Is it 2-bromo-4-methylenehexane or 4-bromo-2-ethylpent-1-ene?
Does the larger number of carbon in a chain take priority, as in the former, or does the carbon-carbon double bond take priority, as in the latter?

The IUPAC recommendations on this matter have been changed. In the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed:

(…), the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second criterion.

For example, a preferred IUPAC name is 3-methylidenehexane (not 2-ethylpent-1-ene) since the longer chain now supersedes a shorter unsaturated chain. (This example is given in the Blue Book.)

Besides, the prefix ‘methylene’ is no longer recommended in IUPAC nomenclature to designate the $\ce{=CH2}$ group. The preferred prefix is ‘methylidene’.
Therefore, the correct name for your example is 2-bromo-4-methylidenehexane.

We are still taught in the eleventh and twelfth grade to give priority to the double bond, smh! Makes it all confusing.

From wikipedia:

Identification of the parent hydrocarbon chain. This chain must obey the following rules, in order of precedence:

It should have the maximum number of substituents of the suffix functional group. By suffix, it is meant that the parent functional group should have a suffix, unlike halogen substituents. If more than one functional group is present, the one with highest precedence should be used.

So the alkene takes priority over the alkane and 4-bromo-2-ethylpent-1-ene is the correct name.

Gotta hate it when the gatekeepers change the rules right out from underneath one.


text:
Why is arsenic more chemically similar to phosphorus than is nitrogen?
Why is arsenic more chemically similar to phosphorus than is nitrogen (to phosphorus)?
I thought that it may be because both phosphorus and arsenic have d orbitals (albeit one has one that is filled and the other has an unfilled one).
Another possibility I thought was that because nitrogen held the electrons very closely, and thus the difference between the second and third period may be more pronounced than the difference between the third and fourth periods.
I'd really appreciate it if anyone could help clarify/ explain. Thanks!

Every element which is in the first place of the group, is pretty much different from the other elements of the same group. The reason is because the first element has ALWAYS the smallest radius, which leads to some changings also in group properties. Due to the small radius they are more stable than the others and they require more energy for any specific reaction. With some research you can see that also other  first elements of other groups such as Li,O,C etc.  differ from the other elements of that specific group.

This answer has the right idea. This simple concept has many interesting and far-reaching implications across the periodic table. A deeper explanation of this effect and many of its consequences can be found in the article "The Role of Radial Nodes of Atomic Orbitals for Chemical Bonding and the Periodic Table". I've made a brief overview of it here.

@NicolauSakerNeto Indeed, the $4d$ transition metals are rather similar to the $5d$, but considerably different from the $3d$; and there are stark differences between the lanthanides ($4f$) and actinides ($5f$). Thank you for the reference, I will be reading it promptly!


text:
Why might the O—H stretch in the IR of MY triphenylmethanol product be less intense than expected?
Why might the O—H stretch in the IR of MY triphenylmethanol product be less intense than expected when compared to the expected IR for triphenylmethanol?
I'm interpreting the IR I took of my product, triphenylmethanol, to the expected IR I obtained from SDBS.
My IR:
SDBS:
We performed a Grignard Reaction in lab using bromobenzene and methyl benzoate in order to synthesize triphenylmethanol.
ANY clues why my triphenylmethanol (O—H stretch) is not as intense? I really don't know.

is your ir instrument properly calibrated? If not that might explain your results

A possible explanation would be the concentration of sample you used. Of course, if you are confident that you had utilized sufficient concentration then there may be another reason for the decreased intensity. However, some of the other peaks are also somewhat decreased in intensity so it is a possibility.
This point is noted here (http://chemwiki.ucdavis.edu/Physical_Chemistry/Spectroscopy/Vibrational_Spectroscopy/Infrared_Spectroscopy/Infrared%3A_Interpretation#The_Origin_of_Peak_Intensities)


text:
Chemical compatibility of polycarbonate and carbonate esters
I am designing a battery box to house lithium ion batteries in an electric vehicle. These batteries use an electrolyte which probably contains carbonate esters (eg. ethylene carbonate, dimethyl carbonate, diethyl carbonate).
If I use polycarbonate as the battery box material, should I be worried about an interaction with the electrolyte in the event of a battery leak?

Yes, you need to verify that the housing of the battery is made of a material that is unaffected by its contents. Polycarbonate is a good material for rigidity and transparency, but may be dissolved or weakened over time by organic solvents (carbonate esters or otherwise) and also by pH extremes and high temperatures that could come into play when the battery is charged and discharged. Thermosetting resins might be a better bet for this type of application.

thanks for the info on the potential reaction with organic solvents. not sure if I made this clear, but the polycarbonate would be forming the battery box that encloses all of the individual cells. each cell has its own casing, which completely encloses the electrolyte and other internal components of the battery.
still, if the cell casings were to rupture, the electrolyte would contact the polycarbonate directly. so I suppose it makes little difference.


text:
What exactly does the term "gram atoms" mean?
The Wikipedia article on the mole states:

 The term gram-atom (abbreviated gat.) has been used for a related but distinct concept, namely a quantity of a substance that contains Avogadro's number of atoms, whether isolated or combined in molecules. Thus, for example, 1 mole of $\ce{MgB2}$ is 1 gram-molecule of $\ce{MgB2}$ but 3 gram-atoms of $\ce{MgB2}$.

Can anybody explain the difference between moles and gram atoms, and how is it that 1 mole of $\ce{MgB2}$ is 1 gram-molecule of $\ce{MgB2}$ but 3 gram-atoms of $\ce{MgB2}$?

The molecule has 3 atoms, so ......

$\ce{MgB2}$ is 1 gram-molecule of $\ce{MgB2}$ but 3 gram-atoms of $\ce{MgB2}$.
Why is this so? Let us consider this problem in the familiar mole:
If we were asked how many moles of the substance (molecule) were in this, we would state 1 mole, if there were $\pu{45.3 g}$ of substance. However, if we were asked how many moles of atoms were in the substance, we would accordingly state 3. Thus, in one molecule, there are 3 atoms, so if we simply consider the number of atoms, there would be the necessity of counting the number. of atoms in a molecule, as a molecule is comprised of more than a single atom
* This is purely hypothetical. One would probably not consider  $\ce{MgB2}$, as J. LS pointed out, it decomposes . However, the concept can be applied to other compounds

There are no molecules of magnesium diboride as it decomposes before you can reach its melting point, which makes this a rather odd example to use here.

@J.LS Thanks for pointing that out! I had simply used the example the asker had given, but I guess it would be nice to be aware of that point.

OP has also asked for difference between moles and gram atoms ,but nobody explained this point?


text:
Why is lithium the most reducing alkali metal, and not caesium?
Caesium has a larger size, and the effective nuclear charge that the valence electron experiences will be far less compared to that of lithium's, right? But lithium is still considered the strongest reducing agent among all the alkali metals, and this is evidenced by its large and negative reduction potential. Why is this so?

I think it has to do with solvation. The more solvated the resulting ions, the more favourable electron loss is. Even though its ionization energy is higher, lithium creates a very small cation which is strongly solvated in water, which drives the oxidation of metallic lithium thermodynamically towards the products.

The trend in the reducing power of the alkali metals is not a simple linear trend, so it is a little disingenuous if I were to solely talk about $\ce{Li}$ and $\ce{Cs}$, implying that data for the metals in the middle can be interpolated.
$$\begin{array}{cc}
\hline
\ce{M} & E^\circ(\ce{M+}/\ce{M}) \\
\hline
\ce{Li} & -3.045 \\
\ce{Na} & -2.714 \\
\ce{K} & -2.925 \\
\ce{Rb} & -2.925 \\
\ce{Cs} & -2.923 \\
\hline
\end{array}$$
Source: Chemistry of the Elements 2nd ed., Greenwood & Earnshaw, p 75
However, a full description of the middle three metals is beyond the scope of this question. I just thought it was worth pointing out that the trend is not really straightforward.

