text:
How do I determine the average carbon oxidation numbers in ethanol and ethanal?
I have an assignment (I quote it word by word):

Determine the oxidation number of carbon in (A) ethanol and (B) ethanal. Answers are (A) -2 and (B) -1.

I reviewed similar question at SE: Oxidation of Carbons
and I found first answer to be clear and links given in this answer to be useful.
I also found a good  Khan Academy video, which basically confirms all in this answer
Now, here is my problem: There are two carbons in ethanol. Oxidation number of carbon which is connected to three hydrogens is $-3$; oxidation number of carbon which is connected to two hydrogens and one oxygen is $-1$. How do we get answer $-2$ out of it then? (Do we add $-3 + (-1)$ and then divide by 2?)
And again there are two carbons in ethanal. Oxidation number of carbon which is connected to three hydrogens is $-3$; oxidation number of carbon which is connected to one hydrogen and one oxygen (but with double bond) is $+1$. How do we get answer $-1$ out of it? (Do we add $-3 + 1$ and divide it by 2?)
Overall, my question does not seem to concentrate on one of the carbons present in ethanol (or ethanal). And answers given look like there is oxidation number of first carbon added to oxidation number of second carbon and then divided by 2. This situation seems to be true for ethanol, and again the same happens for ethanal.
Where can I read about adding two oxidation states and then calculating average (I have never heard of it)? Or maybe these calculations done in some other way? Thank you very much in advance for answering.

This question is really hard to read. If you would, please check out the information here and here about formatting posts with HTML and/or MathJax, and rewrite your question using these tools. Thanks!

Thank you, Brian! I made paragraphs and hid link for Khan Academy video. I would make some other changes, if I would have a clue what exactly I should change . Your advice on this matter will be very much appreciated.

Yes, it's simply average oxidation number. You were unnecessarily verbose here IMO.

Thank you, Mithoron! I asked about link for any source which would confirm that idea. If you know one, please post it. I would be most grateful. I also have a reason to write a detailed question, but thank you for sharing your opinion.

@SleepyHollow Those edits are great! Much easier to read.  Thanks!

About average oxidation numbers one can read here: https://en.wikipedia.org/wiki/Oxidation_state . It is not about carbon, but one can easily imagine, how to apply it to carbon.

What you have been asked to calculate is what is called an average oxidation number. This is often used in inorganic chemistry (not because it is rather powerful, but because you can easily get away with it) but lesser in organic chemistry.
To properly determine oxidation states, you should do the calculation on a per-atom basis. And doing that would give you $\mathrm{-III}$ and $\mathrm{-I}$ for ethanol’s two carbons and $\mathrm{-III}$ and $\mathrm{+I}$ for ethanal’s. This method has a number of advantages:

you are always dealing with integers; no need for any fractional states
it is easy to see exactly which carbon’s oxidation state changed.

The downside of this method is that you cannot use it if you don’t know the structural formula. Consider if, for example, you only saw $\ce{C2H6O}$ or $\ce{C2H4O}$ — especially the first one could represent two different structures with different properties. This is were average oxidation states come in handy: you calculate an average of all the atoms of a single element. You may remember that oxygen is typically given $\mathrm{-II}$, hydrogen $\mathrm{+I}$ (except in peroxides and hydrides, respectively; although oxofluorides should get a special mention). You may also remember that the sum of oxidation states of a neutral compound must be $\pm 0$ that allows you to calculate the average oxidation states as follows:
$$\begin{align}\text{For }\ce{C2H6O}\text{:}&\\
0 &= 2 x + 6 \times (+1) + (-2)\\
0 &= 2x + 6 - 2\\
0 &= 2x + 4\\
-4 &= 2x\\
x &= -2\\
\\
\text{For }\ce{C2H4O}\text{:}&\\
0 &= 2 x + 4 \times (+1) + (-2)\\
0 &= 2x + 4 - 2\\
0 &= 2x + 2\\
-2 &= 2x\\
x &= -1\end{align}$$
Since these are averages, you can also arrive at the same conclusion by taking the arithmetic mean of all carbon atoms’ oxidation state; you will obviously arrive at the same result:
$$\begin{align}\ce{C2H6O}\text{:}&\\
\bar{x} &= \frac{-3 + (-1)}{2}\\
\bar{x} &= \frac{-4}{2}\\
\bar{x} &= -2\\
\\
\ce{C2H4O}\text{:}&\\
\bar{x} &= \frac{-3 + 1}{2}\\
\bar{x} &= \frac{-2}{2}\\
\bar{x} &= -1\end{align}$$


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Can we make chromatography less troublesome?
Thin Layer Chromatography is a very powerful method in analytical chemistry for separating complicated chemical matrices.
As an engineer, I am interested to build device capable of automating TLC such that I can just place a droplet of my sample onto a piece of paper and let the device perform the TLC for me.
How achievable is this?

https://en.wikipedia.org/wiki/High_performance_thin_layer_chromatography

^^ So, to answer your question: It is very achievable. Just don't try to patent it unless you come up with some innovative new addition.

Are there HPLC machines out there that will automatically lower my plate into a solvent front? Would it be possible to automate this process such that a person with no chemistry background would be able to use?

And also, something less than 50 USD?

We can't even get a box of TLC plates for less than 50 USD.

@SendersReagent I have UTLC plates with Silica (http://www.google.ch/patents/US8810789) costing about 1 dollar per cm squared.

