text:
What do these mineral classification symbols mean?
For my birthday I got an awesome mineral specimen case, as pictured below. What piqued my interest was the notations in the lid where the minerals are described with 5 specifications: C (obviously for Colour) then S, H/SG (I'm guessing specific gravity), L and CL.
Can anyone shed any light on what these abbreviations mean? Additionally, do the coloured squares mean anything or are they just for identification (they match colours painted on some of the samples).

You will probably get a much better answer over on Earth Science Stack Exchange.  That is a sweet present.  Makes me want to get one.

My guesses are S for "sheen," H for "hardness," L for "luster," and CL for "cleavage," but I think @CurtF. is right in that this should be on Earth Science.

Definitely a question for Earth Science SE.

I didn't realise there was an Earth Sciences Stack Exchange, I guess there's an S.E. for pretty much everything now. And yes. Best. Birthday. Ever.

In addition to being a non-practicing chemist, I'm also a non-practicing geologist, so look no further for an answer you can trust.
Your minerals are classified as follows: C is for color (sample, gross), S is for streak (color of powdered mineral, essentially, on an unfinished streak plate), H is hardness on the Mohs scale, SG is specific gravity, L is luster, and CL is cleavage (how the sample breaks along planes dictated by the crystal habit of the mineral). See this reference for a printable chart.
The color-coding is certainly used to mark the samples in case they get jostled or mixed up and need to be matched to their appropriate slots: those codes are not related to the mineral classification.


text:
acids, bases, salts, etc
All the chemical compounds can be categorized as acids, bases, and neutral compounds.
There are different theories, namely Arrhenius Theory, Brønsted-Lowry Theory, Lewis Theory, which define the acids and bases. Each successive theory becoming more general; Lewis Theory is the most generalized one of the three.
Neutral compounds could be salts, hydrocarbons like ethanol, oxides, and many more. A salt is always a result of a reaction between an acid and base.
The difference between a base and an alkali is that an alkali is a base which are soluble in water. In other words, all alkalies are bases but not all bases are alkalies.
I'm sure like bases there would be many acidic compounds which are not soluble in water. Please correct me if I'm wrong.
Q1:
Is litmus test really a defining criterion for classification of a compound as an acid or base? I believe that to perform litmus test, the compound should be soluble in water, and we have already agreed that not all bases and acids form aqueous solution. It means that litmus test is not a main criterion for the classification. It is something of secondary importance. Do I make sense?
Q2:
Likewise, general properties given in school chemistry books are also not primary criteria. Stated differently, an acid doesn't always need to taste sour, turn litmus red, and a base doesn't have to feel slippery or turn litmus blue. Is what I'm saying correct?
Boron trifluoride is classified an acid. Boron has three electrons in its outer shell and it forms three covalent bonds with three fluorine atoms which give it total sum of six electrons. It's still deficient of two electrons and therefore it can accept an electron pair so it's Lewis acid.

A lot of the information in this question seems to be irrelevant. Can you actually clarify what the question is and remove some of the unnecessary information (and write a better title).

Have tried to clarify it.

There is a big problem with your first sentence, "All the chemical compounds can be categorized as acids, bases, and neutral compounds." Namely amphoteric compounds that can act as either an acid or a base.  Water is a good example.  In the presence of HCl, water acts as a base. In the presence of ammonia, water acts as an acid.  So the question "Is water an acid, a base or is it neutral?" cannot be categorized as any one of those three options.

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text:
Determination of brominated aromatics by 13C and 1H NMR spectroscopy
How can I identify 4-bromoazobenzene, 4-bromoazoxybenzene, and 4,4’-dibromo-$\ce{N,N}$’-diphenylhydrazine products using $^1\ce{H}$ and $^{13}\ce{C}$ NMR?
I’m really confused by the general shape of NMR peaks of aromatic protons. I want to distinguish each product from the other products, knowing that the three products are synthesized starting by the reaction of 4-bromonitrobenzene with sodium borohydride.

Do you want to distinguish between those substances in a single mixture or one by one?

I doubt the peak shapes are going to help you, since these are all para-substituted aromatics. You could probably distinguish by chemical shift. Try to find authentic spectra of each. They should be available.

one by one in itself

I don't find any experimental spectra online, but I have been suggesting NMRDB.org to students and colleagues. I don't know how accurate the predictions are, but the site is easy to use and for the compounds you're suggesting, rule-based methods should be fairly reliable.

Predict $\ce{^1H}$ NMR
Predict $\ce{^13C}$ NMR

In the 4,4’-dibromo-N,N’-diphenylhydrazine, you should have two sets of "double doublets" (i.e., four peaks in each set of equal height).

In the 4-bromoazobenzene, you break symmetry, so there should be four signals which integrate to two protons, and one additional signal integrating to one proton (i.e., all ~7.3-7.8 ppm, so a big mess).

In the $\ce{^1H}$ spectra of 4-bromoazoxybenzene, the protons closest to the azoxy -N+-O- should be deshielded a bit (maybe ~7.8-7.9 ppm?) compared to the aza analogue. Certainly it's also going to be a fairly congested spectrum.
I'll leave the $\ce{^13C}$ NMR as an exercise (or to others to answer).

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text:
How to calculate the amount of water produced by the incomplete combustion of propane?
I was wondering if someone could walk me through how to do the following two questions?

The complete combustion of pentane has a enthalpy change of $-3509~\mathrm{mol^{-1}}$. Write an equation for the complete combustion of pentane. Write a possible equation for the imcomplete combustion of pentane.

For this question, would the following equation suffice:
$$\ce{ C5H12 (g) + 8O2(g) -> 5CO2 (g) + 6H2O (g)  }$$
For an incomplete reaction, the products will be water, carbon/and or carbon monoxide due to insufficient oxygen, so would this equation be correct:
$$\ce{C5H12 (g) + 11/2 O2 (g) -> 5CO (g) + 6H2O (g)}$$

$17~\mathrm{L}$ of Propane are incompletely combusted in $34~\mathrm{L}$ of oxygen at $25~^\circ\mathrm{C}$ and $100~\mathrm{kPa}$ into only water and carbon. Write a balanced chemical question to represent this including state symbols. What is the mass of the water produced? What observable change would indicate that a chemical change occurred?

Would an equation like this work: $$\ce{C3H8 (g) + 7O2 (g) -> 3CO2 (g) + 4H2O (g)}$$
The ratio for the above equation will be 1:7:3:4.
I am given the volume of the propane to be $17~\mathrm{L}$ and the volume of the oxygen to be $34~\mathrm{L}$. Since the question is a molar calculation, I think I have to convert the $~\mathrm{L}$ into the SI units of grams. This means $17~\mathrm{L} \Rightarrow 17000~\mathrm{g}$ and $34~\mathrm{L} \Rightarrow 34000~\mathrm{g}$.
The I have to calculate the number of moles for propane and oxygen (I think).
$$
\text{number of moles} = \frac{\text{mass}}{\text{molar mass}}\\
\text{number of moles of propane} = \frac{17000}{44.1} = 385.4~\mathrm{\frac{mol}{gram}}\\
\text{number of moles of oxygen} = \frac{34000}{16} = 2125~\mathrm{\frac{mol}{gram}}$$
Finally I have to find the mass of the resulting water, which is the part I am stuck at. I know I have to find the number of moles times the molar mass of water to find the total mass of the water, but I am just somewhat stumped as to how to find the number of moles of water given what I have worked out above.
For chemical changes, I think that bubbles of gas may appear, a temperature change will occur since this is a combustion reaction, and maybe a change of mass/volume?

Yes the complete combustion equation is correct as you stated and the equation of the incomplete combustion is certainly one possibility. A more general formula is:
$$\begin{multline}\ce{C5H12 + $\left(\frac{x}{2}+y+3\right)$\,O2 -> $(5-x-y)$\,C + $x$\,CO +$y$\,CO2 + 6H2O}\\
\left\{x,y\in\mathbb{R}^{+}_0; (x+y)\leq5\right\}
\end{multline}$$
This is assuming all the hydrogen will be converted into water. In some cases you might have incomplete combustion, where ethene or ethyne (ar many others) are formed, but I think this is taking it a bit far at this point.
The above case reduces to your formula for $x=5$ and $y=0$.

For the second question, your equation has a small error, correct is:
$$\ce{C3H8 (g) + 5O2 (g) -> 3CO2 (g) + 4H2O (g)}$$
But this is not the equation you need, as the question explicitly states, that the combustion products are carbon and water. So you working equation becomes:
$$\ce{C3H8 (g) + 2O2 (g) -> C (s) + 4H2O (g)}$$
So you need twice the number of moles of oxygen than propane, to produce four times as many moles of water than you had propane.
You cannot convert a volume of gas into a mass by an easy conversion factor, you have to use the ideal gas law:
$$pV =n\mathcal{R}T$$
You can then use $m =n\cdot M$ to convert to grams, but you don't yet have to do that.
You have been given the temperature $T=25~^\circ\mathrm{C}=298.15~\mathrm{K}$ and pressure $p=100~\mathrm{kPa}$ and the volumes for the reactants, so you can calculate the amount of substance for propane and oxygen.

 $n(\ce{C3H8})\approx 0.7~\mathrm{mol}$ and $n(\ce{O2})\approx 1.4~\mathrm{mol}$

From the reaction equation you can now determine the amount of substance of water, that will be produced.

 $n(\ce{H2O})\approx 2.8~\mathrm{mol}$

You can convert that easily into a mass now.

 $m(\ce{H2O})\approx 50~\mathrm{mol}$

There are two notable changes you will observe. Since you are combusting a gas, obviously no bubbles can appear. If you do the combustion in a see-through container you will see black stains due to the solid carbon that is produced by the reaction and liquid water drops will form on colder surfaces.
You are of course right, that you will also have a rise in temperature.
Note: The here given values are crude approximations only, that might not satisfy the task at hand. You need to do the calculation yourself to find the correct answer.

Hey for the answer, is the unit supposed to be grams and not mols?

It says "What is the mass of the water produced?" in the question, so the answer must be given in grams.


text:
Can acid concentration be determined from Ka values in a titration?
Say I have a sample of vinegar and I want to find the concentration of ethanoic acid in it. So I perform a titration using a $\mathrm{pH}$ meter, and using that data I can draw a $\mathrm{pH}$ curve and find quantities such as the initial $\mathrm{pH}$, the half-equivalence point, the equivalence point, and the $\mathrm{pH}$ at the equivalence point.
I feel like my teacher hasn't explained this very clearly, but given the $K_{\mathrm a}$ value for ethanoic acid and the data collected from the titration, is it possible to determine the initial concentration? If so, how do I proceed?
(I'm aware of the simple stoichiometric method, just curious whether this approach will work.)