The $\ce{M+}/\ce{M}$ standard reduction potential is related to $\Delta_\mathrm{r}G^\circ$ for the reaction
$$\ce{M(s) -> M+(aq) + e-}$$
by the equation
$$E^\circ = \frac{\Delta_\mathrm{r}G^\circ + K}{F}$$
where $K$ is the absolute standard Gibbs free energy for the reaction
$$\ce{H+ + e- -> 1/2 H2}$$
and is a constant (which means we do not need to care about its actual value). Assuming that $\Delta_\mathrm{r} S^\circ$ is approximately independent of the identity of the metal $\ce{M}$, then the variations in $\Delta_\mathrm{r}H^\circ$ will determine the variations in $\Delta_\mathrm{r}G^\circ$ and hence $E^\circ$. We can construct an energy cycle to assess how $\Delta_\mathrm{r}H^\circ$ will vary with the identity of $\ce{M}$. The standard state symbol will be dropped from now on.
$$\require{AMScd}
\begin{CD}
\ce{M (s)} @>{\large \Delta_\mathrm{r}H}>> \ce{M+(aq) + e-} \\
@V{\large\Delta_\mathrm{atom}H(\ce{M})}VV @AA{\large\Delta_\mathrm{hyd}H(\ce{M+})}A \\
\ce{M (g)} @>>{\large IE_1(\ce{M})}> \ce{M+ (g) + e-}
\end{CD}$$
We can see, as described in Prajjawal's answer, that there are three factors that contribute to $\Delta_\mathrm{r}H$:
$$\Delta_\mathrm{r}H = \Delta_\mathrm{atom}H + IE_1 + \Delta_\mathrm{hyd}H$$
(the atomisation enthalpy being the same as the sublimation enthalpy). You are right in saying that there is a decrease in $IE_1$ going from $\ce{Li}$ to $\ce{Cs}$. If taken alone, this would mean that $E(\ce{M+}/\ce{M})$ would decrease going from $\ce{Li}$ to $\ce{Cs}$, which would mean that $\ce{Cs}$ is a better reducing agent than $\ce{Li}$.
However, looking at the very first table, this is clearly not true. So, some numbers will be needed. All values are in $\mathrm{kJ~mol^{-1}}$.
$$\begin{array}{ccccc}
\hline
\ce{M} & \Delta_\mathrm{atom}H & IE_1 & \Delta_\mathrm{hyd}H & \text{Sum} \\
\hline
\ce{Li} & 161 & 520 & \mathbf{-520} & 161 \\
\ce{Cs} & 79 & 376 & \mathbf{-264} & 211 \\
\hline
\end{array}$$
Source: Inorganic Chemistry 6th ed., Shriver et al., p 160
This is, in fact, an extremely crude analysis. However, it hopefully does serve to show in a more quantitative way why $E(\ce{Cs+}/\ce{Cs}) > E(\ce{Li+}/\ce{Li})$: it's because of the extremely exothermic hydration enthalpy of the small $\ce{Li+}$ ion.
Just as a comparison, the ionic radii of $\ce{Li+}$ and $\ce{Cs+}$ ($\mathrm{CN} = 6$) are $76$ and $167~\mathrm{pm}$ respectively (Greenwood & Earnshaw, p 75).

I did not understand the formulas you provided. I hope to be able one day. My Russian textbook says: "lithium ions are much smaller than Na and K ions. That's why they will have stronger electrical fields around them. Hence, more active hydration. This eases the transition of electrons from Li into the solution and explains Li's position in the electrochemical series".

That's pretty much the same as what I wrote, except it's more descriptive and less quantitative. OP's argument is that it is difficult to remove the electron off lithium, therefore lithium should be hardest to oxidise. However, in reality, lithium is coerced to give up its electron by the surrounding water molecules, which are very happy to form ion-dipole interactions with the small ion $\ce{Li+}$.

To decide which is the best reducing agent we only not consider that who has less ionisation energy yet it follows 3 steps:

Metal (solid) to Metal (gaseous state) sublimation energy
Metal from gaseous state to M+ ionisation energy
M+ to M+ (aqueous state) hydration energy

Lithium having more charge density has more sublimation energy and ionisation energy than caesium but hydration  energy is released in  such a big amount that it compensates the S.E and I.E. and caesium's hydration is less than lithium. That's why lithium is good reducing agent.

Well there might be more reasons than these two:

Lithium has a higher reduction potential.
If you also look at the electronegativities of just Lithium and Cesium then you would notice that the shielding effect is more prevalent in Cesium, thereby reducing the electronegativity and affecting the reduction potential. So Lithium however, just compared to Cesium, has a higher electronegativity.

I think these are the two main reasons, please correct me If I am wrong.

(-1) Saying that "lithium is a stronger reducing agent because it has a higher reduction potential" is just like saying "fluorine is reactive because it reacts with a lot of species readily". It's a tautology. Secondly, the greater shielding effect in caesium works to make caesium a stronger reducing agent because it reduces the first ionisation energy. It's exactly the same thing that OP states in the question.


text:
Understanding gas stoichiometry for the reaction of xenon and fluorine

A $\pu{20.0 L}$ nickel container was charged with $\pu{0.500 atm}$ of xenon gas and $\pu{1.50 atm}$ of fluorine gas at $\pu{400 ^{\circ}C}$. The xenon and fluorine react to form xenon tetrafluoride. What mass of xenon tetrafluoride can be produced assuming $100\%$ yield?

Attempt at solution: From the description, the reaction can be written as
$$\ce{Xe + 2F2 -> XeF4}.$$
I first calculated the amount of substance of $\ce{Xe}$ using the ideal gas law:
$$n(\ce{Xe}) = \frac{pV}{RT} = \frac{\pu{0.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.18 mol}$$
Doing the same for fluorine gives:
$$n(\ce{F2}) = \frac{\pu{1.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.54 mol}$$
Then, for every mole of $\ce{Xe}$ we need two moles of $\ce{F2}$. Since we have more than $2 \cdot \pu{0.18 mol} = \pu{0.36 mol}$ of fluorine, xenon is the limiting reactant. The molar mass of $\ce{XeF4}$ is $207.1\,\mathrm{g/mol}$.
So now I would just do:
\begin{align}
n(\ce{XeF4}) M(\ce{XeF4}) &= m(\ce{XeF4})\\
\pu{0.18 mol}  \times \pu{207.1 g mol-1}  &= \pu{37.278 g}
\end{align}
However, the answer at the back of my textbook says it should be $\pu{37.5 g}$. So did I make a mistake somewhere or is this just a roundoff-error?

Yes, this is a rounding error.
Where you calculate:
$$n = \frac{pV}{RT} = \frac{\pu{0.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.18 mol}$$
The actual unrounded answer is $\pu{0.181072885 mol}$ - a little bit higher than the rounded value.
So, in your final equation:
$$\pu{0.181072885 mol} \times \pu{207.1 g//mol} = \pu{37.5 g}$$
I find using the full value of an intermediate step usually eliminates these types of errors in the final calculation.

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


text:
Alkane, alkene, alkyne boiling point comparison
Which of the following has higher boiling points? Alkanes, alkenes, or alkynes? And why?

Also, you do realize that these are a whole family of aliphatic compounds, right? You might be wanting to compare their members relatively, but that isn't clear from here.

I've been looking for answers through internet, what I've learnt, alkanes should have a higher boiling point due to higher Molecular mass. But, what I got Alkyne has higher boiling points.

Remember a Golden Rule. Boiling point is directly proportional to it's molecular weight. The greater the Mol. Wt., the more will be the BP.

@siddharth yadav by directly proportional do you mean it as a general guideline or are you suggesting that there is a one size fits all constant that relates boiling point and molecular weight  as $y=mx$  for every family of compounds?

Disclaimer: All of this "jazz" will be about reaching a mere rule-of-thumb. You can't just compare whole families of organic compounds with each other. There are more factors to consider than below, mostly based on isomerism notions. However, as most of the A grade exams emphasize on the lighter aliphatic compounds, we can understand each other here. :)

It's all about polarizability.

Polarizability is the ability for a molecule to be polarized.

When determining (aka comparing) the boiling points of different molecular substances, intermolecular forces known as London Dispersion Forces are at work here. Which means, these are the forces that are overcome when the boiling occurs. (See here for example)

Source
London forces get stronger with an increase in volume, and that's because the polarizability of the molecule increases. (See the answer to this recent question)
Alkanes vs. Alkenes
In their simplest form (where no substitution etc. has occurred) alkanes tend to have very close boiling points to alkenes.

The boiling point of each alkene is very similar to that of the alkane with the same number of carbon atoms. Ethene, propene and the various butenes are gases at room temperature. All the rest that you are likely to come across are liquids.
Boiling points of alkenes depends on  more molecular mass (chain length). The more intermolecular mass is added, the higher the boiling point. Intermolecular forces of alkenes gets stronger with increase in the size of the molecules.
\begin{array}{|c|c|}\hline
\text{Compound} & \text{Boiling point / }^\circ\mathrm{C} \\ \hline
\text{Ethene} & -104 \\ \hline
\text{Propene} & -47 \\ \hline
\textit{trans}\text{-2-Butene} & 0.9 \\ \hline
\textit{cis}\text{-2-Butene} & 3.7 \\ \hline
\textit{trans}\text{-1,2-dichlorobutene} & 155 \\ \hline
\textit{cis}\text{-1,2-dichlorobutene} & 152 \\ \hline
\text{1-Pentene} & 30 \\ \hline
\textit{trans}\text{-2-Pentene} & 36 \\ \hline
\textit{cis}\text{-2-Pentene} & 37 \\ \hline
\text{1-Heptene} & 115 \\ \hline
\text{3-Octene} & 122 \\ \hline
\text{3-Nonene} & 147 \\ \hline
\text{5-Decene} & 170 \\ \hline
\end{array}
  In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane. The only attractions involved are Van der Waals dispersion forces, and these depend on the shape of the molecule and the number of electrons it contains. Each alkene has 2 fewer electrons than the alkane with the same number of carbons.