Although it may be useful for science facilities with low resources, I am not sure if it will provide anything that is in a way better than more functional chromatography methods: gas chromatography, liquid chromatography, ion chromatography, etc. With these and a combination of detectors (FID, ECD, Mass Spectrometers) we can pretty much separate and detect any mixture of compounds. Furthermore, chemometrics  improves identification by using data generated by these procedures.

If you equate "less troublesome" with more automation of sample deposit and development: http://www.camag.com/en/tlc_hptlc/products/sample_application/automatic_tlc_sampler_4_ats_4.cfm (as an example).

source: Stack Exchange

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What is the difference between a positive and negative electron orbital?
Why are do some electron orbitals have a negative number in the subscript (i.e. f-1), some a positive (f1), and some have a 0 (f0)? What do these numbers tell you about the atom? I know that these orbitals have different shapes, but I was wondering if they had any other effect on the atom/ion and if there was a reason that they were put in the order that they are in now.

There are no such things as positive or negative electron orbitals. The number in the subscript is $m$, known as magnetic quantum number and running from $−\ell$ to $\ell$. It doesn't tell you much about the atom. It is a property of the orbital, not of the atom. Every atom has all these orbitals at once (some of them occupied and some empty); also, there are versions of each for different $n$.
In a free atom, all orbitals with the same $n,\;\ell$ and different $m$ have the same energy, so this number does not mean much for an orbital either (at least unless you put the atom in a magnetic field).
Moreover, the pictures are fake anyway. Orbitals which are the true eigenfunctions of $m$ (that is, correspond to specific $m$) are complex-valued and thus can't be properly represented in a plot. As for the orbitals featured on this diagram (and all over the textbooks), these are the linear combinations of complex orbitals for different $m$, hence each of them does not correspond to any particular $m$. It's not that they are wrong per se (absolutely not!), it's just that you can't point at them and tell which is $p_1$ and which is $p_{-1}$.

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Precipitaiton of aragonite
I'm not positive what this questions is asking.  The prompt says "Write balanced equations for the following processes.  Chemical species should be the dominant ones at the given $\mathrm{pH}$. Show your work in deriving these equations."  Then the actual question is:

Precipitation of aragonite ($\ce{CaCO3}$;  $K_\mathrm{sp} = 4.8 \times 10^{-9}$), at $\mathrm{pH}~8$.

The balanced equation would just be $\ce{CaCO3 -> Ca^{2+} + CO3^{2-}}$ .
I know that in an acid-base reaction I can use the $\mathrm{pH}$ as the $K_\mathrm{sa}$ to find the concentrations of the reactants and products, but I'm not sure where the $\ce{H+}$ would come in here.  I'm also not sure what finding the concentrations has to do with writing a balanced equation.  Does $\ce{Ca(OH)2}$ have something to do with this?
I could change the equation to $\ce{CaCO3 -> Ca(OH)2 + CO3^{2-}}$  and then calculate the concentrations, but is that what this question is even asking?  Thanks!

You need to write the precipitation reaction. If something precipitates, what is happening?

Using a graph to plot a solubility curve or use already obtained data to then determine the solubility of CaCO3 at ph 8 may help?

I doubt that aragonite will precipitate at all unless additional conditions are met (e.g. presence of magnesium ions). Under normal conditions I would expect calcite to be formed since it is less soluble than aragonite and is the thermodynamically more stable modification.

I think the question aims at the composition of carbonic acid solutions. The concentration of carbonate ions ($\ce{CO3^{2-}}$) depends on the pH of the solution. At low pH values the concentration is low, at high pH values the concentration is high. So the question is if the concentration of $\ce{CO3^{2-}}$ at pH 8 is high enough to precipitate calcium carbonate.

$\ce{H2CO3}$: black, $\ce{HCO3^{-}}$: violet, $\ce{CO3^{2-}}$: turquoise, $\ce{H+}$ and $\ce{OH-}$: grey


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Why does chromium metal not react with a solution of Mg^2+ even though their standard cell potentials produce a net positive voltages?

Write net equations for the spontaneous redox reactions that occur during the following or NONE if there is no extensive reaction. Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.

My effort:
Chromium metal is added to a solution of $\ce{Mg^2+}$
\begin{align}
\ce{2Cr^3+ (aq) + 6e- &-> 2Cr(s)}\\
\ce{3Mg(s) &-> 3Mg^2+ (aq) + 6e-}
\end{align}
The net equation is
$$\ce{2Cr^3+ (aq) + 3Mg(s) -> 2Cr(s) + 3Mg^2+ (aq)}$$
Standard reduction potential of $\ce{Cr}$ is $\pu{-0.74 V}$ and the standard reduction potential of $\ce{Mg}$ is $\pu{-2.36 V}$.
$\ce{Cr}$ reaction is cathode reaction and therefore it is $\ce{Cr^3+}$ getting reduced, $\ce{Mg}$ is anode reaction and therefore it is
$$\ce{Mg(s) -> Mg^2+ (aq) + 2e^-}.$$
$$E^\circ_\mathrm{cell} = \pu{-0.74 V} - (\pu{-2.36 V}) = \pu{1.62 V}$$
Yet apparently my homework here says that that is wrong.  Does anyone see anything wrong with how I am going about this problem?

It's the magnesium ion that is being converted into magnesium metal. $\ce{Mg^{2+}}$ is being reduced at the cathode.

In other words, all your equations are written backwards.

Metals that  could theoretically reduce magnesium ions would rather reduce water to hydrogen.