The Henderson-Hasselbalch equation will help:
$$\mathrm{pH} = \mathrm{p}K_{\mathrm a} + \log {[\ce{A-}]\over [\ce{HA}]}$$
Once the $\mathrm{p}K_{\mathrm a}$ is known, the equation will tell you the ratio of acetate to acetic acid (ethanoic acid). The ratio changes at various $\mathrm{pH}$ values.
The $\mathrm{p}K_{\mathrm a}$ is 4.76, so for a $\mathrm{pH}$ below 3.75 the dominant species in solution will be acetic acid; at pH 5.76, the dominant species in solution is the acetate ion. If you want to make a solution at a particular $\mathrm{pH}$ value, use the equation to give you the ratio of acetate to acetic acid needed.

Yes. Your question is basically how to determine $K_\mathrm a$ from $\mathrm{pH}$ of a salt of the acid. Since ethanoic acid is a weak acid, its salt will hydrolyze to give an alkaline solution. You can calculate the $K_\mathrm a$ given that the concentration and the $\mathrm{pH}$. The equation for hydrolysis of sodium ethanoic is
$\ce{CH3COO- +H2O <=> CH3COOH + OH-}$.
$\ce{CH3COO-}$ is the conjugate base of $\ce{CH3COOH}$, therefore its $\mathrm pK_\mathrm b$ equals to $14 - \mathrm pK_\mathrm a$. You can then establish a equation for $K_\mathrm b$ using the concentration of the reactants and products.

How does this find the concentration of the acid?

@jerepierre [OH-] could be found from the pH value at the equivalence point and so does [CH3COOH]. Thus [CH3COO-] can be found using the equation for Kb.

I'm not sure I'm trying to determine the Ka. If I'm not mistaken, the Ka can be found from the titration curve at the half-equivalence point, right? My question is how the concentration of ethanoic acid in vinegar can be found given the Ka and other corresponding quantities that can be found on a titration curve.


text:
Cleaning stains with hydrogen peroxide and bleach
A while back, I discovered that if you have a stain on a white shirt, you can remove it by putting hydrogen peroxide on the stain, then adding some bleach to it.  I have used this for getting out wine, sweat, chocolate, etc.  The reaction produces gases and heat.  What is happening in this reaction and why does it clean stubborn stains?

Peroxide already is a "bleach", and you don't say what the other one is. The gas you see evolving is most likely oxygen (from the peroxide).
Generally (most importantly) bleach oxidises double bonds (which make the colour in many organic compounds). The remaining molecule is also more polar afterwards, because an oxygen-containing group was inserted, and therefore easier to wash out with water.

By bleach, I mean sodium hypochlorite.

Just wanted to add to this that if you see a greater effect from HOOH + hypochlorite than with hypochlorite alone, it is likely because HOOH is in fact a better oxidizer than almost everything else. The oxygen released by its decomposition is also a very strong oxidizer. In general, though, you should be very careful about mixing chemicals with strong oxidizers like HOOH/hypochlorite. They can and do produce toxic and extremely explosive compounds easily.


text:
Spectrophotometric determination of copper concentration
Context:
A typical lab experiment involves the addition of ammonia in excess to solutions of copper(II) to produce a complex ion whose absorbance at roughly 610 nm varies by the concentration of copper in solution.
Question:
Why is it less practical to simply test the aqueous solution of $Cu^{2+}$ rather than complexing it first with ammonia?
Thoughts:
My thoughts on this begin with the Beer-Lambert law: $$A =\epsilon lc$$
If the molar absorptivity of the ammonia complex in the visible spectrum around 600 nm is greater than that of the aqueous complex, [1] then small changes in concentration of copper can be differentiated by large changes in absorbance. This increases the sensitivity of spectrophotometric methods of determining copper(II) concentrations.
References
[1] http://chemed.chem.purdue.edu/demos/main_pages/18.8.html

If you have data that indicate the absorptivity is higher, you answer is correct.  But you might be missing the higher solubilization as a small component to this as well.


text:
Carbon/silicon pentacovalent intermediate during substitution
During SN2 reaction with a silicon compound, a pentacovalent intermediate is formed, while there is no such intermediate formed with carbon compounds (only a pentacovalent transition state). Why is it so? I need an MO explanation (no d orbitals involved).

Stable compounds of even 6-coordinated silicon are common. https://en.wikipedia.org/wiki/Hexafluorosilicic_acid

Bonds involving carbon are typically shorter than bonds involving silicon.  Therefore a pentacoordinated carbon species would be sterically more crowded around the central carbon atom than its pentacoordinated silicon analogue would be. This crowding would raise the energy of pentacoordinate carbon compared to pentacoordinate silicon; which is consistent with the observation that the SN2 reaction with carbon produces a higher energy pentacoordinate transition state, while in the case of silicon a more stable pentacoordinate intermediate can be observed.
\begin{array}{lcc}
\hline
\ce{Bond} & d(\ce{X=C})/\pu{pm} & d(\ce{X=Si})/\pu{pm} \\
\hline
\ce{X-O} & 143 & 163 \\
\ce{X-Cl} & 177 & 202 \\
\ce{X-Br} & 194 & 215 \\
\ce{X-I} & 214 & 243 \\
\hline
\end{array}
Recasting this into MO terms, we would say that both the carbon and silicon pentacoordinate species involve a trigonal bipyramid structure where the two ligands (the "attacking" and "leaving" groups in the carbon case) occupy axial positions in the trigonal bipyramid arrangement.

(source)
The figure above makes it easy to see that we have a hypercoordinated species involving a 3-center 4-electron bond. Since a silicon p-orbital is larger than a carbon p-orbital, the 4 electrons are spread out over a larger volume in the silicon case, reducing electron-electron repulsion.  This makes the pentacoordinate silicon case more stable than pentacoordinate carbon.

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text:
Molecular formula of a hydrocarbon

A is a non cyclic hydrocarbon chain with $\ce{C_{3x-1}H_{6x}}$ as a molecular formula which is mainly used in cars as fuel, its cracking leads to hydrocarbon (B) which belongs to the same family of hydrocarbon A, also two molecules of ethene form the second product in the above reaction.

My effort:
To determine if it is an alkane, we compare the given subscripts on $\ce{C}$ and $\ce{H}$ ($3x-1$ and $6x$, respectively) to those in the general formula for an alkane, $\ce{C_{n}H_{2n+2}}$. Let $n=3x-1$ where $n$ is the number of carbon atoms.
$$6x\buildrel{?}\over{=}2(3x-1)+2=6x$$
and therefore A is an alkane, with the molecular formula satisfying the form $\ce{C_{n}H_{2n+2}}$.

You're right in thinking that it's an alkane. Now I think I shouldn't blame you for getting confused by

$\ldots$ as a molecular formula which is mainly used in cars as fuel $\ldots$

Because

First off, we don't know if $x$ in the molecular formula is an integer or a decimal number.
Second, we don't know what fuel is meant. It's just supposed to be a fuel used in cars. And what are the fuels used in motor vehicles again?

Currently, the majority of motor vehicles worldwide are powered by gasoline or diesel. Other energy sources include ethanol, biodiesel, propane, compressed natural gas (CNG), electric batteries charged from an external source, and hydrogen.

What are these?

The bulk of a typical gasoline consists of $\mathrm{\color{red}{hydrocarbons~with~between~4~and~12~carbon~atoms~per~molecule}}$ (commonly referred to as C4-C12).

Petroleum-derived diesel is composed of about 75% saturated hydrocarbons (primarily paraffins including n, iso, and cycloparaffins), and 25% aromatic hydrocarbons (including naphthalenes and alkylbenzenes). The average chemical formula for common diesel fuel is $\color{red}{\ce{C12H23}}$, ranging approximately from $\ce{C10H20}$ to $\ce{C15H28}$.

Ethanol is an $\color{red}{alcohol}$.

Biodiesel refers to a vegetable oil - or animal fat-based diesel fuel $\mathrm{\color{red}{consisting~of~long-chain~alkyl~(methyl,~ethyl,~ or~ propyl)~ esters}}$.

Propane is an $\color{navy}{alkane}$.

CNG is made by compressing natural gas (which is mainly composed of $\color{red}{methane}$, $\ce{CH4}$), to less than 1 percent of the volume it occupies at standard atmospheric pressure.

Electric batteries and hydrogen are $\ldots$ a bit irrelevant. Let's not make this answer longer.

Thus, the answer would be propane, since it's the only alkane among the common car fuels. But is it? Of course not; since propane cracking isn't in high school syllabuses anywhere in the world. There is a very very good chance the querent had gasoline mind.
Now that we trust our natural instincts (!) that they meant gasoline when they were referring to the fuel, we should remain perplexed and run in circles. Because gasoline isn't consisted of one or two distinctive and well-known substances, and it's a mixture of  It is a mixture of paraffins (alkanes), cycloalkanes (naphthenes), and olefins (alkenes), where the usage of the terms paraffin and olefin is particular to the oil industry. To make matters worse, gasoline composition is obviously not the same around the world.

$\hspace{25ex}$
$\hspace{25ex}~\small{\mathrm{Some~of~the~compounds~commonly~found~in~gasoline}}$
$$\small{source}$$
So, let me rephrase your question into something actually meaningful:

$\ldots$ as a molecular formula which is mainly known to be a really smooth-burning species in gasoline $\ldots$

Now we're talking. The best known gasoline would be one that consists of 100% isooctane, and that's why we have octane rating. It's not a real fuel, just the material in an imaginarily ideal mixture that would burn "real smooth". That's why your question is problematic and scientifically incorrectly asked.
Back to the question, 2,2,4-Trimethylpentane must have been the compound the querent had in mind. Now we have the answer for A.
$\hspace{25ex}$
$\hspace{28ex}$ Isooctane (PIN: 2,2,4-Trimethylpentane)
$\hspace{43ex}$ $\small{source}$
But what would the products of the cracking be?
According to the Wikipedia article on cracking the cracking process in entropy-driven and the radicals formed in certain conditions usually move the reaction forward. But the final products (and even which radicals form) highly depend on the methods of cracking (thermal, steam etc.) and/or the catalysts. There are numerous catalysts which demonstrate the diversity in the mechanism of cracking.
For example, take a look at this paper (It's just an abstract, but it gives you an idea)

$\ldots$ The main reaction route on FAU is hydride transfer followed by β-scission leading to mainly C4 species, while on MFI protolytic scission is responsible for the formation of high amounts of C1−C3 species. $\ldots$

So, we're left with speculations. This isn't a real question. The only answer the examiners want is B's most probably butane, as it must be an alkane.
As this also demonstrates, a very simplistic view is that an octane can break into two ethene molecules and one butane molecule. So, the reaction they must have had in mind is
$$\ce{C8H18 -> C4H10 + 2C2H4}$$
Pfft.
TL;DR:
This isn't a good or correct question to ask at all; at the very least, it's very poorly worded. The answers are supposedly isooctane for A and butane for B, but the OP's confusion was innocent.

Footnote: Take a look at this as a good example of why we're rolling dices trying to predict the products of a cracking process with too less info.

Awesome answer -- I would note that $\ce{H2}$ as a fuel is getting noticed. High potential exists for hydrogen storage and fuel application using metal-organic frameworks, which constitutes the bulk of my current research.