Alkanes vs. Alkynes
As explained, since there is a bigger volume to an alkane than its corresponding alkyne (i.e. with the same number of carbons) the alkane should have a higher boiling point. However, there's something else in play here:
Alkynes, have a TRIPLE bond!
I currently can think of two things that happen as a result of this:

London Dispersion Forces are in relation with distance. Usually, this relation is $r^{-6}$. (See here) The triple bond allows two alkynes to get closer. The closer they are, the more the electron densities are polarised, and thus the stronger the forces are.
Electrons in $\pi$ bonds are more polarizable$^{10}$.

These two factors overcome the slight difference of volume here. As a result, you have higher boiling points for alkynes than alkanes, generally.
\begin{array}{|c|c|}\hline
\text{Compound} & \text{Boiling point / }^\circ\mathrm{C} \\ \hline
\text{Ethyne} & -84^{[1]} \\ \hline
\text{Propyne} & -23.2^{[2]} \\ \hline
\text{2-Butyne} & 27^{[3]} \\ \hline
\text{1,4-Dichloro-2-butyne} & 165.5^{[4]} \\ \hline
\text{1-Pentyne} & 40.2^{[5]} \\ \hline
\text{2-Heptyne} & 112\text{–}113^{[6]} \\ \hline
\text{3-Octyne} & 133^{[7]} \\ \hline
\text{3-Nonyne} & 157.1^{[8]} \\ \hline
\text{5-Decyne} & 177\text{–}178^{[9]} \\ \hline
\end{array}
1: http://en.wikipedia.org/wiki/Acetylene
2: http://en.wikipedia.org/wiki/Propyne
3: http://en.wikipedia.org/wiki/2-Butyne
4: http://www.lookchem.com/1-4-Dichloro-2-butyne/
5: http://en.wikipedia.org/wiki/1-Pentyne
6: http://www.chemsynthesis.com/base/chemical-structure-17405.html
7: http://www.chemspider.com/Chemical-Structure.76541.html
8: http://www.thegoodscentscompany.com/data/rw1120961.html
9: http://www.chemsynthesis.com/base/chemical-structure-3310.html
10: https://chemistry.stackexchange.com/a/27531/5026

Conclusion: We can't fully determine the boiling points of the whole class of alkanes, alkenes and alkynes. However, for the lighter hydrocarbons, comparing the boiling points, you get:
$$\text{Alkynes > Alkanes > Alkenes}$$


text:
Is there a way to turn alcohol into a solid gel?
I'm wondering if there is a way to turn alcohol or another bacteria killing substance into a solid gel form, sort of like the gel you can use to clean a keyboard.

Source: Aliexpress.com Computer Cleaning Compound

I'm wondering if you could put either alcohol or other sanitizer into this sort of gel, without the sanitizer having a "best before" date.
I'm a complete beginner when it comes to chemistry so forgive me if there's an obvious reason why this wouldn't work. If it does work, or partially, could you give me a run down on what I would need or need to test to achieve my goal?

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

I'm not sure how common it is elsewhere, but there is such a thing as alcohol gel which is used as a hand sanitizer. It's composed chiefly of ethanol and a gelling agent, usually a polymer such as polyacrylate. Presumably you could get a more viscous gel simply by stirring in more gelling agent with the ethanol.

Gelled alcohol is sold under Sterno and other brands as a fuel. You can make your own with alcohol and calcium acetate; see http://io9.com/5929133/how-to-make-the-worlds-most-flammable-jelly.

If your alcohol is ethanol, calcium acetate works very well.  I use calcium acetate to make gel in petri dishes, and sprinkle metal salts on it for colored flame demonstrations.  Calcium acetate is also used in some canned alcohols used for warming/cooking foods.
3 grams of calcium acetate dissolved in 10 ml of water will make a saturated solution (some may not dissolve, which is OK).  This is mixed with 75 ml of 95% ethanol, which will then form a colloidal gel identical to that used in canned alcohol warmers.  I do not know if this gel is suitable for your purposes, but it works well for a heat source.  If used as a heat source, a pinch of salt should be added to the aqueous solution before mixing with alcohol to make the flame more visible.

Try mixing styrofoam with gasoline and alcohol. The gasoline will kind of dissolve the styrofoam and form a paste with imbued alcohol.

Isn't that basically napalm?... won't be a firm gel though.

Use Hydroxypropylcellulose (HPC) as gelling agent.

This answer could use a bit more detail.


text:
Is p-bromophenol more acidic than p-chlorophenol?
p-chlorophenol is more acidic than p-fluorophenol due to presence of vacant d orbitals. But then is p-bromophenol more acidic than p-chlorophenol since it has vacant f orbitals also? But I guess that’s wrong, isn’t it? Please give your suggestions.

Disclaimer: This is not necessarily the "correct" answer as many factors play roles and it would be difficult to describe the observation so simply.
I'm not sure I can provide an easily "correct" answer, but I can try to explain some of the many factors that influence the acidity.
Most notably, the factors we should consider are inductive effects and delocalization of charge.
The inductive effects, based on electronegativity, would dismiss the experimental observation, so I'll look at delocalization of charge.
The delocalization of charge, which introduces a negative charge by the halogen is greatest by F>Cl>Br>I. This would reduce acidity, as a more negative benzene ring would not desire to gain another negative charge. Thus, this is one factor that supports the experimental observation.
Thus, we may be able to explain the observation, if only partially, by the delocalization of charge.
I adapted the answer to a similar question on http://www.quora.com/Organic-Chemistry/Which-is-more-acidic-ortho-fluorophenol-ortho-chlorophenol-ortho-bromophenol-or-ortho-iodophenol-Why . I would have preferred more sources but I could not find a very good source.

Experimental data provides the correct answer: 4-bromophenol (pKa = 9.17) is a stronger acid than 4-chlorophenol (pKa = 9.41). Scroll down to "dissociation constant". References are provided.

I looked up the pKa values of each chloro- and bromophenol and compiled them in the following little table:

pKa
chlorophenol
bromophenol

ortho-
8.49
8.45

meta-
8.81
9.03

para-
9.14
9.17

(These are just the first values that came up when I googled the name of the compound plus pKa. I did not go back and check the primary sources.)
Keep in mind that both chlorine and bromine are primarily inductively electron withdrawing (-I), and to a lesser extent mesomerically electron donating (+M). Chlorine is more electronegative than bromine, so its -I effect will be more pronounced. On the other hand, chlorine's d orbitals are smaller and overlap more efficiently with the ring, giving a stronger +M effect than bromine. This, in my opinion, nicely explains all the major trends in the table:

First, keeping the halogen constant, acidity decreases in the order ortho > meta > para, for both chloro- and bromophenols. This makes sense as the -I effect dominates and becomes weaker as the halogen atom moves further away from the hydroxyl group.
For the ortho- and para-compounds, the acidities of the chloro- and the bromophenol are practically equal. When going from chlorine to bromine, the decreased -I effect is almost completely canceled out by the similarly decreased +M effect.
However, meta-chlorophenol is significantly stronger than its bromo-analogue. Again what we would expect, as the mesomeric effect does not affect meta substituents, leaving only the inductive effect which is stronger in chlorine.


text:
Calculate the number of moles produced per hour per liter of air?
A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by
$$
\ce{O3(g) + NO(g)->O2(g) + NO2(g)}
$$
the rate law for this reaction is:
$$
\text{rate of reaction} = k\left[\ce{O3}\right]\left[\ce{NO}\right]
$$
Given that $k = 2.98 \times 10^6\ \mathrm{M^{–1}\cdot s^{–1}}$ at a certain temperature, calculate the initial reaction rate when $\left[\ce{O3}\right]$ and $\left[\ce{NO}\right]$ remain essentially constant at the values $\left[\ce{O3}\right]_0 = 5.34 \times 10^{–6}\ \mathrm{M}$ and $\left[\ce{NO}\right]_0 = 7.99 \times 10^{–5}\ \mathrm{M}$, owing to continuous production from separate sources.
So, I have found that by using the rate of reaction equation
$$
\text{the rate of reaction} = $0.00127\ \mathrm{M/s}
$$
Now I’m wondering how to calculate the amount of substance of $\ce{NO2(g)}$ produced per hour per liter of air.
Right now, my idea is that you can take
$$
\left(1.27\times 10^{-3}\ \mathrm{M}\over\mathrm{s}\right)\left(60\ \mathrm{s}\over{1\ \mathrm{min}}\right)\left(60\ \mathrm{min}\over{1\ \mathrm{h}}\right) = \ce{NO2} \,\text{per hour per liter of air}
$$
Is this the correct way to go about it? I was thinking that the seconds and minutes would cancel leaving just $\mathrm{M/h}$ and then if you multiply by 1 liter it would be $\mathrm{M/(h\times L)}$.

That last equation looks like you are on the right track.  What are those units M?  What is the definition of M?