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Reasonably complete infrared spectral database with data in CSV form?
I'm currently developing a neural network that analyses IR spectroscopy signatures. I've found a very complete data set at The Spectral Database of Organic Compounds, but data is only available as images, which are not ideal for training neural nets.
Does anyone know of a reasonably comprehensive open database of spectroscopy data that offers data in CSV/tabulated form?

First off, I would think you could fairly easily design some data extraction tools for the AIST database. After all, the X and Y axes are standardized (i.e., $4000-400 \mathrm{cm}^{-1}$).
Beyond AIST, you can find machine-readable JCAMP files from the NIST WebBook. There are supposedly ~16,000 spectra, although you'd have to script some way to download them one-by-one.
I think a better bet would be to purchase a database:

Bio-Rad sells an IR database. A quick look now indicates that it has ~235,000 spectra and I'm sure it's in an accessible format.
NIST sells a smaller database ~5000 compounds in JCAMP format.

A quick search also turned up:

IR Spectra.com - claiming ~140,000 spectra in various formats.

I will mention that while I've heard of efforts to create an IR spectral database (e.g., all those new compound published in journals), nothing significant has appeared. (The Open Spectral Database has ~45 spectra.)


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Why there is an Upper Explosive concentration Limit?
If we have a mixture of combustible gas (let's say it's methane) and air, what is the physical explanation for having a upper concentration limit?
I found this document:
Lower and upper explosive limits
For example, methane has 5% and 15%.
Let's say we have a 30% concentration of methane... There is still enough oxygen, so that some of the methane could burn.
There won't be any combustion at all, or there is some combustion, but no explosion?

see also http://chemistry.stackexchange.com/questions/7587/what-makes-substances-explosive/7739#7739

Crossposted from http://physics.stackexchange.com/q/248599/2451

Deleted from Physics

In simple terms, an explosion takes place when the reaction front propagates faster than the speed of sound. If it is slower, the expanding gas cloud can push away the unreacted mixture before it reaches ignition temperature.
With too much methane, the combustion reaction slows down, because the additional material takes up energy for heating, but cannot react.
A very simplistic microscopic picture would be that a hot methane molecule too often does not hit a cold oxygen molecule, with wich it could react to produce more heat, but a cold methane molecule, leaving two lukewarm methanes, than won't react any further.


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What is more important, to have the parent chain be the cycle or the chain that has an hydroxyl group
I know how to name alcohols, I just don't know what I should chose as the parent chain. Should I chose the cycle or the chain with the hydroxyl group. Here is an example:

My current algorithm tells me that this is 1-(propan-2-olyl)cyclohexane, because it always chooses the cycle with the most skeletal atoms for a parent chain no matter what else is present.
So my questions are:

In what situation an acyclic chain is chosen for a parent chain over a ring?
If this is one of those situations would the choice be affected if we add on hydroxyl group to both the ring and the chain?
If this is not one of those situations would the choice be affected if we increase the length of the acyclic chain and/or add more hydroxyl groups?
Something on a slightly different topic: is 1-(propan-2-olyl)cyclohexane a correct name or should it be 1-(propan-2-ol-1-yl)cyclohexane? I am not sure if the 1 is omitted or not.

P.S. Is there a place where I can find the Blue Book for free? I only have the 2004 draft and I am not sure how much of it is changed in the 2013 version?

The compound given in the question contains a characteristic group $(\ce{-OH})$. Since there is only one characteristic group, the seniority order of classes is not relevant in this case; thus, the $\ce{-OH}$ substituent corresponds to the principal characteristic group that is expressed as a suffix (‘ol’).
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the senior parent structure has the maximum number of substituents corresponding to the principal characteristic group.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)

Therefore, the correct name for the compound given in the question is ‘1-cyclohexylpropan-2-ol’ (not: ‘(2-hydroxypropyl)cyclohexane’). (Note that in ‘(2-hydroxypropyl)cyclohexane’, the locant ‘1’ of ‘1-(2-hydroxypropyl)cyclohexane’ is omitted since, according to Subsection P-14.3.4.2, the locant ‘1’ is omitted in monosubstituted homogeneous monocyclic rings.)

If there is still a choice (for example, if the ring and the chain each have a substituent corresponding to the principal characteristic group), the ring or the chain can be the senior parent structure; however, for the preferred IUPAC name, the ring is always selected as the senior parent structure.

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.
(…) For selection of a preferred IUPAC name, see P-52.2.8.

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Is the characteristic group always cited as a prefix and not a suffix if it is in a substituent and if so, does that mean that it has the same priority for low locants as everything else?

@K.Smith Yes. In names of simple substituent groups, the presence of a free valence formally derived from the loss of one hydrogen atom from a parent hydride is denoted by the suffix ‘yl’. Since only one suffix can be placed at the end of a name, any further characteristic group of the substituent group is cited as prefix.

I see. I thought "yl" was an infix.


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Binding Energy and Enzyme Rate
I was preparing for the Canadian Chemistry Olympiad and I ran into the following question on one of the past exams:

If enzyme A binds to the substrate 25 times stronger than enzyme B, what is the ratio of the catalytic rate between enzyme A and enzyme B if the energy of the two transition states is identical? What is the difference in activation energy between the two reactions?

I feel as if I can make a qualitative prediction, but cannot derive a quantitative prediction. I suspect, since enzyme A bind more strongly to the substrate than enzyme B, it would have a greater activation energy, and thus result in a slower reaction.