"The best known gasoline would be one that consists of 100% isooctane," – Not really, there are various substances that have better knock resistance than isooctane so that octane numbers > 100 can be reached.

You have done the first step in the solution.
So we know now that A is an alkane and after cracking it leads to a molecule B and $\ce{2C2H4}$. But $\ce{2C2H4}$ contains $4$ carbon atoms so the number of carbon atoms in A must be $5$ or more (i.e. $3x-1\geq5 \Longrightarrow x\geq3$)
let us start trying:

$x=3$ will lead to $\ce{C8H18}$
$x=4$ will lead to $\ce{C11H24}$
.
.
.
.

And I suppose that you as a student only studied alkanes till octane ($\ce{C10H22}$)
so the answer will be $\ce{C8H18}$.
I know that is not a good reasoning but I am sure that it is the true answer.

" . . . as a student only studied alkanes till octane ($\ce{C10H22}$) . . . " That's decane, not octane.


text:
A thorough benchmark of DFT functionals without dispersion corrections
So, I'm looking for a benchmark of DFT functionals not including dispersion corrections. Something similar to Grimme's and coworkers GMTKN30 database, but for benchmark calculations done without dispersion corrections.
As in the case of GMTKN30 database, I'm interested in the same properties: general main group thermochemistry & kinetics. But

I already have calculations done many years ago without dispersion corrections since this feature simply was not implemented in most of the packages that time.
My workhorse is G3/G4 method which relies on geometry obtained without dispersion corrections anyway.
I quickly run few calculations including dispersion corrections and found no noticeable difference. Don't have time to redo all the work anyway.

I basically need to justify the choice of just a few DFT functional for my own benchmarks based on thorough literature reviews.

Maybe you could ask Prof. Grimme if he knows sth like that.

I would guess that Truhlar has some papers like this on DFT thermochemistry. Maybe in the M06 era?

@GeoffHutchison, good hint. The M06 paper + references there in seems to be the thing I'm looking for.

There is a benchmark of the GMTKN30 database with and without  the D3 dispersion correction. See Figure 4 here: "Effect of the damping function in dispersion corrected density functional theory"
However, your statement

I quickly run few calculations including dispersion corrections and found no noticeable difference.

is not supported by this.
Another example is this paper: "Improving intermolecular interactions in DFTB3 using extended polarization from chemical-potential equalization" where we compare PBE to PBE-D3 for 11 different datasets (see the SI). For many dispersion-dominated complexes, DFT (no D3) is qualitatively wrong, and in many other cases, the error is comparable to the mean interaction energy.
In general, I would strongly recommend against DFT without an empirical dispersion correction unless you are only focusing on electronic properties, but then you wouldn't be very interested in GMTKN30 anyway.
Since you say you are using G3/G4 you might also be interested in these recent benchmark papers:

"Benchmarking Compound Methods (CBS-QB3, CBS-APNO, G3, G4, W1BD) against the Active Thermochemical Tables: Formation Enthalpies of Radicals"
"Accurate reaction barrier heights of pericyclic reactions: Surprisingly large deviations for the CBS-QB3 composite method and their consequences in DFT benchmark studies"

Many thanks for the references! I know that today it it generally not recommended to do DFT without a dispersion correction, but as I said I have a huge pile of calcs done few years ago at G4 level. :( And what I mean by saying that there is no noticeable differences is that overall G4 numbers agrees with, say, M062x-D3. So, in other words I do not compare DFT against DTF-D3, rather G4 against DTF-D3. But I have to investigate few non-covalently bonded complexes more closely...


text:
Reduction of butadiene with Na/NH3 - stereoselectivity
Why is the major product of the reduction of butadiene with sodium in liquid ammonia cis-2-butene (60:40) when it is thermodynamically less stable than trans-2-butene?

This page discusses the same reaction (scheme 8) but it doesn't speak of the stereoselectivity. This paper seems to explain it in detail (i.e. the ratio is temperature-dependent) but is behind a paywall.:\

This question is from Ian Fleming - Molecular Orbitals, so I believe an MO explanation is desirable.

Page 96. question no. 6.

At first ... I don't like books that don't give proper solutions to their questions$^{*0}$. This doesn't help anyone ... or at least not me.
Old sources
I started my search on Reaxys and there are two old papers that I found when searching for $$\text{buta-1,3-diene} \xrightarrow{\ce{Na}} \text{but-2-ene}$$
They are:

K. Ziegler, F. Häffner, H. Grimm, Liebigs Ann. Chem. 1937, 528, 105
W. Hückel, H. Bretscheinder, Liebigs Ann. Chem. 1939, 540, 157

In both (german$^{*1}$) papers they work at very low temperatures around $\mathrm{-70\ldots-60~^\circ C}$. Both say that they get mainly the Z-isomer and discuss somehow why this happens. They think about their reaction as follows: $$\ce{dien + 2 M + 2R2NH -> dien.H2 + 2R2N.M}$$
They think that the butadiene is somehow pre-oriented towards the sodium surface from which the preference for the cis-isomer follows.
Old but not soooo old sources
Through the paper that M.A.Ramezani mentioned (W. Reeve, D. R. Kuroda, J. Org. Chem. 1980, 45 (12), 2305–2307), I found another paper (N. L. Bauld, J. Am. Chem. Soc. 1962, 84 (22), 4347–4348) that clarifies some of the mysteries. It is about the stereochemistry of metal addition to conjugated systems, esp. dienes. One of the first things that is shown is the following table.

$$\small\begin{array}{ccc}
\hline
\text{Diene} & \text{Temp., }\mathrm{^\circ C} & \text{% cis-Olefin}\\
\hline
\text{Butadiene} & -33 & 13\\
\text{Butadiene} & -78 & 50\\
\text{1,3-Pentadiene} & -33,\,-78 & 68\\
\hline
\end{array}$$

It contains the cis-yields from v.p.c. analyses for two different temperatures for the reduction of butadiene with sodium in liquid ammonia. As one can see, there is less Z-but-2-ene produced when the temperature is higher$^{*2}$ (13 %) and more when the temperature is very low (50 %).
What the author also mentions is, that there is in every case more Z-product produced than one would expect, because there exists only about 3-7 % of Z-butadiene at room temperature (which must be even lower at lower temperatures). So somehow the metal is included, wich is described by Bauld with a cyclic complex of sodium and the radical anion that is supported by the experiments of Reeve and Kuroda.

As there are many possibilities how the sodium ion "binds" to E- and Z-form$^{*3}$ and there is also the possibility that there are two (or maybe none?) sodium ions involved, this is far off from the scope of this answer.
Besides that, I roughly calculated the structures from the mechanism that is schown in Scheme 8 from the other link by M.A.Ramezani --which, as I found out in this moment, is also from Prof. N. L. Bauld--. Below are the corresponding $\pi$-MO's$^{*4}$ -- you can imagine the mechanism with involved sodium ion(s) to be similar but somehow with the above cyclic intermediate.

TL,DR
The question in Flemings book lacks some informations. The reaction in liquid ammonia does not only depend on temperature (which was not stated!) but also on the rests that are applied to the diene. For Buta-1,3-diene in $\ce{Na/NH3}$, the Z-isomer is favored when cool temperatures are applied and therefore yield the kinetic product of the reaction. This follows from a cyclic radical anion intermediate with the sodium ion. But when the reaction temperature is rised, the thermodynamically preferred E-product is favored, as is shown by Bauld (see the table).
To get to know more about what happens when different methyl and ethyl groups are added, read the paper by Reeve & Kuroda.

$^{*0}$ Dare you, Szabo and Ostlund ಠ_ಠ
$^{*1}$ Old german papers are a joy to read ... usually the older the better.
$^{*2}$ $\mathrm{-33~^\circ C}$ is the boiling point of ammonia
$^{*3}$ In my case it seems to be purely vdW-interactions but it is gas-phase, this is a small basis set and even if the functional fits there was no dispersion correction.
$^{*4}$ be aware ... those are Kohn-Sham orbitals

Thank you for the comprehend answer and I totally agree with your first statement.


text:
Gibbs energy and chemical equilibrium : false proof of spontaneous generation, but why?
Let’s assume we have a reaction medium containing only simple molecules such as $\ce{H2O}$, $\ce{CH4}$, $\ce{NH4+}$ (and maybe some more).
Now we consider the following equation ;

$\Delta G_\mathrm r = \Delta G_\mathrm r^{\circ} + RT \ln Q_\mathrm r$

were $\Delta G_\mathrm r$ is the Gibbs energy change associated to a chemical reaction occurring in this reaction medium, $\Delta G_\mathrm r^{\circ}$ the standard Gibbs energy change of this reaction and $Q_\mathrm r$ the mass action ratio of the reaction in the reaction medium at a given time.
Let’s consider the reaction consisting of the synthesis of a very complex molecule from the simple molecules of the reaction medium (for example, a protein). Whatever the complex molecule is, it is so complex that we are sure that it is absent from the reaction medium, i.e. its concentration is exactly 0. Also, the standard Gibbs energy change of this reaction must be very high (positive), but not infinite.
If the concentration of the complex molecule starts at 0; then $Q_\mathrm r$ is 0; then $RT \ln Q_\mathrm r$ is $-\infty$ ; as $\Delta G_\mathrm r^{\circ}$ is finite, $\Delta G_\mathrm r$ is $-\infty$. So in these conditions, the synthesis of at least one copy of our very complex molecule is infinitely exergonic. This reasoning could be applied to any molecule whose concentration in the reaction medium is 0 at a given time and whose synthesis from the reaction medium is theoretically possible. In the thermodynamics courses I attended, a negative Gibbs energy change was associated with the “favourableness” of a reaction, so those reactions would be “infinitely favourable”.
Of course, this is absurd. But why?

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The equation you use is only valid under equilibrium conditions. So, at the start of the reaction when there is no product present in the medium, the equation you quoted does not apply. Only after equilibrium has been established you can use it. And in an equilibrium there is always at least some tiny amount of product in the mixture otherwise there wouldn't be a reaction. Thus, no infinite $\Delta G$ can arise.

Within the context of the equation in question, $\Delta G$ refers to the change in free energy between two specific thermodynamic equilibrium states:
State 1:  Stoichiometric quantities of pure reactants in separate containers at temperature T and individual specified pressures
State 2:  Corresponding stoichiometric quantities of pure products in separate containers at temperature T and individual specified pressures
In the case of $\Delta G^0$, the specified pressures are all 1 atmosphere.  This means that Q = 1, and $\Delta G =\Delta G^0$.
If the specified pressures in states 1 and 2 happen to correspond to the partial pressures of an equilibrium mixture of reactants and products, $\Delta G$ will be equal to zero.  In this case, if one had an equilibrium mixture of the reactants and products, the partial pressures in the equilibrium  mixture could also be held in equilibrium through individual semipermeable membranes with the gases in the separate containers of reactants and products (with no tendency for the reactants or products to enter or leave the reaction mixture through the semipermeable membranes).


text:
Is methoxide a better nucleophile than hydroxide?
What is more nucleophilic, $\ce{OH-}$ or $\ce{CH3O-}$?
When I look it from the viewpoint of steric hindrance, the answer seems to be $\ce{OH-}$, but when I see it from the viewpoint of extra electron density provided by hyperconjugation of $\ce{CH3}$, the answer seems to be $\ce{CH3O-}$.