That is the correct way to go about it. In your calculations the rate of the reaction should include L because M is defined as moles/L. (1.27×10^-3 moles/Ls)(3600s/Hour) = 4.572 moles/Lh NO₂ produced.


text:
How to determine whether the enthalpy of solution is positive or negative by calorimetry?
How can we determine if the enthalpy of solution of a compound is positive or negative?
In order to make a clear idea about what I mean, take the following problem:
A 3.5 g sample of $\ce{NH4NO3}$ was added to 80. mL of water in a coffee cup calorimeter of negligible heat capacity. As a result, the temperature of solution decreased from $21.6 \ ^\circ\text{C}$ to $18.1\ ^\circ\text{C}$. Calculate the enthalpy change of the solution ($\Delta H_{\text{sol}}$ ). (specific heat of the solution = $4.18 \frac{\text{J}}{\text{g}\cdot ^\circ\text{C}}$; density of water $1.00 \frac{\text{g}}{\text{mL}}$)
Answer: $+28 \frac{\text{kJ}}{\text{mol}}$
Why it is positive despite there is a decrease in temperature?

Possible duplicate of http://chemistry.stackexchange.com/questions/27159/how-to-determine-the-sign-of-the-enthalpy-of-formation

This question is not actually about enthalpy of formation (not a duplicate). It is about enthalpy of solution and the calorimetry experiment one might do to determine said enthalpy. I have edited the question to make it clearer.

The sample takes in the heat from the water, so it takes in energy to "form" the solution and the water thereby loses it, becoming a lower temperature. You need to realize the reference point is not the water, but rather the sample in this case. The heat transferred (lost) by the water is to the sample, so it is the negative of the heat lost by water, making it positive.


text:
Effect of different electron withdrawing/transition state stabilising groups in nucleophillic aromatic substitution
I have been asked the following question in my problem sheet:

The only differemce between reaction A and reaction B is the electron withdrawing group that stabilises the anionic transition state changes from $CN$ in reaction A to $NO_2$ in reaction B. I can see no glaringly obvious reason why there should be a different rate. The problem is really (at least in my mind): what is the better electron withdrawing group out of $CN$ and $NO_2$. Is this a reasonable interpretation? If so, what reasoning can I bring to the answer to suggest that one is a better EWG than the other?

Your reasoning is correct.  The relative reaction rates would probably be quite similar if we didn't have the methyl groups on both sides of the substituent X.
In order for the substituent to stabilize the transition state through resonance, the substituent must be coplanar with the aromatic ring.  This isn't a problem for the linear $\ce{CN}$ group, but the $\ce{NO2}$ group with the two oxygens protruding out will not be able to remain planar with the aromatic ring.  The two adjacent methyl groups create steric interference with the planar nitro group; this steric interaction is minimized when the nitro group is twisted out of the aromatic plane.  When the nitro group is twisted out of plane it cannot resonance stabilize the transition state.  Hence the nitro compound reacts more slowly than the cyano compound.
While the underlying reaction is different, the thinking behind this question is identical to that found in this question from yesterday.


text:
Truly Understanding the Second Law of Thermodynamics and Entropy
There are some seemingly contradictory things I’d like cleared up:
Second Law states: $\mathrm dS_\text{univ} = \mathrm dS_\text{sys} + \mathrm dS_\text{surr}$, and $\mathrm dS_\text{univ}$ is always positive in a spontaneous process.
First off, what does “spontaneous” truly mean? “It means the reaction occurs without any external energy input” – I’ve heard this one before, but it doesn’t make sense. In an equilibrium there is a spontaneous direction (the side that is favored) but the reverse reaction also occurs, though at a slower rate, than the spontaneous reaction. What is the most fundamental definition, intuitive and universally applicable definition of spontaneous?
Moving on, according to ChemWiki the Second Law states that the entropy change of the universe can never be negative, even though my textbook says that this is only true for spontaneous processes. Who is right and why (explain in quantum terms, since I understand that in the long run of course the entropy of the universe increases by the laws of probability)?
An example to work through:
In a living cell, large molecules are assembled from simple ones. Is this process consistent with the second law of thermodynamics?
Book’s solution: To reconcile the operation of an order-producing cell with the second law of thermodynamics, we must remember that $\mathrm dS_\text{univ}$, not $\mathrm dS_\text{sys}$, must be positive for a process to be spontaneous. A process for which $\mathrm dS_\text{sys}$ is negative can be spontaneous if the associated $\mathrm dS_\text{surr}$ is both larger and positive. The operation of a cell is such a process.
First off, are we to consider the operation of a cell in this problem spontaneous? As I recall from biology, life processes require ATP, the carrier of energy, so these would technically “require external energy input,” would they not? However, the book treats it as a spontaneous process. Again, this confuses me because the definition of spontaneity is inadequate as I now semi-understand it.
Sorry for such a long post, but these are very important and fundamental ideas in spontaneity, entropy, and free energy, and I don’t want to read on until these ideas are well ingrained, and correctly so.

+1 for the research. Spontaneity means a reaction will happen without any, say, intervention. Also, please take a look at this question which explains the [seemingly deviating] contradiction between life and the second law.

Gibbs enthalpy is for spontaneity - it seems your textbook is wrong

Oh, I think I get it, "It means the reaction occurs without any external energy input" - it's closed system in definition and that's all.

Be careful in citing equilibrium as an example, as it fully integrates entropy in its definition. See this article for the details, but in a spontaneous reaction the entropy is increasing, and part of reaching equilibrium is that the energy gain and loss among products and reactants is balanced. Also, most experimental reactions are (nearly) closed systems for short periods of time, and if no energy is gained or lost (again, for short periods), you can have an equilibrium where there is no change in entropy. If you let your experiment sit and the heat of the system changes (for example), the equilibrium will change as well. And eventually you end your experiment, clean out your flask, and thus increase the entropy of the universe again!
@MARamezani's comment pointing to the article on Biology looks like a good answer to your biology questions. Earth is most definitely not a close system with that giant energy source 93 million miles away.

Some concepts in thermodynamics are unclear due to historic issues.
At first: "In an equilibrium there is a spontaneous direction (the side that is favored) but the reverse reaction also occurs". Thermodynamics arose as a macroscopic subject, in the XIX century two elemental reactions in opposite directions were not obvious. (Also, analyze your statement cautiously).
You should understand "spontaneous" in a macroscopic sense. Of course, you may find contradictions in the literature (because of the mixing of the macroscopic and microscopic worlds because of pedagogical reasons?)

that the entropy change of the universe can never be negative, even though my textbook says that this is only true for spontaneous processes

The first one is referring entropy changes associated to processes that are real. I mean, non considering results of calculations for arbitrary processes (which are very unlikely to have happened).
About the example. As most theories, thermodynamics has applicability limits. Specially the restricted version treated in most text books. The example is invalid because of it is completely incorrect to apply the thermodynamic of you book to a non-homogeneous system (or a system composed by homogeneous parts). There is a strange tendency to apply thermodynamic to anything. (I do not know exactly what is your book but I can be sure of the kind of treatment because of your cites).
In the other side, there the author is trying to relate entropy with order/disorder. This is only valid for total energy constant processes (I'm referring to the tendence to disorder). My advice is: be patient till you take a rigorous course (statistical physics or statistical thermodynamics or thermodynamics from another perspective), learn to repeat the examples for your test and forget it all as fast as you can. Or remember this while keeping in mind that maybe it is not exactly this way.
If you want to start with a relatively simple but more serious thermodynamics book, you can check out this:
Callen, Herbert B. (1985). Thermodynamics and an Introduction to Thermostatistics (2nd ed.). New York: John Wiley & Sons. ISBN 0-471-86256-8.

First off, what does “spontaneous” truly mean? “It means the reaction occurs without any external energy input” – I’ve heard this one before, but it doesn’t make sense. In an equilibrium there is a spontaneous direction (the side that is favored) but the reverse reaction also occurs, though at a slower rate, than the spontaneous reaction. What is the most fundamental definition, intuitive and universally applicable definition of spontaneous?

"Spontaneous" is being replaced in some syllabi by "thermodynamically favoured", which is a mouthful, but maybe more accurate! It means that $\Delta G < 0$ for the forward reaction, in other words, that we move to a lower $G$ if the forward reaction runs and, overall, converts some reactants to products.
We often consider this under standard conditions, where all concentrations are $1M$ so that concentration effects on the favourability of a reaction are ignored and the intrinsic stabilities of the reactants and products can be seen. This also gives us an idea of where an equilibrium will lie, to the right (mostly products) gives a large $K$ or to the left (mostly reactants) gives a small $K$. Overall the relationship is given by $$\Delta G^o = -RTlnK$$
In other words. A process will be thermodynamically favoured ($\Delta G^o < 0$) under standard conditions ($Q = 1$) if the equilibrium position is to the right of standard conditions (more products than reactants, $K > 1$).
To me, this is the real definition of "spontaneous". If a system is to the left of the equilibrium position (more reactants compared to products than we see at equilibrium) then it will naturally happen that the system will shift towards equilibrium as the forward reaction runs. This involves a decrease in $G$ as you move towards the more stable, equilibrium composition.

http://people.cst.cmich.edu/teckl1mm/PChemI/Chm351Ch9aF01.htm
Interestingly, this phenomenon can also be seen in terms of $S$ rather than $G$. The difference is that an equilibrium position involves the highest $S_{TOT}$ as this actually corresponds to the lowest $G$. Another way of thinking of $G$ is the opposite of the entropy of the universe. Processes tend to involve $G$ decreasing and $S_{TOT}$ increasing. Either one of these can be thought of as a statement of the Second Law.
Your analysis of the kinetics is correct - if a system is shifting to the right to come to equilibrium, then $$forward~rate > reverse~rate$$ this is by definition and it explains, from a kinetics point of view, what is happening. However the overall thermodynamics depend only on the state before and after, not the path taken in terms of rates.