I suspect that it is necessary to use the equation
$$ \begin{align}
k = \frac{kT}{h} e^{-\Delta G^\ddagger/RT}
\end{align} $$
But the relationship between the two activation energies does not seem simple
$$\begin{cases}
\Delta G^\ddagger_A = \Delta G_{E + S \rightarrow T} - 25\Delta G_\text{bind} \\
\Delta G^\ddagger_B = \Delta G_{E + S \rightarrow T} - \Delta G_\text{bind}
\end{cases}$$
Plugging this into the the original equation, and solving for $k_B/k_A$, I get the following:
$$ \frac{k_B}{k_A} = \frac{e^{\Delta G_\text{bind}/RT}}{e^{25\Delta G_\text{bind}/RT}} = e^{-24\Delta G_\text{bind}/RT}$$
Since $\Delta G_\text{bind} < 0$, $k_B/k_A > 1$, supporting my qualitative prediction. Yet this does not give me a useful ratio, since it is dependent on the binding energies of enzymes A and B. This makes me suspect I am missing/overlooking something, any insight would be appreciated.

The question as written is a little unclear but if we assume that the two enzymes are initially at the same energy and that the transition state from species enzyme + substrate, [ES] to its product is also at the same energy, say $E_{TS}$, then possibly an answer can be found. ( I assume that by 'catalytic rate', 'catalytic rate constant ' is meant).  An equilibrium constant in a reaction only informs us about the ratio of rate constants and free energy change $\Delta G^0 $. It does not on its own help us find activation energies or individual rate constants. In this question the transition states are of equal energy, which additional information we can use. I have assumed that the schematic of the energy levels shown below represents the levels in the question. The dotted lines are only schematic. The activation energy is from ES to TS in each case.

Let $K_B$ be the binding equilibrium constant for species B, then $25K_B$ is that for A. The species [ES] for A is then at free energy $\Delta G^0_A=-RTln(25K_B)$ and B at $\Delta G^0_B=-RTln(K_B)$, which means that the activation energy (from [ES] to its transition state towards products) is greater for A by the difference between these values which is $RTln(25)$. The ratio of rate constants is the ratio of Arrhenius equations each of which we assume has the form $k=A_0exp(-(E_{TS}+\Delta G)/RT)$ (where $A_0$ is the pre-exponential term assumed equal). Most terms cancel giving the ratio of rate constants as 25 with B being faster.

Your intuition that the enzyme binding more strongly to substrate will be slower is correct, this is called ground-state stabilization.  I think where you might be making an error is how you handle the factor of 25.  The factor of 25 probably refers to the ratio of equilibrium constants between E and ES for the two enzymes, not the ratio of binding energies.  So you need to start with delta G = - RT ln Keq, and compare for 2 different Keq's.  Then use your equation to estimate the difference in rate constants.


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Is there any reason to fear personal exposure to rain with a pH of 3.1?
Where I live now there can be significantly acidic rain. The government posts measurements for many cities and there can be stretches for days or longer with reported pH of 3.2 or 3.1. I even saw 2.9 reported once.
While this can have quite serious environmental effects depending on a number of factors including total rainfall, geology, geography, soil composition just to name a few, my question is about the direct effect of rainfall this acidic on people and their possessions.
Just how dangerous or deleterious to people and their immediate possessions (e.g. clothing, other objects) would getting a bit of pH 3.1 rain on oneself actually be? Would it sting at all if it got in the eye? Discolor hair? Is there any basis at all to being collectively phobic of being touched by this rain specifically because of its pH?
I've tried searching the internet but there is a jumble of information there. As far as I can tell, pH 3.1 seems to be somewhere between a soft drink, and tomato juice, but this is from unreliable sources.
So I thought I would ask in Chemistry SE for a solid, scientific answer. I'm not looking for an opinion, or a reasoned answer (e.g. "if it was that dangerous, then..."). I'd prefer factual data, or references related to safety and effects of topical exposure to (mildly) acidic water.

Well I guess rain falls on you from time to time and it seems you didn't notice anything... And comparisons with juice or drink are OK

If you google "pH of common substances site:.gov" (no quotes), you'll get several reliable(ish) links that confirm your understanding that 3.1 is about as acidic as a soft drink, and much less acidic than lemon juice.
3.1 seems really acidic because it's closer to pH 0 than to pH 7, but pH is a logarithmic measurement, so it's really not as bad as it seems.

@barrycarter I have heard of logarithmic scales before, thanks. pH of 3.1 means there is almost one H${}^+$ (or H${}_3$O${}^+$) ion for every thousand water molecules, and I think that's a lot. But my question here is rather specific - I'm asking about the chemical effect of said pH on people or personal belongings, not if it sounds big to them or not.

According to a United States Environmental Protection Agency report on the effects of acid rain on human health:

Walking in acid rain, or even swimming in a lake affected by acid rain, is no more dangerous to humans than walking in normal rain or swimming in non-acidic lakes. However, when the pollutants that cause acid rain $\ce{—SO2}$ and $\ce{NOx}$, as well assulfate and nitrate particles— are in the air, they can be harmful to humans.
$\ce{SO2}$ and $\ce{NOx}$ react in the atmosphere to form fine sulfate and nitrate particles that people can inhale into their lungs.  Many scientific studies have shown a relationship between these particles and effects on heart and lung function, such as asthma and bronchitis.