Related: How to determine order of nucleophilicity among commonly used nucleophiles?

I have seen all of that. All say that a nucleophile is good if it has more charge and is less bulky. But both of those factors are competing against each other here.

Alkoxides tend to be stronger bases than the hydroxide ion. Adjacent atoms in an alkane chain will donate electron density to oxygen, allowing it to be more polarizable. An increase in polarizability will allow oxygen to more easily donate electrons.
The oxygen in a hydroxide ion does not experience the inductive effect. Therefore its valence electrons do not "reach" out as far.
Methoxide is a relatively unhindered base, as is hydroxide. But as explained above, methoxide experiences induction whereas hydroxide does not. So one would expect methoxide to be a better nucleophile than hydroxide because it is a stronger base than hydroxide and still remains unhindered.

Stronger base is not automatically the same as stronger nucleophile.  Be careful.

I love that you used the word polarizable, this answer kind of ties up my question too.

Methanol is more acidic than water. So shouldn't methoxide be a weaker base?

When negative charge is present on same type of atom in different nucleophiles, the order of nucleophilicity is determined by taking the strength of their conjugate acids.  If the acid is strong its nucleophilicity is less.  $\ce{CH3OH}$ is a slightly more acidic ($\pu{pKa = 15.5}$) than $\ce{H2O}$ ($\pu{pka = 15.74}$).  Therefore hydroxide ion is slightly more nucleophilicity than methoxide ion

My answer would be same as the above mentioned.  I would just add that of the reduction in nucleophilic character of methoxide due to steric factors is 10 then an increase due to inductive factor is 40 which makes a net +30 better than hydroxide.
Note- above values have no physical significance they are just made up to explain the comparative difference.

10 what? As you said, these numbers are meaningless and come from nowhere - so they are not useful as an explanation.


text:
Can the steric nature of a carbon be quantified?
With respect to this question as suggested by a user (he asked me to ask this as a separate question), I wanted to know if there is any way in which steric hindarance of an organic carbon be quantified, as to a value that would indicate how bulky a compound is.

This has nothing to do with biochemistry. Please don't mistag on purpose.

Can you be more specific about what you mean by bulky? Are you referring to the steric bulk of different functional groups or some sort of measure of bulk of a molecule as a whole?

Bulky as in more bulky a compound more likely it is to react

@Rishika I'm afraid that doesn't make much sense. Generally steric bulk hinders a reaction by reducing the rate.

okay so steric bulk hindrance is a quantitative property?

Energy density or charge density quantify this, in some regard.

@khaverim can you please elaborate? in some regard as in?

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text:
Experimental melting-point of Arm-and-Hammer sodium bicarbonate is remarkably high(?)
Online literature reports the melting point of $\ce{NaHCO3}$ as ~50 deg. C.
Using a standard Mel-Temp II melting-point apparatus and a standard alcohol thermometer, I didn't observe any melting of any sample lower than 150 deg. C. I don't think hydration of the salt can even explain this. Is there some other material in commercial baking soda that would alter the melting point this much?
As a raw calibration, I found that some Bayer Aspirin powder (possibly containing cellulose or other binding agent) had a melting point of 120 deg. C., a good deal below the reported ~135 deg. C.
Does anyone know what may be present in commercial Arm-and-Hammer baking soda besides sodium bicarbonate? The box reports 100% purity but I have trouble believing it now.
I've heard rumors that some baking-soda products contain aluminum and this has raised some alarm about ingestion hazards. Is this valid?

See the MSDS for Arm and Hammer Baking Soda here, from the manufacturer (Church & Dwight Co.) (it will link to a PDF) in which it is stated:

CONDITIONS TO AVOID: Temperatures above 65°C (150°F).
INCOMPATIBILITY WITH OTHER MATERIALS: Reacts with acids to yield carbon dioxide. May also yield free caustic in presence of lime dust (CaO) and moisture (i.e., water, perspiration). Dangerous reaction with monoammonium phosphate or a sodium-potassium alloy may occur.
HAZARDOUS DECOMPOSITION PRODUCTS: Heating above 100°C may cause dangerous levels of carbon dioxide gas to be present in confined spaces. Yields sodium oxide if exposed to temperatures above 850°C. Avoid inhalation, eye and skin contact with sodium oxide.

The ingredients are listed as 100% sodium bicarbonate with CAS number 144-55-8.
Your choice of thermometer is questionable in the present case. See this Wikipedia entry which reliably reports that the upper limit on temperature measurement from an alcohol-based thermometer is 78 $^{\circ}$C. A mercury thermometer has a higher upper temperature range (see this Wikipedia article) and is preferable to the alcohol thermometer. Alternatively, use a gallium thermometer or a digital model with an operating temperature range that suits your needs.
You ask:

I've heard rumors that some baking-soda products contain aluminum and this has raised some alarm about ingestion hazards. Is this valid?

Probably not in the case of the specific brand in question here, based on the MSDS.

It may not be an alcohol thermometer, but it is accurate up to 150 deg C. Its the typical red-fluid kind. Thanks for your response. Regardless, the samples of baking soda I had simply did not melt at >150 deg. C.

It has much less to do with the thermometer than the fact that I heat the sample substantially and it doesn't actually melt.

Interesting - maybe you're rapidly driving off the $\ce{CO2}$ and have something that's not exactly baking soda leftover, which exhibits the high melting point.

Found this. $\ce{2 NaHCO3 → Na2CO3 + H2O + CO2}$, upon heating >50 deg. C

Online literature reports the melting point of $\ce{NaHCO3}$ as ~50 deg. C.

No!!!
On-line literature reports that $\ce{NaHCO3}$ decomposes at ~50 deg. C. to $\ce{NaCO3} + \ce{H2O}$. Both are solids. So under magnification you'd see the grains change shape and some water droplettes above, but you'd still have a white solid, not a clear melt.

authors: غيث الزيتاوي


text:
Can a high-spin complex be diamagnetic?
There are two nickel complexes: one is octahedral and the other square planar. One complex is diamagnetic, the other is paramagnetic. Both are high-spin.
Now, $\ce{Ni^2+}$ that is octahedral should be paramagnetic, as should $\ce{Ni^2+}$ that is square planar. Actually any high-spin complex should be paramagnetic. Is it possible that the octahedral $\ce{Ni^2+}$ is paramagnetic, and square planar $\ce{Ni^2+}$ diamagnetic (achieved by not filling the last orbital even tho that goes against high-spin concept)?

It is highly unlikely for square planar complex to be high-spin, otherwise the benefit of square planar coordination is too low. That said, addressing your title question: it is possible for metal in high-spin state to form covalent bonds, like in chromium diacetate molecule, forming diamagnetic molecule.

What would be the spin state of the low-spin complex if the high spin is S=0?

A high-spin complex cannot be diamagnetic. By definition, a high-spin system has multiple unpaired electrons. The names "high-spin" and "low-spin" refer to the total electron spin. In the following example ($d^6$ octahedral), the high spin state $S = 2$ and the low spin state $S = 0$.

In an electron configuration such as $d^8$ the outcome is independent of whether electrons were filled by pairing in the $t_{2g}$ orbitals first, or by filling all orbitals and pairing last. You end up with the same configuration. In this case, there is no such thing "high-spin" or "low-spin" states.

The same is true for $d^0$, $d^1$, $d^2$, $d^3$, $d^7$, $d^9$ and $d^{10}$. Octahedral is really the only geometry with both high- and low-spin states, as this is the only geometry where the splitting energy $\Delta$ is comparable to the spin-pairing energy.

Both are high-spin.

No, square planar complexes are nearly always low-spin. With a $\mathrm{d^8}$ configuration, the lowest four orbitals would be filled. That would make the complex diamagnetic.
For the octahedral $\mathrm{d^8}$ complex, regardless of whether it's high- or low-spin, it will have the same electronic configuration $\mathrm{(t_{2g})^6 (e_g)^2}$. By Hund's first rule, the two electrons in the $\mathrm{e_g}$ orbital will both be unpaired, so the complex will be paramagnetic.
Here's a quick sketch (energy levels not to scale):


text:
Using helium to make home brewed beer
I want to brew beer using helium instead of $\ce{CO2}$.
Most beers use about 2.5 grams of $\ce{CO2}$ per litre except stout like Guinness which uses nitrogen as it allows bubbles for much longer hence a frothy head on the stout.
So, perhaps a blend of helium and nitrogen would work? Helium is more soluble and if used in place of $\ce{CO2}$, as soon as the seal broke on a can it would escape rendering the cartoon voice idea useless.
How much helium can I squeeze into a litre of beer with or without compounding it to get the chipmunk voice?

Since you are not inhaling your beer, I don't think you'll get a cartoony voice.

Oh I know some who seem to inhale their beer

Might be better on http://homebrew.stackexchange.com/  Still don't get why you want to use helium

That would be a waste of good helium!

CO2 is needed to draw beer from a tap on the counter, but not for brewing. Brewing makes carbon dioxide!

To correct one point in the question, helium has the lowest water solubility of all gases. At one atmosphere and $\mathrm{25^oC}$, $\ce{CO2}$ is 100 times more soluble than $\ce{He}$ and $\ce{N2}$ is about twice as soluble as $\ce{He}$.  This is why deep-divers prefer helium to nitrogen.
At 45 psi, which is 3.1 atmospheres pressure, you could dissolve about 0.005 g $\ce{He}$ or roughly 30 mL when fully degassed.  A typical adult breath is around 500 mL, and of course the $\ce{He}$ will not degas instantly.
The bottom line is that pressurizing beer with helium will not make you talk like Donald Duck.  And the biggest difference you would see between $\ce{N2}$ and $\ce{He}$ would be in your bank statement.


text:
Exceptions to the Madelung rule for electronic configurations
The electronic configuration of platinum is
$$\mathrm{[Xe] 4f^{14} 5d^9 6s^1}$$
and not $\mathrm{[Xe] 4f^{14} 5d^{10} 6s^0}$ or $\mathrm{[Xe] 4f^{14} 5d^8 6s^2}$.
I see this question has already been answered, but here is what Wikipedia says:

Other exceptions to Madelung's rule:
  For example, in the transition metals, the 4s orbital is of a higher energy than the $\ce{3d}$ orbitals; and in the lanthanides, the $\ce{6s}$ is higher than the $\ce{4f}$ and $\ce{5d}$.