Moving on, according to ChemWiki the Second Law states that the entropy change of the universe can never be negative, even though my textbook says that this is only true for spontaneous processes. Who is right and why (explain in quantum terms, since I understand that in the long run of course the entropy of the universe increases by the laws of probability)?

$\Delta S_{TOT}$ is always positive for a spontaneous change in isolation. However, by coupling a change that has a negative $\Delta S_{TOT}$ (non-spontaneous) to another process that has a more positive $\Delta S_{TOT}$ (spontaneous) we can get a situation where a "non-spontaneous" change is carried out but the overall process, including both changes, is spontaneous. You can argue the same point from the point of view of $\Delta G$ except that the signs are reversed.
See if the example that you gave to work through makes any more sense now. You have all the pieces needed in order to understand it.


text:
Understanding a passage related to salt buffer
Passage from my textbook:

Are reactions possible among any of the solution components; is so, what is their stoichiometry?
Suppose that you are asked to calculate $[\ce{OH-}]$ in a solution that is prepared to be $\pu{0.10 M}~\ce{NaOH}$  and $\pu{0.20 M}~\ce{NH4Cl}$. Before you answer that $[\ce{OH-}]=\pu{0.10M}$, consider whether a solution can be simultaneously $\pu{0.10M}$ in $\ce{OH-}$ and $\pu{0.20M}$ in $\ce{NH4+}$ . It cannot; any solution containing both $\ce{NH4+}$ and $\ce{OH-}$ must also contain $\ce{NH3}$. The $\ce{OH-}$ and $\ce{NH4+}$ react in a $1:1$ mole ratio until $\ce{OH-}$ is almost totally consumed:$$\ce{NH4+ +OH- ->NH3 + H2O}$$
You are now dealing with the buffer solution $\pu{0.10M}~\ce{NH3}-\pu{0.10M}~\ce{NH4+}$

I am having trouble understanding the third last sentence in the paragraph that starts with "2." The statement reads: "It cannot; any solution containing both $\ce{NH4+}$ and $\ce{OH-}$ must also contain $\ce{NH3}$. The $\ce{OH-}$ and $\ce{NH4+}$ react in $1:1$ mole ratio until $\ce{OH-}$ is almost totally consumed"
Can someone please help me out with understanding this passage?

To see if you really understand it, you should probably try some similar problems, such as what the pH is if 0.1M NaOH and 0.1M NH4+ are added in equal concentrations, and do the problems in the book. It will help you realize that the assumption is only momentary, and that there are more parts to the problem

There is no such thing as  $\ce{NH4OH}$ . They would react to form water and ammonia. Thus, to say that both exist in the solution at such high concentrations would be mistaken. They would react. Consider the reaction between ammonium and hydroxide.
$\ce{NH4+ + OH- -> NH3 + H2O}$
However, do note that at low concentrations, they can exist according to dissociation. However that is not what your passage is saying
*On your test, it may be good to assume the reaction goes to completion and then solve for the dissociation of ammonia. That is most likely what is expected of you.
*I thought of a more understandable analogy:
Think about $\ce{OH-}$ and $\ce{H+}$ in solution at 25 celsius. If one increases, the other decreases! According to ${Kw}$. Adding a lot of acid and base together would simply neutralize, they cannot exist at any higher than ${Kw}$

So if we were asked to calculate the [OH-] in a solution that is prepared to be 0.10M NaOH and 0.2M NH4Cl, is saying that [OH-] = 0.1M wrong?

Yes, you should assume that the base completely reacts with NH4Cl to create a solution of .1M NH4 and the reacted NH3. SInce both are identical, you could use the relation pH=pKa. But the pOH would not be 1. This is closely related to the concept of a buffer, which you could look into

I see! But if the OH- and the NH4+ are reacting in a 1:1 mole ratio, how is it possible that they can continue reacting until the OH- is totally consumed, if they are in equal concentrations?

Well, in that case, we "pretend" they completely react because by observing the K of ammonia, we see ammonia is more favored so more of it would exist. Then, we would need to use the K value to calculate it. So we first assume it goes to completion and use that to find what happens "afterwards", which is more close to the experimental value

Saying that "the $OH^-$ is totally consumed" is wrong and misleading. Most of the $OH^-$ is consumed and very little remains (I'm guessing less than $10^{-4} M$). This is because water can disassociate to produce more $OH^-$.


text:
In entropy change of surroundings, what is T?
This question is simpler than the previous one about intuition of entropy; I just need to know whether the $T$ term in $dS_{Surr} = \frac{-dH}{T}$ refers to the temperature of the surroundings or the system. I think it needs to be both, no? For example, for water to boil, the system must have $T \geq$ 100 degrees C. But can it be said that at high temperatures the decrease in entropy of the surroundings due to endothermicity of the boiling reaction is less significant than at low temperatures, as per division by T, so at high temperatures of the surroundings boiling occurs spontaneously?
It seems like both explain the phenomenon, especially since boiling occurs over time in both a sample of water at 110 degrees C in a 32 degrees C environment until it cools and boiling eventually occurs in a 110 degree C furnace if the temperature of the water starts at 32 degrees C.

dS_Surr means how the surroundings' entropy changes due to the energy released/absorbed by the system, so I think it means the temperature of the surroundings.


text:
Iterative method for calculating pH of a weak acid / base
The problem asks to determine the $[\ce{H+}]$ in a $0.20~\mathrm{M}$ solution of $\ce{Na3PO4}$. The $K_\mathrm{a}$ of $\ce{HPO4-}$ was given as $4.5\times 10^{-13}$, which then allows one to calculate the corresponding $K_\mathrm{b}$ as $2.22\times10^{-2}$.
As any acid-base problem, I simply set up the expression $\frac{x^2}{0.20~\mathrm{M}}=2.22 \times 10^{-2}$ with $x$ being $[\ce{OH-}]$ and solved for $[\ce{H+}]$.
However, the solution involves an interesting iteration which I have never seen before for this type of problem:

Can anyone shed some light on this?

This method of repeating iterations is interesting, though I would suggest something perhaps a little more straightforward (and rigorous) than doing something until you are "acceptably close."
This method I learned is called a RICE diagram.
Reaction
$$\ce{PO4_{(aq)}^{3-} +H2O_{(l)}<=>HPO4_{(aq)}^{2-} +OH^{-}_{(aq)}}$$
Initial (concentrations)
$$[\ce{PO4^{3-}}]=0.20~~~~~[\ce{HPO4^{2-}}]=0~~~~~[\ce{OH-}]=0$$
Change (in concentrations)
$$[\ce{PO4^{3-}}]=-x~~~~~[\ce{HPO4^{2-}}]=+x~~~~~[\ce{OH-}]=+x$$
Equilibrium (concentrations)
$$[\ce{PO4^{3-}}]=0.20-x~~~~~[\ce{HPO4^{2-}}]=x~~~~~[\ce{OH-}]=x$$
We can write the reaction expression as:
$$K_\mathrm{b}=\frac{[\ce{HPO4^{2-}}]\cdot[\ce{OH-}]}{[\ce{PO4^{3-}}]}$$
We substitute in the equilibrium concentrations and the $K_\mathrm{b}$:
$$2.22\times10^{-2}=\frac{[x]\cdot[x]}{[0.20-x]}$$
And solve for $x$:
$$2.22\times10^{-2}=\frac{x^2}{0.20-x}$$
$$4.44\times10^{-3}-2.22\times10^{-2}\cdot x=x^2$$
$$0=x^2+2.22\times10^{-2}\cdot x-4.44\times10^{-3}$$
Use the quadratic formula:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-(2.22\times10^{-2})\pm\sqrt{(2.22\times10^{-2})^2-4(1)(-4.44\times10^{-3})}}{2(1)}$$
$$x\approx-7.87\times10^{-2},~5.65\times10^{-2}$$
Obviously the negative solution is not applicable to our problem, so $x=5.65\times10^{-2}$.
Since $x=[\ce{OH-}]$, $[\ce{OH-}]=5.65\times10^{-2}$.
We know that:
$$[\ce{H+}]\cdot [\ce{OH-}]=1\times10^{-14}$$
so:
$$[\ce{H+}] \cdot [5.65\times10^{-2}]=1\times10^{-14}$$
$$[\ce{H+}] =\frac{1\times10^{-14}}{5.65\times10^{-2}}\approx1.77\times10^{-13}$$
If you have access to a calculator with a numerical solver function, I would recommend that instead of using the quadratic formula, but if not, this is how it is done.