Thank  you for the attention to my question and the EPA link! I noticed that the EPA page What is acid rain? puts the pH of acid rain in the general area of 4.3 plus "Acid rain usually has a pH between 4.2 and 4.4." - I've seen values as low 2.9 reported by the national weather service here, and 3.1 much more often, thus the title of the question. I hadn't known about the serious risks associated with breathing sulfate and nitrate particles (as opposed to gas) and I'll read further about that. Thanks!

No problem. As an aside, I did come across a story about acid rain after volcanic events that had rain with pH ~2 that would irritate people's eyes.  But that's some really acid rain!


text:
What means can be used to distinguish different sorts of plastic?
So, different sorts of plastic usually have different marks on them (the resin identification code, like 1/PET for polyethylene terephthalate etc). However, sometimes one can encounter a piece of plastic (or film; it can be even non-plastic like cellophane) without identification code or whatever sign. What ways to distinguish plastic types can you propose?
It is desirable that these methods can be applied without lab hood and chemical hazard when applied to ordinary types of plastic (at least PET, HDPE, PVC, LDPE, PS, PP, cellophane) unless there's a simple way to eliminate some variants which would cause hazard (like "don't use this for white non-transparent plastics and that otherwise). Ideally this shouldn't require reagents which are hard to accuire but if you have some ideas, throw them anyway.

FT-IR spectroscopy would be the simplest choice, but probably not what you had in mind. You could make a chart with tests like "reaction to acetone" (crazes, dissolves, nothing), "flame test" (keeps burning, smoke, ...), "mechanical" (breaks, bends, ...). I think you should be able to distinguish most common polymers that way, but there'll certainly be others behaving the same. And many properties vary with processing, additives, etc, so i would not bet on any results from that.

Thanks for this @Karl, I can't "+1" you comment yet, but that's a start. Yeap, I'd prefer to avoid methods requiring set up available in labs only like FT-IR or NMR spectrometers.


text:
Experimentally finding order of a reaction
For the reaction of bromocresol green and bleach, I was supposed to conduct an experiment to find the order of the reaction with respect to bleach. To do this, I used a fixed amount of bromocresol green and varied the amount of bleach in each trial. I recorded the amount of time it took for each reaction to finish.
To find the order, I was told that if the order = x, $\frac{time_1}{time_2} = (\frac{[bleach]_2}{[bleach]_1})^x$. However, I'm having trouble understanding why this is true. For example, say we begin with $[bleach]_1=y$ and $[bleach]_2=2y$. If $\frac{time_1}{time_2} = 2^x$, this means that the rate of reaction 2 is always $2^x$ times faster than the rate of reaction 1.
In the beginning, this is definitely true, since rate = $k[bleach]^x$. Initial $rate_1 = ky^x$ and initial $rate_2=k(2y)^x$. But at a time $dt$ later, $[bleach]_1=y-ky^x\cdot dt$ and $[bleach]_2=2y-k(2y)^x\cdot dt=2y-2^xky^x\cdot dt$. At this time, $\frac{rate_2}{rate_1} = (\frac{2y-2^xky^x\cdot dt}{y-ky^x\cdot dt})^x$, which does not equal $2^x$ unless $x=1$.
Thus, I'm not understanding why $\frac{time_1}{time_2} = (\frac{[bleach]_2}{[bleach]_1})^x$ is true for x other than 1. Any clarifications would be much appreciated.

It's only true for the initial rate.

Which of the reagents is in excess, to begin with?

@IvanNeretin bleach

Then maybe its concentration does not change much throughout the reaction?

There are a couple of odd things here.

the time for a reaction to be complete is usually infinite in theoretical kinetics, so I don't see how it could be useful and/or relevant. One usually looks at the initial rate of a reaction.
if bleach is in large excess, you may have a degenerate order. That means that the concentration of bleach barely varies during the reaction and can be considered a constant.
you use an expression with an infinitesimal amount of time ($\mathrm{d}t$), when $\mathrm{d}t$ is everything but infinitely small...

Let me note bleach $\mathrm{ClO^-}$, and bromocresol green "BCG".
I assume that the reaction is monitored through the changes in absorbance, that is, the concentration of BCG. Its rate $r$ is the rate at which BCG is disappearing.
$$r = - \frac{\mathrm{d}[\mathrm{BCG}]}{\mathrm{d} t} = k \times [\mathrm{ClO^-}]^\alpha \times  [\mathrm{BCG}]^\beta $$ where $\alpha$ and $\beta$ are the partial orders.
If you use two different concentrations of bleach ($C_1$ and $C_2$), the ratio of rates will be :
$$\frac{r_1}{r_2} = \frac{k \times (C_1)^\alpha \times  [\mathrm{BCG}]^\beta}{k \times (C_2)^\alpha \times  [\mathrm{BCG}]^\beta} = \left(\frac{C_1}{C_2}\right)^\alpha$$
A simple way to get the initial rate is to measure the slope of the tangent to the graph of  ${[\mathrm{BCG}]}$ along time (dotted line).

This line crosses the time axis at $t=\Delta t$.
In this case,
$$ r_i(t=0) = \frac{\mathrm{[BCG] (t=0)}}{\Delta t_i}$$
Therefore, one can write  :
$$\frac{r_1}{r_2} = \left(\frac{C_1}{C_2}\right)^\alpha =  \frac{\Delta t_2}{\Delta t_1}$$
which kind of looks like the formula you were given.


text:
Bond Type of Lead: Metallic or Network Covalent?
I know that lead is classified as a metal, from its appearance, nonexistent band gap, and its position on the periodic table... etc. However, all of the elements above it (carbon, silicon, germanium, and tin) usually form covalent network bonding with four surrounding atoms. Why is lead suddenly different, engaging in delocalized metallic bonding when it has enough electrons?
Or is it different? Lead has a low melting point for a metal (600 K), and a lower melting point than, say, copper.