I don't understand the reason behind this.
Shouldn't it be that the $\ce{4s}$ orbital is of a lower energy than the $\ce{3d}$ orbitals (that's why it is filled first) and in the lanthanides, the $\ce{6s}$ is lower than the $\ce{4f}$ and $\ce{5d}$?
It also states that in heavier elements the inner electrons move with a speed approaching the speed of light. So why don't we consider the great amount of kinetic energy associated with the electron? If we consider this energy then the inner orbitals should not be filled first owing to their great amount of energy?
There is a list of exceptions arising from electronic configurations of lanthanides and transition series.
The exceptions in the transition series are explained by the fact that half and fully filled orbitals are more stable (although my text doesn't state why they are stable)?
So similarly do we have a rule that governs the electronic configuration of lanthanides? I believe for someone who is in his first year Hartree-Fock calculations are not a good thing to start with. So do I need to just memorize them all?

A lot of questions in here.

It also states that in heavier elements the inner electrons move with a speed approaching the speed of light. So why don't we consider the great amount of kinetic energy associated with the electron? If we consider this energy then the inner orbitals should not be filled first owing to their great amount of energy?

In the Schrödinger Equation, $\hat H\Psi = E\Psi$, the Hamilton operator $\hat H$ is generally considered to consist of a kinetic term $\hat T$ and a potential term $\hat V$. In general, $\hat T$ is defined via an electron’s impulse $\vec p$ rather than its velocity $\vec v$, meaning that we end up with a term $$\hat T = -\frac{\hbar^2}{2m}\nabla^2$$
Within this operator, the kinetic energy of an electron is considered and since $\hat T$ forms part of $\hat H$, we use an electron’s kinetic energy in the original Schrödinger equation and thus it is considered in the calculations.
Using relativistic quantum mechanics will complicate the calculations but the general idea will remain the same.

The exceptions in the transition series are explained by the fact that half and fully filled orbitals are more stable (although my text doesn't state why they are stable)?

That’s … overly simplistic. In reality, it is a series of considerations we could make at the drawing board than when calculated leads to one electronic configuration being preferred over another. However, this question already covers that part.

Other exceptions to Madelung's rule:
For example, in the transition metals, the 4s orbital is of a higher energy than the 3d orbitals; and in the lanthanides, the 6s is higher than the 4f and 5d.

I don't understand the reason behind this.
Shouldn't it be that the 4s orbital is of a lower energy than the 3d orbitals (that's why it is filled first) and in the lanthanides, the 6s is lower than the 4f and 5d?

While your sentence is correct according to the aufbau principle or Madelung’s rule, this is no longer true when it gets to everything that’s not a neutral atom. Indeed, as soon as you turn scandium into a scandium(I) ion, the 4s orbital becomes less favourable than the 3d orbitals because the latter are still ‘closer to the nucleus’ and experience more stabilisation from an overall net positive charge. In general, once you go from neutral atoms to cations, the aufbau principle loses its relevance and a more ‘logical’ 3d-4s-4p order is favourable.

So similarly do we have a rule that governs the electronic configuration of lanthanides? I believe for someone who is in his first year Hartree-Fock calculations are not a good thing to start with. So do I need to just memorize them all?

Lanthanides tend to be rather subtle in their electronic effects if I recall my lecture correctly; as far as I am aware there is no explicit rule of any kind for them. But as for memorising: yes, you would have to do that.


text:
What is the difference between wet paint, dry paint, and in between those stages?
A day after painting a closet, I set a piece of wood on the wall of the closet that was painted a day ago as well. Within minutes, both of their layers of paint fused together (please correct me if there is a better term than 'fused').
I'm assuming that the paint approaches a state of being a solid, but the process is never stopping, (as I've heard:) similar to how concrete is much stronger thousands of years after it is made rather than a couple of years.
So I understand that it started off as a liquid with a high viscosity, and liquids with high viscosity's turn into a solid when spread thin, but why does that even happen. As well, was I right when I explained that it is constantly 'drying', explaining why it was able to fuse with the other layer of the same paint even a day later?

Ullmann’s Encyclopedia of Industrial Chemistry lists numerous components of commercial paints:

dyes, pigments, extenders
organic solvents, water, coalescing agents
binders
plasticizers
paint additives (levelling agents, film-formation promoters, wetting agents, dispersants, antisettling agents, antifoaming agents, catalysts, antifloating and antiflooding agents, antiskinning agents, matting agents, neutralizing agents, thickening agents, preservatives, corrosion inhibitors)

The application method, drying and hardening behaviour of a paint is essentially determined by the binder. Binders are macromolecular products. The low molecular mass binders include alkyd resins, phenolic resins, polyisocyanates, and epoxy resins. These binders must be chemically hardened after application to produce high molecular mass cross-linked macromolecules in order to form acceptable films. The higher molcular mass binders include cellulose nitrate, polyacrylate, and vinyl chloride copolymers, which are readily suitable for film formation.
As the paint dries, a firmly bonded film is formed. Drying and film formation take place physically and chemically. Physical drying involves evaporation of the solvents (organic solvents and/or water). This is the main process for paints with high molecular mass polymer binders. Chemically drying paints contain binder components with a relatively low molecular mass that react together on drying to form cross-linked macromolecules. This process can occur by polymerization, polyaddition, or polycondensation.
In practice, physical drying and the various mechanisms of chemical drying can proceed concurrently or consecutively, depending on the composition of the binder system.


text:
How do I find the expiration time of drug in the following example?
An optically active drug has one chiral center and only dextrorotatory isomer is effective. Moreover, it becomes ineffective when original activity is reduced to 35 % of original. The rate constant is  $1 \cdot 10^{-8}\ \mathrm s^{-1}$. Find the expiration time of drug in years.
I considered the reaction as $$\ce{A -> B}$$ $$a-x \rightarrow x $$
so the, $\left(a-2x\right)r = 0.35$ and then I got $x$ and then I used
$$ t=\frac{2.303}{k}\log\frac{a}{a-x} $$ I got answer 3.65 years. But the answer given is 1.37 years.

Is my answer right ?

could you post your calculation? did you turn seconds to years right?

Your general approach to the calculation looks great.  My only question is, what is the meaning of $x$ in your calculation?
Here's how I would do the problem.
$$\ce{A->B}$$
$$A = A_0 e^{-kt}$$
$$\frac{A}{A_0} = e^{-kt}$$
$$0.35 = e^{-(~{\mathrm 1 \times 10^{-8} {\mathrm s^{-1}}})t}$$
$$ \ln{0.35} = -(~{\mathrm 1 \times 10^{-8} {\mathrm s^{-1}}})t$$
$$ t = \frac{-\ln{0.35}}{(~{\mathrm 1 \times 10^{-8} {\mathrm s^{-1}}})}$$
If I did the units conversion from $\mathrm{s^{-1}}$ right, then the rate constant in units of $\mathrm{{yr}^{-1}}$ is 0.316.  (I just multiplied by 60×60×24×365.25).
$$ t = \frac{-\ln{0.35}}{(~{\mathrm 0.316~{\mathrm {yr}^{-1}}})}$$
When I plug in the numbers, I get 3.2 years.  It's possible I did something wrong, but my answer is way closer to yours than to the given answer.
Here are some possible mistakes that the original writer of the question could have made to give the wrong, lower, answer.

The question has a bunch of superfluous information about the chirality of the drug.  Mistakenly thinking that this is important to the question, perhaps somebody divided by 2 to "adjust" for chirality.  But this is not the correct approach.  We're given the activity of the drug and the decay rate of activity, so the structure of the molecule is completely irrelevant to the problem.
Another possible error is misinterpreting "is reduced to 35 % of original" mistakenly as "is reduced by 35%".  That leads to the calculation for $t$ being $\frac{-\ln{0.65}}{0.316}$ and gives the expected (but wrong) answer.

The bottom line is I think you mostly did the calculation correctly and the given answer seems incorrect, at least given the wording of the question as you supplied it.


text:
Which alkene has a higher Cahn-Ingold-Prelog priority in (2Z,5E)-hepta-2,5-dien-4-ol?
How do I determine whether the central carbon of this compound is (R) or (S)?

On both the sides of carbon, the groups are same. They differ only in the relative position of hydrogen atoms. On what basis do I assign priority in these groups?
Does the R/S nomenclature for the middle carbon really depend on E/Z nomenclature of the alkenes, in such cases of stereoisomerism? Or is there an entirely different argument in such a case?

The other way round, but related: How to assign E/Z configuration according to the Cahn-Ingold-Prelog rules when subsituents differ only by stereochemistry

cis > trans

In order to arrive at an unambiguous description of stereoisomers, the priority for the groups attached to the chirality center is established by the application of ‘Sequence Rules’. In the hierarchical Sequence Rules given in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), we find

P-92.1.3.3 Sequence Rule 3
When considering double bonds and planar tetraligand atoms ‘seqcis’ = ‘Z’ precedes ‘seqtrans’ = ‘E’ and this precedes nonstereogenic double bonds.

In the given example, however, the stereochemical configuration at the central carbon atom is not defined. Note that solid wedges should be used to indicate bonds that project above the plane of the paper and hashed wedges to indicate bonds that project below the plane of the paper.

Maybe it's a Fischer projection.

According to IUPAC Recommendations (2006): “The use of Fischer projections in non-carbohydrates is problematic. Any drawings of this type are necessarily ambiguous since it is unclear whether the author of the drawing intended to make a Fischer projection or simply happened to draw a diagram in that orientation. (…) Accordingly, (…), similar depictions of non-carbohydrates should be used only when the viewer of the diagram is certain to understand that the diagram was intended to represent a Fischer projection. Fischer projections should be avoided where ambiguity is possible, (…)”


text:
Approximating strength and hardness of alloy
I do not have any formal education in metallurgy, but I am trying to make a game mechanic based on it.
I need to be able to alloy together different metals and alloyable materials such as carbon in different percentages, and get an approximation of the strength and hardness of the resulting alloy. If anyone has any ideas how what I would need to do to calculate such a thing I would be very appreciative.

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Looks too broad - big part of metallurgy - even about steel there are lots of books.

Hmm, @Mith I think the answer could boil down to introducing a general formula $\ldots$ and it doesn't need to be something specific. Are you sure there's no way for a short answer on this?

@inɒzɘmɒЯ.A.M Looks like Todd managed to do this generally enough ;)

Look at the free resource provided at MatWeb.
Under the Physical Properties tab, you can choose up to 3 of them from a rather extensive list, including hardness. For example, I chose "Carbon" and "Hardness, Rockwell R," and 25 results were returned.
Under the Alloy Composition tab, you can choose up to 3 elements and percentages for your alloy, and see if you get a hit. I chose "Nonferrous Metal" and carbon (minimum 5%, maximum 75%) and silicon (minimum 5%, maximum 75%), and got one hit.
I believe for your purposes, you can assemble a list of materials by hardness and strength (I'm not a metallurgist either, so you might have to do some research on which of the hardness and strength metrics you'll need to choose) and/or by selecting by composition (again, you might need to research which 2- and 3-element combinations make sense and which ones don't).


text:
Electroplating reaction
In electroplating, can a galvanic cell substitute the direct current supply? The process is to change iron metal to a silver one, but a galvanic cell is connected to the two electrodes.