The RICE table (or ICE table) is pretty much what the book used. However, both the book and the questioner had done the approximation that x is insignificant, which would lead to large deviations. Thus, successive iterations could lead to a close answer. Alternatively, the x could be included as significant and the quadratic could be solved, as is demonstrated here. I think the book was trying to make a point about checking your approximations, but not approximating is just as well, or perhaps even better.

Just to add to what everyone is saying, the reason that the iterations are shown is to arrive at a more accurate answer. Your method was also correct as long as you assume that $0.20 - x \approx 0.20$, but the problem's provided solution was just more accurate.
A good way to remember this is that if after you find the concentration of $\ce{x}$ (not doing the quadratic method), divide it by the original concentration of $\pu{0.20 M}$ and calculate the percent dissociation.
If $x \leq 5\%$ then your approximation is acceptable, and you don't have to worry about solving for $x$ using the quadratic equation and can cancel out the insignificant $x$ in the denominator.
However, if you want a more precise answer, it is better to use the quadratic equation to solve for $x$.

The correct equation for determining the value of $x$, as noted by ringo, is $$2.22\times10^{-2}=\frac{[x]\cdot[x]}{[0.20-x]}$$ With a $K$ value sufficiently low, you can ignore the $x$ value on the bottom of the fraction, as it is negligible compared to the initial concentration of the species present ($0.20$).
However, in this case the $K$ value of $2.22\times10^{-2}$ is actually quite close to the initial concentration of the $\ce{PO4_{(aq)}^{3-}}$ of $0.20$ (only around $10$ times less). As a result, we cannot ignore the $x$ in the denominator, as that would cause a significant change in the ultimate value of $x$. This is shown by the iterations you provided, the initial (approximate) $x$ value being about $\pu{0.01 M}$ different from the final. The iterations then are just a method of compensating for ignoring the bottom $x$ value from the very beginning. I was taught the same method that ringo outlined, and it is the more accurate method, but the iterations seem to work as well.

I was also taught this, but I didn't  like the wishy-washiness of being "sufficiently small." I was more for the method that would give me the right answer no matter the size of the $K_\mathrm{a}$ (or $K_\mathrm{b}$ as the case may be). This method does work for some cases, however, so really I say it comes down to preference.


text:
Is 1.5M NaOH a strong enough base to deprotonate benzyl alcohol?
Is 1.5M NaOH a strong enough base to deprotonate benzyl alcohol? I realize that NaOH generally does not deprotonate benzyl alcohol, but we performed an extraction in a Fischer esterification lab with benzyl alcohol an acetic acid to synthesize benzyl acetate. Acetic acid was added in excess, and the final ester product was purified with extraction techniques. I know that addition of NaOH removes acetic acid from the organic, diethyl ether layer, but would 1.5M NaOH also remove any remaining benzyl alcohol?

Only a very small fraction of the benzyl alcohol ($\mathrm{p}K_\mathrm{a} = 15.4$) would be removed by sodium hydroxide. Acetic acid is about 10 orders of magnitude more acidic than benzyl alcohol with a $\mathrm{p}K_\mathrm{a}$ of $4.8$, so it is completely deprotinated to form sodium acetate and water.
For the purposes of your experiment you can assume that there is effectively no reaction between benzyl alcohol and sodium hydroxide.


text:
Household Melting Point Standards
I'd like to accurately calibrate a digital thermometer I'm building in the range of $50^\circ$ to $120^\circ$ Celsius.
I'm aware that one can buy various chemicals that have melting points in this range, but are there any readily available household / pharmaceutical items I could use instead?
From what I have read, it looks like the two easiest candidates would be vanillin or Sodium acetate (Thanks Curt F.), but I'm not sure how one would go about creating a pure enough calibration standard at home.

The boiling point of water is a good calibration point.
Sodium acetate trihydrate has a well-known melting point at 58 °C but I don't know if you count that as "household".

@CurtF. So one would just boil vinegar and baking soda, then cool and reheat to find the melting point? That sounds pretty household to me.

Related: Looking for a chemical to calibrate a temperature sensor

nubis: true but (i) it would take a lot of boiling and (ii) I worry about purity.  Without perfectly pure compounds, the melting point is not exact.

According to this cooking site, Crisco brand vegetable shortening melts between $\mathrm{47^oC-48^oC}$ (yes, I know $\mathrm{48^oC}$ is less than $\mathrm{50^oC}$, but it seems pretty close to me). The best way to use such a non-certified melting point "standard" like this is to use your thermometer side-by-side with a properly calibrated thermometer.
Vanillin may be your best bet (melting point of $\mathrm{81^oC - 83^oC}$).  Again, using a calibrated thermometer to check the actual melting point would be the ideal situation.
Although your question asks about calibrating based on melting point measurements of household items, household items just aren't usually of the purity required for accurate melting point based thermometer calibration. In the absence of a calibrated thermometer as discussed above, the best way to calibrate your thermometer is probably to use the freezing and boiling points of water (don't forget to adjust boiling point for altitude if you don't live at sea level), then check the melting points of Crisco and vanillin if you want. Good agreement with these stated melting points would be a good indicator of proper thermometer calibration, while poor agreement could either be due to improper thermometer calibration or impure material.


text:
Finding pH of a buffer
What is the pH of a solution prepared by mixing 25.00 mL of 0.10 M methylamine, $\ce{CH3NH2}$,
with 25.00 mL of 0.10 M methylammonium chloride, $\ce{CH3NH3Cl}$? Assume that the volume of
the solutions are additive and that Kb = 3.70 × 10-4 for methylamine.
1) First, I multiplied 0.025 L with 0.10 M and received 0.0025 moles. I then divided 0.0025 moles with 0.050 L (The total volume of the solution) and received 0.05 M, the molar concentration of the methylamine and its conjugate acid in this mixture.
2) I then created an ICE Chart (my reactant was the methylamine so that I would not need to find the Ka) and I solved for "x" (which would give me the concentration of $\ce{OH}$ ions at equilibrium). First, I solved for "x" but I assumed that is was insignificant. In the end I received 10.57 as the pH, which was the wrong answer. Then when I solved for "x" using the quadratic formula (assuming that "x" was significant), I got 0.0003646 as my "x".
3) When I took the negative log of 0.0003646, I got a pOH of 3.46. I subtracted this number by 14, and I still received a pH of 10.57
4) I even attempted to use the Henderson Hasselbalch equation and still received 10.57 because the ratio of the base and acid is 1, meaning that the log is 0.
I don't understand what I am doing incorrect. The answer is supposed to be 10.87, can someone please assist me?

Didn't you say the answer is 10.57 and you got it? Twice?

Sorry, that was a typo. The answer is 10.87.

Your approach is correct and the result you obtain is right. Checked here.
As you correctly affirm, since the ratio of the base and acid is $1$, log will be zero and therefore $\mathrm{pH}=\mathrm{p}K_\text{a}$, where $\mathrm{p}K_\text{a} = 10.57$ (derived from the value of $K_\text{b}$ you provide).


text:
Why are inert gas (especially Xenon) compounds powerful oxidizing agents?
I am curious as to why compounds with inert gases, such as $\ce{XeF4}$, $\ce{XeF2}$, and $\ce{XeO3}$ are considered powerful oxidizing agents. I would attribute the phenomenon to the highly oxidized state of the noble gas which would be unstable but I am not too sure about it.

http://chemistry.stackexchange.com/questions/13602/how-many-argon-compounds-are-theoretically-and-experimentally-available, http://chemistry.stackexchange.com/questions/7764/are-there-natural-conditions-that-could-enable-the-formation-of-noble-gas-compou

Let's be clear here. Xenon by itself is not an oxidizing agent. Instead, certain compounds of xenon tend to be strong oxidizing agents.
First, let's consider an analogous case - the nitro group. Molecules containing nitro groups ($\ce{-NO2}$) tend to be explosive - the nitro groups contain oxygen which facilitates combustion. Remember, combustion is just a form of oxidation.
Back to xenon!
Xenon, being a noble gas, shouldn't like bonding too much. It already has a complete octet of outer shell (valence) electrons.
So it is likely the case that compounds of xenon are only weakly held together. Compounds of xenon can therefore said to be relatively thermodynamically unstable. Therefore, it is also likely that compounds of xenon tend to react in the presence of other molecules - especially ones containing atoms to which fluorine can form stronger bonds.
We verify the above by examining bond enthalpy data: the Xe-F bond has a bond energy of about 130 kJ/mol. Compare this to a typical C-F bond with a bond energy of 485 kJ/mol.
It is no wonder therefore that compounds such as xenon difluoride are used in fluorinating (oxidizing) organic molecules.
Here are some examples from Wikipedia. Why the nitro group in the second photo has been reduced to an amino group - I have no clue - and I suspect it's just an error. Further updates soon ...

Why does $\ce{XeF2}$ reduce the nitro group to an amine group in your second example?