As you proceed down Group 14 the metallic bonding structure tends to become more stable than the tetrahedrally bonded structure we see in diamond, Si and Ge.  Tin is actually a transitional case.  At typical room temperature it is metallic (white tin) but upon chilling it transforms to the tetrahedral allotrope (gray tin).  By the time we get to lead we have fully gone over to the metallic phase.
Because the tetrahedral bonding structure requires each atom to contribute four valence electrons, this tetrahedral --> metallic transition may be seen as a form of the "inert pair" effect that emerges in heavy atoms.

As more shells are added, electrons are shielded from the nucleus, and become freer to wander -- i.e., behave as metals. Note even the chalcogens, starting with oxygen, wind up as semimetals by the time one reaches polonium. See http://sciencenotes.org/printable-periodic-table/

And indeed, grey tin is a semi-metal. Liquid silicon and germanium are metallic. Every time you start thinking that the 'periodic' table should be nice and predictable you come up against an 'anomaly' that seems weird.

Thanks Jon, I know about liquid Si and Ge.  If I am not mistaken, the melting of those elements going along with the change in bonding means there is a "collapse" like that in water (ice) when it melts.  So Si and Ge, like water, contract upon melting.  Tin undegoes this contraction while still solid.


text:
How to calculate the pH of a redox reaction between nickel(IV) oxide and silver?

For the following reaction:
\begin{gather}
\ce{NiO2(s) + 4 H+(aq) + 2 Ag(s) -> Ni^2+(aq) + 2H2O(l) + 2Ag+(aq)}\\
E^\circ = \pu{2.48 V}
\end{gather}
Calculate the $\mathrm{pH}$ of the solution if $E = \pu{2.23 V}$ and $[\ce{Ag+}] = [\ce{Ni^2+}] = \pu{0.023 mol/l}$.

I know that there is the Nernst equation:
$$E = E^\circ-\left(0.0592/n\right)\log Q$$
Where $E^\circ$ is the standard cell potential, $E$ is the cell potential for non-standard conditions, and at non-standard pressures or concentrations,
$$Q = \frac{[\text{products}]}{[\text{reactants}]}.$$
I know that for this problem, $n(\ce{e-}) = 4\ \mathrm{mol}$, and that
$$Q = \frac{[0.023]^3}{[\ce{H+}]^4}.$$
So:
\begin{align}
  2.23
  &= 2.48 - \frac{0.0592}{4}\log\left(\frac{0.023^3}{[\ce{H+}]^4}\right)
  \tag{1}\\
  0.25
  &= \frac{0.0592}{4}\log\left(\frac{0.023^3}{[\ce{H+}]^4}\right)
  \tag{2}\\
  1.00
  &= \frac{0.0592}{1}\log\left(\frac{0.023^3}{[\ce{H+}]^4}\right)
  \tag{3}\\
  1.00
  &= 0.0592\log\left(\frac{0.023^3}{[H^+]^4}\right)
  \tag{4}\\
  \frac{1.00}{0.0592}
  &= \log\left(\frac{0.023^3}{[\ce{H+}]^4}\right)
  \tag{5}\\
  \frac{1.00}{0.0592}
  &= \log(0.023^3)-\log\left([\ce{H+}]^4\right)
  \tag{6}\\
  \frac{1.00}{0.0592} - \log(0.023^3)
  &= -\log\left([\ce{H+}]^4\right)
  \tag{7}\\
  \frac{1.00}{0.0592} - \log(0.023^3)
  &= -4\log\left([\ce{H+}]\right)
  \tag{8}\\
  \frac{\frac{1.00}{0.0592} - \log(0.023^3)}{4}
  &= -\log\left([\ce{H+}]\right)
  \tag{9}\\
  \mathrm{pH} &= -\log\left([\ce{H+}]\right)
  \tag{10}\\
  \frac{\frac{1.00}{0.0592} - \log\left(0.023^3\right)}{4}
  &= 5.45
  \tag{11}
\end{align}
Therefore, $$\mathrm{pH} = 5.45\tag{12}.$$
Yet, the online assignment here says my answer is wrong.

It even gives me practice versions for other variations of this problem, and yet I still always get a wrong answer using the same methods.  Would someone be so kind as to point out any errors I may have made while calculating this answer?

For the following reaction:
\begin{gather}
\ce{NiO2(s) + 4 H+(aq) + 2 Ag(s) -> Ni^2+(aq) + 2 H2O(l) + 2 Ag^+(aq)}\\
E^\circ = \pu{2.48 V}
\end{gather}
Calculate the $\mathrm{pH}$ of the solution if $E = \pu{2.23 V}$ and $[\ce{Ag^+}] = [\ce{Ni^2+}] = \pu{0.023 M}$.