Chroumium metal is involved in the galvanic cell

Anything with an electrical current, whether from a galvanic cell or battery etc., is fine for electroplating. It would help if you were more specific.


text:
Why does SO2 have a larger bond angle than H2O, H2S, and NH3
So we have to find which of the follwing compounds has the smallest bond angle:
$\ce{H2O}$, $\ce{H2S}$, $\ce{NH3}$, $\ce{SO2}$.
So L $\propto \frac{1}{BA}$
where $L$ is the number of lone pairs and $BA$ is bond angle
So the number of lone pairs is:-
$\ce{H2O}$: 2
$\ce{H2S}$: 2
$\ce{NH3}$: 1
$\ce{SO2}$: 2
Since $\ce{NH3}$ has the fewest lone pairs, it must have the highest bond angle, so it is out.
Also $BA \propto ENC$ where $ENC$ is the electronegativity of the central atom.
The electronegativiy of the O atom in $\ce{H2O}$ is the highest, so it is out.
Also $BA \propto \frac{1}{ENS}$ where $ENS$ is the electronegativity of the surrounding atom. The electronegativiy of the H atom in $\ce{H2S}$ is the lowest of the remaining two. The only one left is $\ce{SO2}$ which must have the lowest bond angle but when I checked on google it had the highest bond angle. How is that possible?

Here are the bond angles for each molecule (data from wikipedia):
\begin{array}{|c|c|}\hline
\mathrm{Molecule} & \mathrm{Bond \space Angle \space (^\circ)} \\ \hline
\ce{H2S} & 92.1 \\ \hline
\ce{H2O} & 104.5 \\ \hline
\ce{NH3} & 107.8 \\ \hline
\ce{SO2} & 119 \\ \hline
\end{array}

So $L \propto \frac{1}{BA}$ where $L$ is the number of lone pairs and $BA$ is bond angle.

This is true but only in very specific situations; when dealing with molecules that have a central atom in the same period and outer atoms of the same element (e.g $\ce{CH4}$, $\ce{NH3}$, $\ce{H2O}$). It breaks down as soon as you start comparing molecules with central atoms from different periods (e.g $\ce{PH3}$ has fewer lone pairs than $\ce{H2O}$ but a smaller bond angle) or when you compare molecules with different outer atoms (e.g $\ce{NF3}$ has fewer lone pairs than $\ce{H2O}$ but a smaller bond angle). For the reasons behind this I direct you to an excellent previous answer from @ron.
Additionally, I think you need to reconsider the number of lone pairs in $\ce{SO2}$, which can be described by these two resonance structures.

As you can see there is only one lone pair, but unfortunately this doesn't help us very much as we have no other molecule to compare it to.

Also $BA \propto ENC$ where $ENC$ is the electronegativity of the central atom.

This is incorrect. In fact the reverse is true - that $BA \propto \frac{1}{ENC}$ - but only in the same situations as mentioned before. In this case the trend is only really pronounced in the second period; it is very slight in the third period and virtually non existent in the fourth. In fact, this 'rule' doesn't really have anything to do with electronegativity (that's just a coincidence) and it's essentially just a result of the first rule.

Also $BA \propto \frac{1}{ENS}$ where $ENS$ is the electronegativity of the surrounding atom.

This is sort of true - it's known as Bent's rule and it can be very useful but it's not really applicable here. It's been discussed many times on this site but here and here are some good introductions.
So how should you answer this question?
The first thing to note is that $\ce{SO2}$ only has three 'groups' on the central atom (sometimes called 'effective electron pairs' in VSEPR theory) - two bonds intermediate between a double and a single bond and a lone pair - whereas all the other molecules have three. Therefore we expect $\ce{SO2}$ to have the largest bond angle of the four molecules, and this is indeed the case. $\ce{H2O}$ and $\ce{NH3}$ are hydrides of the same period so we can use the first rule to determine that $\ce{H2O}$ has a smaller bond angle. Now we just have to decide whether $\ce{H2O}$ or $\ce{H2S}$ has a smaller bond angle. We can apply the hybridisation arguments given by @ron in the answer I linked earlier to determine that $\ce{H2S}$ has the smallest bond angle, and indeed we find that it is almost unhybridised with a bond angle very close to $\mathrm{90~^\circ}$.

"$BA \propto ENC$ where $ENC$ is the electronegativity of the central atom": What probably the OP was referring to is the trend of decreasing bond angle in similar compounds of the elements of p-block down a group, for example, from ammonia to stibine and from water to hydrogen selenide.


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Why are alkanes also called paraffins?
Why are alkanes also called as paraffins?   Is it because they are mainly found from petroleum and natural gas?

I don't have concrete proof for this, but as my teacher said it is because it has a wax like texture.

The origin of the word is historical, and has to do with the observed properties of the class of hydrocarbons we know refer to as alkanes. The name originated in the Latin for "little affinity."
From M. P. Crosland's text Historical Studies in the Language of Chemistry:

The term paraffin was coined by Reichenbach (from Lat. parum + affinis = little affinity) to denote a particular substance, a mixture of hydrocarbons (Journal fur Chemie und Physik, 59 (1830), 455). The term was extended to cover any hydrocarbons of the series $\ce{C_{\rm n}H_{\rm 2n + 2}}$ by Watts (Dictionary of Chemistry, 1868, vol. iv, p. 344 and Supplement, 1872, p. 705).

Reichenbach, referred to above, is Carl Reichenbach, a multifaceted scientist who did pioneering work in isolating (and thus naming) hydrocarbons in the early 1830's.

As already noted in Todd Minehardt's answer, the name "paraffin" is based on the Latin words parum + affinis = "little affinity".
The "little affinity" behind their Latin name referring to their relatively low reactivity. They have "little affinity" for other elements or compounds and a consideration of their chemical structure will make you see why.

Well, it's not a great answer, and it's certainly not an expansion of anything. But it's not VLQ IMO.

I can see semantics is more important than assistance here. This was a simple development of an answer already given, using facts that perhaps should have been included in the first place for clarity. The questioner must be perplexed at these developments and I - as a very experienced educationalist who thought your site was an excellent idea and opportunity to help - certainly am. Rest assured I shall not trouble you or the site further.

Petrol is a word which refers a combination of all hydrocarbons in that major part is paraffins (alkanes)(as the chemical structure shows that all the single bonded joins are called as alkanes). So paraffins are also called as alkanes. But the fact tell us (any of the series of saturated hydrocarbons including methane, ethane, propane, and higher members.)

Since alkanes are saturated, their reactivity seems to be less, hence the name parafin due to less affinity.

Paraffins means little reactive and alkanes are little rective due to presence of single bond. Therefore  they are also called paraffins

A rehash offering nothing novel to the answers already present.


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Safety/health hazards: working with 2-aminobenzimidazole, iminoesters, and 1,3,5-triazine
How dangerous is it working with 2-aminobenzimidazole, iminoesters and 1,3,5-triazine? Has there been any cases of health problems recorded concerning those three compounds?

You should edit your question immediately - eliminate personal info and state problem more precisely.

I wanted to do it myself but wasn't sure how to state it. IMO, if promoter gives you such work, you shouldn't worry too much - you need to be cautious in lab on daily basis and probably worked with more dangerous stuff not even knowing about it.

Your previous revision stated that you studied some stuff about their hazard $\ldots$ could you provide a link to what you studied?

Uhm. This is really really weird. You want to quit your chemistry study because you worry about lab safety? Laboratories are generally safer than freeways and should pose no hazard when being properly guided, which is in principle what a university is for. I'm not sure whether I believe the story of yours. It's also none of my business. Just saying that we are not here to help you synthesize your own pharmaceuticals...

@user19185 - If you are very concerned about how to proceed if you believe your research has become hazardous to your health, you could ask a question at Academia. I would avoid personal details and the specifics of your research so that the question could be general enough to be useful. Also, be clear that you are not looking for legal or medical advise. The internet is a terrible place to find either.

Looking up the chemical you're working with is always a good idea, but you need to look closer and evaluate the specific danger those chemicals present to you. What you want to look at are the danger symbols and the hazard and precautionary statements. Those will tell you the specific dangers a chemical presents.
If you look up your first substance, 2-aminobenzimidazole, it looks pretty tame for organic chemistry standards. It has an exclamation mark as danger symbol, which doesn't tell you much, but the absence of a skull and crossbones and a health hazard symbol tell you that this probably won't kill you immediately. This doesn't mean the stuff is harmless, but it is also not extremely toxic. If you look at the hazard statements of this chemical you'll find that you shouldn't get it into your eyes, on your skin or breathe it. All of that is a good idea when working with any chemical, and this is exactly what you should learn when studying chemistry. This is a really tame chemical compared to a lot of stuff organic chemists work regularly with, but you still should avoid exposing yourself to it.
Your second item is an entire category of chemicals, I'll skip it for this reason. The third one is a bit more dangerous than the first one, it has a corrosive and a health hazard danger symbol. You want to avoid contact with it, but that's what all your protective equipment like safety goggles, your lab coat and your fume hood are for.
I'm not saying those chemicals are harmless, they're not. But they are less dangerous than many other common chemicals in an organic chemistry lab.

I've answered this in part based on an older revision of the question, as the newer one doesn't have enought context.


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What typically happens to H2 molecules released into atmosphere (with no spark source)?
I'm wondering what happens when $\ce{H2}$ leaks from say a transmission pipe in an unenclosed area. No immediate source of ignition, I know it rises more quickly than helium (x2 I believe) and dissipates in concentration quickly but does it react with gases in atmosphere to form methane or other compounds?
Short of ignition and burning causing more $\ce{H2}$ to dissociate what can happen, especially in upper atmosphere with more UV and so on. Methane has reaction in the upper atmosphere giving it a half life of ~7 years for example.

The covalent radius of a neutral hydrogen atom is 0.0371 nm, smaller
  than that of any other element. Because small atoms can come very
  close to each other, they tend to form strong covalent bonds. As a
  result, the bond dissociation enthalpy for the $\ce{H-H}$ bond is relatively
  large (435 kJ/mol). $\ce{H2}$ therefore tends to be unreactive at room
  temperature. In the presence of a spark, however, a fraction of the $\ce{H2}$ molecules dissociate to form hydrogen atoms that are highly reactive. Source: chemed.chem.purdue.edu

Hmm, UV can readily dissociate $\ce{Cl2}$, but I'm not really sure about hydrogen gas. It's known to be reacting vigorously, but only if some good amount of activation energy is provided.

interesting, activation energy in the form of radiation or kinetic agitation?

Activation energy can be kinetic or potential. Basically, you need the reacting species with correct orientation and enough energy to collide. See collision theory and The Arrhenius Law.

The most common, reactive stuff in the atmosphere are $\ce{-OH}$ radicals (Wikipedia). So I guess they're the natural antagonist for $\ce{H2}$ molecules.
Btw., as soon as it's diluted in the surrounding air, it will no longer rise. Only bulk masses have buoyancy. The g gradient in the atmosphere is much too small to separate gases.
In the end, hydrogen however does diffuse upwards and is ultimately lost into space. Hydrogen and helium are light enough to be able to reach escape velocity in the upper atmosphere.