@bon good catch and I have no clue. Picture was taken from Wikipedia


text:
jury-rigging an automotive catalytic converter for use in the Ostwald Process?
Would it be feasible to jury-rig an automotive catalytic converter for use in the Ostwald Process for the production of Nitric Acid? Given reasonable safety precautions, would it be safe enough to be practical for a home lab?
A Catalytic converter is a device used in cars to reduce Nitric oxide into nitrogen and Oxygen, as well as oxidizing Carbon monoxide and unburnt hydrocarbons into Co2 and H20.
What I'm proposing is feeding ammonia and oxygen into it at an appropriate temperature and pressure to produce the Nitric Oxide necessary to create Further oxidize into Nitrogen Dioxide, which is reacted with water to produce Nitric acid and Nitric Oxide.
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) (ΔH = −905.2 kJ)
2 NO (g) + O2 (g) → 2 NO2 (g) (ΔH = −114 kJ/mol)
3 NO2 (g) + H2O (l) → 2 HNO3 (aq) + NO (g) (ΔH = −117 kJ/mol)
So my Question is whether a Catalytic Converter salvaged from an automobile is a suitable substitute for a column packed with Platinum gauze.

The Ostwald process requires high pressures and produces gaseous nitric acid. This is industrial chemistry, not home chemistry.

Why are you trying to produce nitric acid anyway? Just go and buy some.


text:
What happens to chemical shifts of protons close to aromatic systems in ¹H-NMR?
I have two questions regarding chemical shifts on aromatic systems.
First, what does being near an aromatic system do to protons? If we look at the spectra of 1-napthalenemethanol and then 9-anthracenemethanol, the peak of the two aliphatic protons of the benzylic methylene group get shifted way up. Can someone explain why adding in that third aromatic ring does this?
The spectra:

9-anthracenemethanol
1-naphtalenemethanol

I was also under the impression that protons near electronegative elements would be deshielded and be shifted higher. However in the spectrum of Benzil, the protons para to the carbonyl group are higher shifted than the protons meta to the carbonyl. Why is this? Is it to do with conjugation of the carbonyl with the aromatic ring? I'm having trouble finding a proton nmr in the literature, so there's nothing to compare it to.

@Martin Thanks for your effort. It is much more easy to read now and hence more likely to get an answer.

The aromaticity found in certain ring systems (e.g. benzene, see Hückel's rule) causes a magnetic field because of the circular motion of the electrons in the ring (see picture).

(Source: Clayden et al. Organic Chemistry 2nd edition, p. 277)
This additional field enforces the magnetic field of the NMR machine inside the ring, while countering the field outside the ring, causing the signals of the outer hydrogens to shift downfield (i.e. to the larger shifts).
Now for your other questions:

First, what does being near an aromatic system do to protons? If we
  look at the spectra of 1-napthalenemethanol and then
  9-anthracenemethanol, the peak of the two aliphatic protons of the
  benzylic methylene group get shifted way up. Can someone explain why
  adding in that third aromatic ring does this?

Lookig at the difference in shift between the benzylic methylene protons of 1-Naphthalenemethanol and 9-Anthracenemethanol we must conclude that the induced magnetic field of the second is stronger at that position than that of the former, as there is no other appreciable difference. Intuitively this seems to make sense as the field of the 9-Anthracenemethanol is more uniform at and closer to (with respect to the "right" side of the molecule as in the pictures) the discussed proton. Of course this could/should be confirmed by quantum mechanical calculations, but I do not have the tools or the knowledge to do that (if someone has, please feel free to expand this answer!)

(1-Naphthalenemethanol Proton NMR)

(9-Anthracenemethanol Proton NMR)

I was also under the impression that protons near electronegative
  elements would be deshielded and be shifted higher. However in the
  spectrum of Benzil, the protons para to the carbonyl group are higher
  shifted than the protons meta to the carbonyl. Why is this? Is it to
  do with conjugation of the carbonyl with the aromatic ring? I'm having
  trouble finding a proton nmr in the literature, so there's nothing to
  compare it to.

Yes, conjugating effects and inductive effects can counter each other, but there is no easy general way to say which will win out in a particular situation, although usually it is the conjugating effect.

Jori’s answer does a nice job in explaining why aromatic rings generally shift proton shifts downfield, and why this is not restricted to the aromatic protons themselves; see the pictures at the beginning of his post for explanation (and replace the benzene proton with a $\ce{CH2}$ group in the right-hand side picture for further clarity).
The difference in chemical shift $\Delta \delta(\ce{C\mathbf{H}_2OH}) = 0.47~\mathrm{ppm}$ which is a very small, almost neglegible difference. Compare:

the difference between naphthalene-1-ylmethanol and benzyl alcohol: $\Delta \delta(\ce{C\mathbf{H}_2OH}) = 0.41~\mathrm{ppm}$
the difference between benzyl alcohol and methanol: $\Delta \delta(\ce{C\mathbf{H}_{$n$}OH}) = 1.12~\mathrm{ppm}$

(Note that we cannot strictly compare the chemical shift difference between naphthalene-1-ylmethanol and anthracene-9-ylmethanol since the latter has been measured in $\ce{(CD3)2SO}$ rather than $\ce{CDCl3}$. However, the general trend works)
The big difference in shifts occur when we add an aromatic ring to our methyl alcohol; the number of six-membered aromatic rings does not really matter. This is because the first ring is added in close vicinity to the centre in question while the other rings are further away — we might compare this to the difference in chemical shifts between ethanol and 2-phenylethanol: $\Delta \delta (\ce{C\mathbf{H}_2OH}) = 0.09~\mathrm{ppm}$ (even smaller). However, this last comparison is not strictly valid due to the flexibility of 2-phenylethanol being larger than that of the polyaromatic systems.
Summed up in a nutshell: The chemical shift difference between naphthalene-1-ylmethanol and anthracene-9-ylmethanol can be attributed to the presence of a third aromatic ring in relative proximity of the group in question.

Your second (unrelated, and probably should have been posted separately) question deals with how groups outside of a phenyl ring influence the chemical shifts of protons belonging to that ring.
In the case of benzil, we are dealing with an electron-withdrawing carbonyl group, and we might as well choose the simpler benzaldehyde system for comparison. You can picture the electron-withdrawing effect of the carbonyl group by a set of enolate resonance structures including charge separation: We can relocate the positive charge to the ortho- and para-positions of the phenyl ring, but not to the meta-positions, resulting in a lower electron density at ortho and para and thus greater deshielding and a downfield shift.
Similarly for electron-donating groups such as the methoxy group in anisole: The ortho- and para-positions are shifted upfield because the electron-donating group supplies their positions with a greater electron density and thus more shielding. The meta-groups are rather unaffected by that.
This can be generalised to the (not necessarily always true!) statement:

Groups in benzylic positions generally leave meta-positions unaffected and affect mainly the chemical shifts of ortho- and para-positions.

Therefore we expect a downfield shift for ortho- and para-protons in benzil and no shift change for the meta ones.


text:
Do I have to consider the molecular mass of the oxygen atom or the diatomic oxygen molecule when determining the empirical formula of an iron oxide?
In determining the empirical formula of a compound which constitutes a diatomic molecule, should we calculate the molar mass of the divalent molecule by multiplying the molar mass of the element by 2, or should we leave it as it is. Because, I happened to come across a problem, where I had to calculate the empirical formula for an oxide of iron. A similar question has been answered before, but it doesn't seem to clear my doubt - Why not to consider hydrogen and oxygen moles to determine an empirical formula. Hence, I restate it:

If an oxide of Iron has $69.9\%$ iron and $30.1\%$ dioxygen by mass, what is the empirical formula?

Now, if I assume the mass of compound to be $100~\mathrm{g}$, then it contains $69.9~\mathrm{g}$ of iron and $30.1~\mathrm{g}$ of oxygen. If I calculate the no. of moles,
$$\text{Moles of }\ce{Fe} = \frac{\text{given  mass}}{\text{molar mass}} = 69.9/58.5 = 1.25~\mathrm{mol}$$
Moles of oxygen - this is where the problem comes. If I take molar mass of oxygen as $32~\mathrm{g}$, as in $\ce{O2}$, I get $0.94~\mathrm{mol}$. It will not make any sense, as the ratio of iron to oxygen will not be in any reasonable quantity. However, if I take the  molar mass to be $16~\mathrm{g}$, I get $1.88~\mathrm{mol}$ of oxygen, and I can divide both the numbers by $1.25$ to get the ratio as $1.88/1.25: 1.88/1.88 = 1:1.5 = 2:3$,
hence the compound will be $\ce{Fe2O3}$, which makes perfect sense. But the problem is, why should I calculate the no. of moles from the molar mass of $\ce{O}$ rather than $\ce{O2}$? After all, its is the oxygen molecule that will exist, and no the oxygen atom right?
Also, the solution is given as $\ce{Fe2O3}$, so I need to take the molar mass of oxygen to calculate the number of moles as $16~\mathrm{g}$. The question is, Why? Why not $32~\mathrm{g}$?
Does it have something to with the fact that the question states "dioxygen" rather than "oxygen"? If yes, what is the difference between dioxygen being present in $30.1\%$ or oxygen present in $30.1\%$? I am in a real mess.