The important thing to understand that was not immediately obvious to me and a lot of other people was the fact that $n = 2$, and this changes how you calculate your answer.
$\ce{Ni}$ does not change oxidation state in its transition from $\ce{NiO2}$ to $\ce{Ni^2+}$, as its oxidation state in $\ce{NiO2}$ is $+2$.  The 4 $\ce{H^+}$ ions also do not change oxidation state in their transition from $\ce{H^+}$ to $\ce{H2O}$, oxidation state in $\ce{H^+}$ is $+1$ and oxidation state in $\ce{H2O}$ for $\ce{H}$ is $+1$ for 2 $\ce{H}$ atoms.  That leaves us with the only $\ce{e^-}$ transfer occurring to $\ce{Ag}$.  There are 2 moles of $\ce{Ag}$ atoms going to oxidation state $+1$.  Therefore, 2 moles of $\ce{e^-}$ are transferred.  Hence, $n=2$
The Nernst Equation:
$$E = E^\circ  -  (0.0592/n)\log Q$$
Where $E˚$ is the standard cell potential, $E$ is the cell potential for non-standard conditions, and at non-standard pressures or concentrations, $$Q = \frac{[\text{products}]}{[\text{reactants}]}.$$
For this problem, $n(\ce{e-}) = \pu{2 mol}$, and that
$$Q = \frac{[0.023]^3}{[\ce{H^+}]^4}.$$
Following the above calculation with this difference leads to $\mathrm{pH} = 3.34$, which is the correct solution.
\begin{align}
\tag1
2.23 &= 2.48
- \frac{0.0592}{2}\log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag2
0.25 &= \frac{0.0592}{2} \log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag3
0.50 &= \frac{0.0592}{1} \log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag4
0.50 &= 0.0592\log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag5
\frac{0.50}{0.0592} &= \log\left(\frac{0.023^3}{[\ce{H^+}]^4}\right)\\
\tag6
\frac{0.50}{0.0592}
&= \log\left(0.023^3\right) -\log\left([\ce{H^+}]^4\right)\\
\tag7
\frac{0.50}{0.0592} - \log\left(0.023^3\right)
&= -\log\left([\ce{H^+}]^4\right)\\
\tag8
\frac{0.50}{0.0592} - \log\left(0.023^3\right)
&= -4\log\left([\ce{H^+}]\right)\\
\tag9
\frac{\frac{0.50}{0.0592} - \log\left(0.023^3\right)}{4}
&= -\log\left([\ce{H^+}]\right)\\
\tag{10}
\mathrm{pH} &= -\log\left([\ce{H^+}]\right)\\
\tag{11}
\frac{\frac{0.50}{0.0592} - \log\left(0.023^3\right)}{4} &= 3.34
\end{align}

Nice work!  We like questions where the asker is actually engaged.

For your future assignment it is always worth work with two half-reactions: oxidation and reduction: Given is:
$$\ce{NiO2(s) + 4 H+(aq) + 2 Ag(s) -> Ni^2+(aq) + 2 H2O(l) + 2 Ag^+(aq)}\quad
E^\circ = \pu{2.48 V}$$
Oxidation half-reaction:
$$\ce{Ag (s) <=> Ag+ + e-} \tag1$$
Reduction half-reaction:
$$\ce{NiO2 (s) + 4 H+ + 2e- <=> Ni^2+ + 2H2O (l)} \tag2$$
To cancell electrons, add $(1) + 2\times(2)$:
$$\ce{NiO2(s) + 4 H+(aq) + 2 Ag(s) -> Ni^2+(aq) + 2 H2O(l) + 2 Ag^+(aq)}\quad
\tag3$$
Now you know the number of cancelling $\ce{e-}$ is $2$. Thus, $n = 2$ for the Nernst equation. Thus, you would never miss it again.

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


text:
Why is the magnesium(II) ion preferred over other ions in chlorophyll?
Chlorophyll has a $\ce{Mg^2+}$ ion. Why is it preferred over other ions? For example, what happens if there is $\ce{Zn^2+}$ or $\ce{Ca^2+}$ or any other (divalent) cation instead of $\ce{Mg^2+}$?

$\ce{Mg^2+}$ is preferred because the entire huge and complicated molecule is designed just for it. With any other ion, the delicate machinery of photosynthesis will not work.

Why is the delicate machinery wont work? That is, perhaps, my question.

An excellent book is Kaim, W.; Schwederski, B.; Klein, A. Bioinorganic Chemistry: Inorganic Elements in the Chemistry of Life, 2nd ed.; Wiley: Chichester, U.K., 2013.; pp 59-67 discuss in detail the physical processes occurring in the photosynthetic reaction centres. From my brief skim through of it it is similar to what porphyrin has provided in his answer.