Please visit [this page](http://chemistry.stackexchange.com/help/notation " latex="" at="" chemistry.se""="), this page and this one on how to make your future posts better. || I like the "prey" analogy, but I believe the $\ce{H2}$ will be the prey and hydroxide anions the predator.

Your point about buoancy is not really true on geologic time scales.   Early in Earth's history, large amounts of hydrogen (H2) gas were lost to space.  The process continues today albeit at far lower rates.  Even so, the rate of H2 loss to space far larger than say N2 or O2 loss to space.

https://en.wikipedia.org/wiki/Atmospheric_escape - Yeah hydrogen and helium do escape albeit in little amounts

Has nothing to do with buoancy. The speed of H2 molecules( or He) in the upper atmosphere is above the escape velocity. He and H2 have a net flow upwards because of diffusion in the ensuing concentration gradient.

methane rises to the upper atmosphere where it decays and it is heavier than H2 (obv.). does buoyancy apply to gases in air? (As opposed to gases contained in solids on or in water).

No, it just diffuses up there. As it can't diffuse back (it's destroyed up there), there is a net flow upwards. An entropy gradient drives it up, not one in potential energy.

Hydrogen won't last long in a thunderstorm either.  Lightning discharges break up the molecules or produce hydroxyl radicals that later eat the hydrogen.


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Why some materials are transparent, but other ones aren't?
We have two materials; say glass and wood. Light passes through glass but it doesn't pass through wood. Is it caused by atoms and molecules arranged in glass and wood such that they interact differently with light?

I heavily modified your question, you can decide if it's OK for you.

There is an excellent video featuring Dr. Phil Moriarity on the Sixty Symbols channel about this very question.  I'm sure someone here (or perhaps on Physics SE) will give you a good answer, but that is a very accessible video that doesn't require much background.

Check out Quora

A brief summary: It is caused by the electrons in the particles in the substance. When electrons are orbiting the nucleus of an atom they can move to higher and lower energy levels. They do, however, need energy in the form of photons to do this, this means that when a ray of light hits the atom, the electrons in it will absorb some of the energy from the light ray. It needs a specific amount of energy to move up to the next energy level so will only absorb light from a specific part of the spectrum. This means that in some materials most of the light can pass through it and it may only absorb light waves not of visible light.
So wood, to some frequencies of light, is see through but not to the light waves we can see. That is why we can have mobile and radio signal from inside houses but only see through glass.
(Electrons are actually much more complicated and theoretical than this suggests but this)
Try watching this:  https://www.youtube.com/watch?v=Omr0JNyDBI0

Doesn't scattering play a role too?


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Why are (3-methylbut-2-en-2-yl)benzene and (3-methylbut-1-en-2-yl)benzene not stereoisomeric?

My answer key says neither product exhibits stereoisomerism, but both products are substituted. Why is it that I cannot assign E/Z configuration? I thought the only time you cannot assign E/Z is when the product is either mono or unsubstituted.

Even "unsubstituted" would have hydrogens connected... It's about symmetry not no. of substituents

in an unsubstituted alkene, there would be H's on all four sides of the double bond, so when assigning priority to determine E/Z, it would be impossible because nothing has priority over anything else...everything is H. The same would be true of a monosubstituted alkene- there are to many H's to assign priority. So only things that are tri or tetra substituted can be assigned E/Z. The above products should be able to be assigned E/Z...i am just wondering why my book says they are not stereoisomeric.

Because they aren't - if you have only 2 substituents connected to 1 atom or three identical groups they can't be

Try assigning a label to each one and you will see why it is not possible.

I thought the only time you cannot assign E/Z is when the product is either mono or unsubstituted.

Well, you cannot assign E/Z when it doesn't make any sense.
tl;dr: When on at least one side of double bond two identical groups or atoms (e.g., Hydrogen), there's only one way to arrange substituents near double bond.
E/Z is used to explain how substituents on different sides of double bond located, which excludes 1,1-substituted ethylenes (which is the case with minor product on your figure).
When on one side of double bond is two identical groups, e.g., methyl (which is the case with major product) there's no E/Z isomerism too, because there's only one way to arrange substituents. You can actually think of first case as of the second, minor product has 2 identical atoms at one side of double bond.
Hope I haven't made more confusion than there was.
Also, I would suggest to play with some ball and stick models (They could be at the uni library), I think that building couple of examples and trying to rearrange them to stereoisomeric form will help with understanding stereochemistry in general.

I see what you mean. If there are two identical substituents, then it doesn't make sense to assign E/Z because there would be no way to differentiate between high priority and low priority. Thanks!


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Derivative of Gibbs Energy wrt Temperature at constant Internal Energy
I came across with the following solution from Sandler's book: $$\begin{eqnarray}\left( \frac{\partial G}{\partial T} \right)_U &=& \frac{\partial (G,U)}{\partial (T,U)} \\ &=& \frac{\partial (G,U)}{\partial (T,V)}\cdot \frac{\partial (T,V)}{\partial (T,U)} \\ &=&\frac{(\partial G/\partial T)_V (\partial U/\partial V)_T-(\partial G/\partial V)_T (\partial U/\partial T)_V}{(\partial U/\partial V)_T} \end{eqnarray}$$
Please help me understand these steps. What happens in line 1? I know that line 2 is by using Chain Rule by then how did line 2 becomes line 3?

Maybe the quotient rule is used?

The starting point is the two equations $$dG=\left(\frac{\partial G}{\partial T}\right)_VdT+\left(\frac{\partial G}{\partial V}\right)_TdV$$ and $$dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV$$From the second equation, you determine the partial of V with respect to T and constant U by setting dU=0.  I leave the rest of the math up to you.
Chet


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Types of bonds in a molecule
For example in dinitrogen pentoxide, $\ce{N2O5}$, covalent as well as coordinate bonds (type of covalent bonds) are present, but it appears that it contains only covalent bond.
What is a proper method to find out which type of bonds are present in a molecule?

related http://chemistry.stackexchange.com/questions/21709/why-does-coordinate-covalent-bond-form

Electrovalent bonds are easiest to identify. If a compound is made up of a metal and non-metal/non-metallic radical (like carbonate), then, 99.99% times, it contains electovalent bond. If a compound is made up of 2 or more non-metals/non-metallic radicals, then it contains covalent bond. Coordinate covalent bonds appear mostly with compounds containing Hydrogen element. To identify the coordinate covalent bonds, you can draw the branched structural formula of the compound and see if the shared pair of electrons are coming from the same molecule.

What are you even talking about? @Kartik Watwani Please don't accept such low quality answers


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Does a gold detector exist?
I see many sites that claim to sell working gold detectors but I am not convinced that they work. Also, does a metal detector detect gold?

Can you be more specific about what these gold detectors are and how they are claimed to work. Otherwise your question is too vague to answer.

You ask:

Does a gold detector exist?

Absolutely. Here's a metal detector that is fine-tuned for gold detection, available ${\it via}$ Amazon.
That said, if you are looking for something that detects only gold, I'm not sure the best way to proceed in that case. This site appears to be a comprehensive review of some popular precious metal-tuned detectors on the market. The site is maintained by a person with an active interest in prospecting and metal detection. It is stated on the linked page that:

A quick note to those who know nothing about these machines. These are metal detectors. There is no such thing as a "gold only" detector. These detectors will also find lead, copper, aluminum, and other metals. These units are best used to look for relatively larger pieces of gold at relatively shallow depths. Concentrations of gold dust are not detectable. Some of these units can hit gold that weighs as little as a grain (480 grains per ounce) or less but only at an inch or two. Only the larger nuggets can be found at depths exceeding a foot. Only world class nuggets weighing many ounces can be detected at over two feet. The vast majority of nuggets found are found at inches, not feet.

In any event, you can purchase a device which has been engineered to discriminate base metals from precious ones (such as the model I have linked here).

Also, does a metal detector detect gold?

Yes - see the reviews for the product offered at Amazon (and many others like it) as well as the informational site I've linked (above). Also see the Wikipedia page on metal detectors for more information, including the issues relating to fine-tuning the devices to discriminate between metals with similar/different properties "detected" by the devices.


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Mechanism of carboxylic acid and amide dehydration with phosphorus pentoxide
Dehydration of carboxylic acids with phosphorus pentoxide ($\ce{P2O5}$) yields acid anhydrides, and in a similar reaction, amides dehydrated with phosphorus pentoxide yield cyanides. Can an arrow pushing mechanism be given for these reactions? If there isn't a mechanism, what, at least, gives phosphorus pentoxide the ability to dehydrate carboxylic acids and amides?
Phosphorus pentoxide itself reacts with water molecules to form phosphoric acid. That enables it to remove free water molecules by a simple hydrolysis reaction, but here of course there must be some other electron movement since the water being removed is not free water.

Here is a mechanism for the dehydration of an acid to an anhydride using phosphorus pentoxide.

The $\ce{OH}$ group is not a very good leaving group.  Even in simpler reactions like the dehydration of an alcohol to an olefin we often first convert the hydroxyl group into a better leaving group by protonation with acid or conversion to an inorganic ester (for example using thionyl chloride or phosphorus pentoxide).  The same thing takes place here - we are converting the acid's hydroxyl group into a better leaving group.
Why is $\ce{HPO3}$ such a good leaving group? Look at all the resonance structures you can draw for the anion $\ce{PO3-}$.  It's stability makes the reaction exothermic and strongly drives the dehydration process to the product side.
For completeness, here is a link to a drawing and discussion of the mechanism for the dehydration of an amide to a nitrile using phosphorus pentoxide. Same mechanism and principles (good leaving group, exothermic) as discussed above.

Sorry but that's wrong :( phosphorus (V) oxide is actually P4O10.

Actually Mithoron I think the diagram is OK.   Phosphorus(V) oxide is not really P4O10, either, at least not in its most stable state.  It's a polymer.   So showing it as P2O5 isn't 100% accurate but doesn't matter for the mechanism.  Wikipedia has more about the various polymorphs of the compound.

This might sound nonsensical, but I was just thinking - considering OH needs to be made a better LG, which can be accomplished by protonation - an acidic medium would suffice as well. The carboxylic acid itself being acidic, it can provide the H+. What prevents it really from turning into an anhydride by itself?

@Charles Good question.  Typical carboxylic acids are not as ionized as $\ce{HCl}$ or $\ce{H2SO4}$, so the concentration of $\ce{H^{+}}$ will be much, much lower with the carboxylic acid.  This low proton concentration likely slows the reaction down to an imperceptible rate.


text:
Precipitate Sodium Hypochlorite from tap water before distillation
Is there a fairly simple way to precipitate Sodium Hypochlorite from tap water before I distill it? Or, second best, a way to neutralize it?

Hmm, is this unclear? Could the CV'ers enlighten me on what detail this is missing? I think John is looking for a process that precipitates Chlorinated agents.

Yes, that would be ideal.