The question is badly worded. Iron oxide does not contain "dioxygen" in its structure and–even if it did–this would be irrelevant for calculating the mass percent of oxygen.

When determining the empirical formula of a compound, you only consider the molar mass of the element. In the compound itself, it is only the atom (or ion) that exists.
If you would assume the molar mass of dioxygen, then you would also assume, that the iron oxide is made out of iron and dioxygen molecules.
I do not know, and also do not understand, why the question states "dioxygen" as that is - in my opinion - false.


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The nuclear Overhauser effect
I am having some trouble with understanding the NOE. The carbon NMR spectra can't be integrated, because the protons attached to the carbon atoms will make primary carbon atom relaxation faster than tertiary, so their peaks will have bigger intensity. With a bigger acquisition time, the peaks will be more proportional. In proton NMR, nearby protons will cause the target proton relaxation to be faster. When NOE is applied, those nearby protons will be irradiated and the target proton peak will have a slightly bigger intensity. Why will it have a bigger intensity?

Consider the simplest possible NOE experiment, with two dipole-coupled protons, the
target I and nearby proton S.
Before the experiment there are four populated spin states: $\alpha_{I}\alpha_{S}$, $\beta_{I}\alpha_{S}$, $\alpha_{I}\beta_{S}$ and $\beta_{I}\beta_{S}$, where $\alpha$ indicates $m_{i} = \frac{1}{2}$ and $\beta$ indicates $m_{i} = -\frac{1}{2}$. The relative populations of the states at thermal equilibrium (from Boltzmann distribution) are:

$\alpha_{I}\alpha_{S}: 1 + 2\Delta $
$\beta_{I}\alpha_{S}$,  $\alpha_{I}\beta_{S}: 1$
$\beta_{I}\beta_{S}: 1 - 2\Delta $

S is irradiated so that the population difference between the $m_{S} = \pm \frac{1}{2}$ states goes to zero. This results immediately in the following populations:

$\alpha_{I}\alpha_{S}, \beta_{I}\alpha_{S}: 1 + \Delta $
$\beta_{I}\beta_{S}, \alpha_{I}\beta_{S}: 1 - \Delta $

What happens next typically for small molecules is fast "flip-flip" cross-relaxation, restoring the $\alpha_{I}\alpha_{S}, \beta_{I}\beta_{S} $ states to their equilibrium populations. So we have the populations:

$\alpha_{I}\alpha_{S}: 1 + 2\Delta $
$\beta_{I}\alpha_{S}: 1 + \Delta$
$\alpha_{I}\beta_{S}: 1 - \Delta $
$\beta_{I}\beta_{S}: 1 - 2\Delta $

The population difference between the pairs $\alpha_{I}\beta_{S}, \alpha_{I}\alpha_{S}$ and  $\beta_{I}\beta_{S}, \beta_{I}\alpha_{S}$ has now increased by 50%, from $2\Delta$ to $3\Delta$. This corresponds to a 50% increase in intensity of the peak. Transition between these pairs obviously corresponds to spin-lattice relaxation.


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How to calculate "accurate" surface area of a microporous material using Micromeritics equipment?
How do I get an accurate surface area from a nitrogen isotherm?
I've been doing some surface areas on MOFs and upon showing my boss the isotherms he asked me to convert them to show the "range", which will give the most accurate surface area.
I'm using Micromeritics ASAP-2020 software and I believe that you measure this BET range by using points from relative pressures between 0.05 and 0.3. However, this seems to be unsuitable for some microporous compounds?


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Why does iodine have different colors in different media?
It is violet in trichloromethane, but brown in water. why? In the Shriver Atkins book, they tried to explain this using Molecular Orbital Theory which was not clear to me

Some interesting information can be found in this related question.

color is result of electron transfer between orbitals and anything that affect energy of these orbitals can change color of that specie, so iodine in different mediums have different interactions and because of that iodine orbitals have different energies in different medium,so iodine have different color in different mediums.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. Could you elaborate a little bit more, I cannot follow your thoughts. How do the interactions with the medium influence the energies of iodine?


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Exo- or endothermic? Clarification required regarding a few cases
Are the following processes exothermic or endothermic?
a) When solid $KBr$ is dissolved in water, the solution gets colder.
b) Natural gas ($CH_4$) is burned in a surface.
c) When concentrated $H_2SO_4$ is added to water, the solution gets very hot.
d) Water is boiled in a teakettle.
Attempt at solution: For a), I would say that it is exothermic, since the solution gets colder, which means that it releases contained heat to the surroundings.
b) is exothermic evidently.
c) By the same reasoning, the solution is getting very hot, so it must be endothermic since it absorbs heat.
d) Not sure about this one. I would say endothermic, because the absorbed heat makes it boil.
This is what I answered, and apparently I made atleast two mistakes. So can someone explain to me the flaws in my reasoning?

I think you are confusing terms:
exothermic means that it releases heat, that is energy, because the energy of the products is lower than the one of the reagents.
endothermic means that it absorbs heat (energy) because the products are at a higher level of energy than the reagents.
When a solution becomes hot, it's because it's releasing heat (energy), and likewise if it gets cold it's because the reagents inside are absorbing energy from the surronding environment.
The case of boling water: does it need heat to become steam? Or does it release heat? In other words, do you have to provide heat to the teakettle to make the water boil or it provides heat by itself?
I hope you can get to your answers by yourself now.


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What makes a metal more inclined to bond covalently?
Many metals form compounds with s-block elements that are essentially entirely ionic. However, elements like silver, copper, lithium and so on, can form bonds that are significantly covalent in character. What are the factors that govern whether a particular metal will bond with significant covalent character? Is this linked to hardness and softness?

http://chemistry.stackexchange.com/questions/429/why-do-magnesium-and-lithium-form-covalent-organometallic-compounds?rq=1, http://chemistry.stackexchange.com/questions/27735/why-does-silver-form-bonds-with-covalent-character?rq=1

That depends on compounds with which they create bonds and oxidation numbers.

http://chemistry.stackexchange.com/questions/9038/how-would-one-compare-the-magnitude-of-covalent-character-between-sncl4-and-snf2?rq=1, http://chemistry.stackexchange.com/questions/16261/metal-compounds-that-bond-covalently?rq=1


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Mechanism of enolether formation with ethyl orthoformate in estr-4-ene-3,17-dione (Djerassi's norethisterone synthesis)
The second step of Djerassi's synthesis of 19-nor-17α-ethynyltestosterone (norethindrone), involves the protection of the ketone in position 3 via enolether formation, as in the picture:

What is the mechanism of this reaction? I couldn't find it anywhere. I suppose it should be similar to an ether synthesis, but I'm not sure about the precise mechanism.
(On a side note, I assume that the other ketone in 17 remains intact because of controlled reaction conditions.)

Usually a catalytic amount of acid is used in these reactions and they are run in the presence of alcohol (ethanol if you want to prepare the ethyl enol ether). Pyridine is typically added at the end of the reaction to recapture the acid catalyst.
The first step involves protonation of the carbonyl oxygen.  Protonation of the C-3 oxygen produces a resonance stabilized carbocation where the charge is delocalized over the carbonyl and the adjacent double bond (protonation at the C-17 oxygen does not have an adjacent double bond,  hence protonation at the C-3 oxygen is favored).  The nucleophilic alcohol oxygen reacts at the carbonyl carbon (highest positive charge density) to produce a hemiketal.  Loss of water gives rise to the enol ether product.  The eliminated water is captured by the orthoformate and generates more alcohol.

$$\ce{HC(OEt)_3 + H2O -> HCO2Et + 2EtOH}$$
Some argue that the reaction actually proceeds through the ketal (rather than the hemiketal) since ketals can be isolated in these reactions.  Others argue that just because the ketal is formed, doesn't prove it is actually involved in the enol ether formation reaction.  IMO, lacking definitive evidence, I opt for the simpler hemiketal mechanism.


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Why does solvent leveling occur?
Solvent leveling means that strong acids completely dissociate in that solvent so it is impossible to distinguish between two acids that completely dissociate in it. A solvent levels and acid if the $K_\text{a}$ of that acid is greater than 1. However, that’s not completely dissociated. An acid can have a $K_\text{a}$ of 2, 3 or 25 what’s the problem with being able to distinguish between them? In truth, whatever the $K_\text{a}$ it’s not actually COMPLETELY dissociated.

Your final statement:

in truth, whatever the $K_a$ it's not actually COMPLETELY dissociated

is exactly right.
Strong acid is very often stated to be "completely dissociated", but that's just wrong.
For example, nitric acid has a $K_a$ of 24, so 1M nitric acid about 96% dissociated. If the concentration is higher, it will be less dissociated and if the concentration is lower it will be more dissociated.
I would think of the solvent leveling effect this way:
Suppose you have 1 liter of 0.1M ethanol in water and you add 0.1 moles of sodium amide thinking that it will deprotonate the ethanol.  That's not going to happen because most of the protons that protonate the amide will come from the solvent (water) rather than the ethanol.