Sodium hypochlorite is not carried over into water by distillation.  Instead, at high (~100°C) it decomposes, primarily to sodium chloride (i.e. table salt) and oxygen gas, but also possibly to trace amounts of chlorate.
You could neutralize the hypochlorite by mixture with stoichiometric amounts of hydrogen peroxide, which cause immediate release of oxygen gas, a reaction which occurs even at room temperature.

This is helpful. If no one knows how to precipitate the solution, I will mark this as the answer. Thank you!

You should be aware that if you use excess H2O2 instead of the sociometric amount, then you will have a problem with H2O2 contamination, both in the crude water and possibly in the distillate.


text:
How is the unit of measure "equivalent" abbreviated?
I usually use "eq" but I have been told to use "equiv." in order to avoid confusion with "equation". Is there a reference I can check?

Santiago's answer is authoritative but if you want to invent your own abbreviation that is (i) shorter than equiv and (ii) avoids ambiguity with equation, I suggest "eqv".

Yes, and to @CurtF.' comment - whatever you choose to do - be consistent.

The first place to check would be standards that are used frequently globally, such as CAS. According to the American Chemical Society page CAS Standard Abbreviations & Acronyms, the abbreviation for the unit of measurement 'equivalent' is equiv, while that of the adjective 'equivalent' is equiv. (with a dot).
Also, the ACS page Standard Abbreviations and Acronyms, shows equation abbreviated to eq and equivalent as equiv.
However, some journals, such as the Journal of Biological Chemistry page Table I: Abbreviations of units of measurement and of physical and chemical quantities state that the abbreviation of equivalent is as you used, eq, so if you are intending to publishing, just be aware of what the journal expects in terms of abbreviations.


text:
Rate of exothermic/endothermic reactions
Is it necessary that all endothermic reactions be slow and all exothermic reactions fast? This is based on the fact that endothermic reactions require more activation energy than exothermic reactions to initiate a chemical reaction. Or  are there some endothermic reactions that can be fast without the presence of a catalyst?

Why should endothermic reactions have a higher activation energy?

A reaction being endothermic or exothermic depends on thermochemistry while the rate of a reaction depends on its chemical kinetics. Hence, there is no relationship between the activation energy of a reaction and its enthalpy change. Also there is no relationship between the change enthalpy and the rate of a reaction. A rate of reaction is dependent on the concentration of the reactants which are to a certain power which is multiplied by a certain constant, k, which is called the reaction rate constant. For example, for the general reaction: $$\ce{A + B -> C}$$ the rate of the reaction is given by the following equation: $$ rate = k[A]^a[B]^b$$
The reaction rate constant is not dependent on whether a reaction is exothermic or endothermic. However is does depend on the temperature and and also the activation energy of the reaction, given by the equation: $$k = Ae^\frac{-E_a}{RT}$$ So you are right to say that a reaction will be slower if it has a higher activation energy, but you are wrong to say that endothermic reactions have a higher activation energy than exothermic reactions.
There are actually several endothermic reactions that are fast, for example, dissolving common table salt (NaCl) occurs pretty quickly, but it is actually a endothermic reaction. Meanwhile, the spontaneously decomposition of diamond to graphite is actually a exothermic reaction, however it is extremely slow. It is so slow that it not even observed after several decades.

But if an exothermic reaction is giving off heat and increasing the temperature of the solution, doesn't that speed up the reaction?

The activation barrier is cause by energy needed to form the transition state between reactants and products. While it is generally true that the activation energy, and hence rate constant, is not related to how exothermic or endothermic the reaction is that is not the whole story.
If a reaction is endothermic by $+ \Delta E$ then the activation barrier has always to be at least this large for the products to be reached. Of course the barrier could be larger. (Although complicated protein unfolding is an example of a reaction in which the overall activation energy is large and only slightly greater than the endothermicity. Since many proteins spontaneously fold, the barrier to folding is small.)
In an exothermic reaction (by $-\Delta E$), while there can be, and often there is, a large barrier it is possible for there to be no barrier, i.e. the reaction can be activationless. In this case, in solution or gas, the rate constant is limited by how fast the reactants can diffuse together.  Many electron transfer reactions are barrierless.


text:
Understanding Gibbs free energy
I want to verify a thought I had about Gibbs free energy. In some ways, I'm actually wondering why we are using enthalpy instead of the internal energy $U$. Right now, I'm thinking that enthalpy is mostly useful because reactions are done at constant pressure (in an open beaker). If we were to do our reactions at constant volume, we would actually use Helmholtz's free energy instead of Gibbs. My logic:
The definition of internal energy:
$\Delta U = Q_\text{heat} - p\Delta V$
If we inject $Q_\text{heat}$ without holding the volume constant, we might change $U$ or $V$, which is a problem because we'll want to be able to describe our system as fully as possible. If we define the enthalpy:
$\Delta H = Q + V\Delta p$
Then, by keeping the pressure constant, we know that the heat added goes straight into this "enthalpy".
Afterwards, to know if a reaction is spontaneous, we need to see if the enthalpy-heat from a reaction overthrows the energy associated with the change in entropy.
Example: a molecule of two atoms wants to break apart. The bond has potential energy $Q_\text{heat} = \Delta H_0$ but it will increase the entropy by $Q_\text{entropy} = T \Delta S_0$. So, we check if:
$ Q_\text{heat} - Q_\text{entropy} > 0 $
$ \Delta H - T \Delta S > 0$
And thus Gibbs free energy was invented as:
$\Delta G = \Delta H - T \Delta S$
BUT, if we wanted to use a closed beaker, we would keep the internal energy U, and then to see if we had a spontaneous reaction, we would use Helmholtz's free energy:
$ \Delta F = \Delta U - T \Delta S $
I'm not sure about any of this and I'm mostly looking for confirmation/information of my logic. Please tell me if my intuition is wrong somewhere.

Sounds right to me.  But I'm not 100% what your question is.

Yes your logic is basically correct.
Gibbs free energy is really derived is from the Clausius in equality which is: $$dS\geq \frac{dq}{T}$$ Basically is states that in a reaction, the change in entropy must always be equal (in the case of an ideal gas) or greater than the transfer of heat energy divided by temperature. Therefore, for a reaction to be spontaneous, the following relation must be satisfied: $$dS-\frac{dq}{T}\geq0$$ Now, lets consider two cases so we can simplify this relation.
Constant Pressure
Enthalpy is given by: $$\Delta H_p = \Delta U + p\Delta V$$
and internal energy by: $$\Delta U = q + w$$
Now by substituting $\Delta U$ in the expression for enthalpy we get: $$\Delta H _p= q + w + p\Delta V$$ Now, since w = $-p\Delta V$, the above equation simplified so that: $$\Delta H_p = q$$ So, now in the Clausius inequality, we can substitute dq with dH and multiply by $T$ to get: $$TdS \geq dH$$ Now by integrating both sides and putting all the variables on one side, we get the question for gibbs free energy: $$\Delta G = \Delta H - T\Delta S$$ and hence this is why $\Delta G$ has to be negative for a reaction to be spontaneous
Constant Volume
At a constant volume, there is no expansion work done, therefore $dU = dq$
Therefore by using the same processes as before, we are able to get the expression for Helmholtz energy, A, which is defined as: $$\Delta A = \Delta U - T\Delta S$$ Here, for a reaction to be spontaneous, $\Delta A$ must be less than 0

Just to be sure, the definition of enthalpy isn't:
$\Delta H = \Delta U + p\Delta V + V \Delta p$ ?

Yes it is, as dH = dU + d(pV). Then using the product rule, this becomes:       dH = dU + Vdp + pdV. The reason why I wrote $\Delta H = \Delta U + p\Delta V$ was because at constant pressure Vdp = $0$


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Mechanism and type of polymerization of chloroethene
Ethene reacts with hydrogen chloride to form a monomer used in the production of a polymer. This monomer is chloroethene.
How is the polymer formed from the monomer?
Is this an addition or a condensation polymerization?

You got it wrong - monomer is chloroethene (vinyl chloride) and nothing is eliminated in addition polymerization in contrary to polycondensation. This monomer is made via addition of chlorine molecule to ethene and elimination of hydrogen chloride. Alternative methods are addition of hydrogen chloride to acetylene or oxychlorination - in this reaction $\ce{HCl}$ and $\ce{O2}$ react with ethylene to form the monomer and water.

But how is a chlorine molecule added?

Huh? Simple electrophilic addition reaction.


text:
How does water defy gravity in paper towels?
How can water be absorbed into a paper towel such that it continues to be absorbed up into the material against the force of gravity? What property of paper towels/water causes this? I can understand that cohesive forces cause water molecules to follow other water molecules, but why do the molecules begin this climb in the first place?

surface tension...

What are you even asking about? Why is cellulose hydrophilic?

Thanks to Jon Custer, I think I found the explanation. It's called capillary action, and it's a property of water that allows it to travel against gravity in porous substances like paper. In short it's due to the properties of both paper and water.

Yep, you found it.

This is simply an example of capillary action expanded to a sheet of many tiny capillaries. In a narrow glass tube immersed in water, water will climb the inside of the tube to a level higher than that of the water outside because it is energetically favourable for the water to wet the surface due to adhesive forces and the cohesive forces of the water pull the bulk of the water up with the contact line of the water and the glass. Glass and water have strong adhesive forces as glass has many silicate groups which can interact with water through hydrogen bonding and other polar interactions. As the liquid rises, the volume of water above the surrounding liquid increases, thus increasing the weight of the column. When the weight of the column equals the adhesive and cohesive forces drawing the water upwards, the system is at equilibrium and the water will rise no further.
This can be calculated from: $$h = \frac{\gamma\cos\theta}{\rho gr}$$ where h is the height of the column, $\gamma$ is the liquid-air surface tension, $\theta$ is the equilibrium contact angle between the glass and the water, $\rho$ is the density of water, g is the acceleration due to gravity, and r is the radius of the tube. As you can see, increasing the surface tension (which comes from water's cohesive forces) or decreasing the contact angle (the lower the contact angle, the stronger the adhesive forces between liquid and surface) will increase the height the column will reach, and conversely a denser liquid or larger tube will decrease the height (greater increase in weight for a given rise).
I won't do the whole calculation but for a 1 mm radius glass tube with water ($\theta \approx 50°$, $\gamma = 72\ \mathrm{mN/m}$), the equilibrium height is 4.7 mm. Now, if we shrink the tube much further to approximate something like a pore in a piece of paper towel, we can make the liquid go much further. (it's obviously not perfectly cylindrical, but close enough for illustration and the cellulose-water contact angle seems to be close enough to glass' as cellulose's abundance of hydroxyl groups is similarly adhesive to water) At a diameter of 10 µm, we get a height of 47 cm.
When we expand this concept to a large sheet of interconnected pores, we find that capillary action can draw a relatively large volume of water across considerable distances, which in addition to being great for cleaning up household spills, opens the possibility of using paper as a substrate for analytical devices that don't need pumps, including the lateral flow assay (most commonly seen as the pregnancy test stick) and paper microfluidic devices.