Chlorophyll in plants has two main functions. First, it facilitates energy transfer of an absorbed photon's energy via the many other chlorophylls that constitute the antenna pigment-protein complex to the reaction centre. This means that with $> 95~\%$ efficiency the energy of a  single absorbed photon reaches the reaction centre which consists of a dimer of chlorophyll molecules (the special pair), two pheophytins ($\ce{2H}$ replace $\ce{Mg}$) and two other chlorophylls.  The second role is in electron transfer. Electron transfer begins from the excited special pair and the electron is passed to other nearby chlorophyll and pheophytin molecules and then to quinones/iron sulfur complexes depending on the particular type of reaction centre. The exciton interaction between the two chlorophyll in the special pair means that it can be treated as a single entity. This interaction lowers its excited state energy and so traps the energy arriving from the antenna.
It is possible to make chlorophylls with other metals instead of magnesium, but if these contain heavy or paramagnetic ones they will competitively quench the chlorophyll's excited state (via a spin-orbit coupling or 'heavy atom effect') and reform the ground state. Consequently very little if any energy transfer will occur. This means that insufficient energy will reach the reaction centre and so the efficiency of photosynthesis will be dramatically reduced.
Magnesium chlorophyll has a unique property which is that the absorption and emission spectra overlap exceptionally well meaning that energy transfer to nearby molecules by the Forster (resonance) mechanism is very favourable. As shown in the figure below the energy may move among $\approx 100\ \ce{Chl}$ molecules and reaches the reaction centre in a few ($ < 10\ \mathrm{ps}$) picoseconds. This means that light harvesting antenna have very poor fluorescence yield since most of the absorbed energy reaches the special pair, where is quenched, before a molecule has had a chance to fluoresce because the Chl fluorescence lifetime is $\approx 5\ \mathrm{ns}$. Changing the central metal will change the absorption and emission spectra, most likely in a way to be less effective at energy transfer simply because magnesium chlorophyll has almost the best possible overlap of absorption and emission spectra of any molecules.
The energetics of electron transfer are very delicately balanced. The electron has to move from the excited special pair to the final acceptor (say a quinone) through the steps $\ce{(Chl)_2^* -> Chl -> Phe -> Quinine }$ with as high an efficiency as possible (the whole process takes only $\approx \pu{250ps}$ in bacteria, and ten times faster in plants), but the rate of each back reaction on each step has to be as low as possible. Other metals in Chl will change the redox and even a change of $0.1~\mathrm{V}$ will be important, and so make this process less efficient. The reason for this is that in normal photosynthesis electron transfer is almost at the peak of the Marcus rate vs. free energy curve, thus increasing or decreasing the exothermicity will make the electron transfer less efficient.
All of this illustrates that photosynthesis is, via natural selection, highly adapted, and so small changes can have dramatic effects. Under different conditions, it may be possible for alternative schemes to do the same job.
The figure shows the antenna (light harvesting) chlorophylls in the protein complex of plant photosystem 1 and the reaction centre. (The protein residues have been removed for clarity). The reaction centre is at the centre, shown within the dotted line with the special pair shown edge on. The final electron acceptors are $\ce{FeS}$ complexes but they are not shown and would lie directly below the special pair. The molecules to either side of the special pair are the accessory chlorophylls through which the electron from the special pair travels.  The structure is modified from PDB entry 1JBO.

Magnesium ions tune the electronic properties of the tetrapyrole ring of chlorophyll. Why would the delicate photosythetic machinery fail if magnesium was substituted with calcium or zinc ions?
My bet is that magnesium has been selected over millions of years of evolution because of its availability and solubility: it is five times more abundant  than calcium in sea water. Zinc is much more scarce.
Keep in mind that plants contains loads of chlorophylls, so as to harvest all the photons they can.

I think you should provide a reference for this.

How Mg2+ ions tune the properties of tetrapyrrole ring? What is exactly happening if it is there or if it is not there?

@adianadiadi The chelated ion perturbates the energy level of the chlorophyll, which is reflected in its absorbtion spectrum, its reduction potential, etc. For example, chlorophyll without Mg is called pheophytin, and it is a crucial cofactor in reaction centers.

So there are two parts to this story:
1) If the magnesium in chlorophyll is replaced by some other metal it may damage the plant

The
  chlorophyll–heavy metal complexes may cause an impairment of the photosynthetic function and
  this, as a final consequence, may lead to death of the plant

although, that paper only looks at in vivo replacement. They also conclude that copper actually replaces the magnesium, wheras zinc may not:

Both copper and zinc form complexes with chlorophyll, but with different affinities.
  While the Cu–Chl complex is predominantly formed by replacement of
  the central Mg-atom of chlorophyll (“central complex”) with copper, zinc is much
  more engaged in the formation of a cyclic 6-membered peripheral chelate (“peripheral
  complex”), playing a coordinating role between two O-atoms

2) There are other tetrapyrrole compounds that bind to other metals. Haems bound to iron, corrins bound to cobalt, etc. Chlorophyll-c actually has a porphyrin ring, rather than a chlorin ring.
There's at least one study of a mutant that has zinc-bacteriochlorophyll where they report that:

Spectroscopic measurements reveal that Zn-BChl also occupies the H sites that are normally occupied by bacteriopheophytin in wild type, and at least 1 of these Zn-BChl molecules is involved in electron transfer in intact Zn-RCs with an efficiency of >95% of the wild-type RC

however, they report that the mutant does not grow, perhaps because there are not enough of these Zn-BChl molecules formed. This research comes out of the lab of J. Thomas Beatty, and the latest paper is here - they describe a crystal structure : http://www.rcsb.org/pdb/explore.do?structureId=4n7k.
I am not aware of any naturally occurring non-Mg chlorophylls.

The other answers have hit on some key details, but haven't given all the details needed to properly answer your question which is the functional importance of $\ce{Mg^2+}$ in chlorophyll. I'll address that after first reiterating the importance of ion size for binding.
Ion Size and Structure
Looking at the data presented in MaxW's answer, it's quite easy to see that the only two feasible ions are $\ce{Mg^2+}$ and $\ce{Zn^2+}$ based on their radii. This is because there is only so much space in between the tetrapyrrole ring for an ion to fit.
So why not zinc then?
Zinc is fairly inert because it has a full $\ce{3d}$ shell and thus likely would not bind very strongly or at least as strongly as $\ce{Mg^2+}$, thus I would expect magnesium to out-compete zinc for the binding site.
Functional Importance of $\ce{Mg^2+}$
The point which hasn't been addressed here is the fact that chlorophyll must fulfill a specific purpose. This purpose is to absorb photons and transfer this energy to photosystems which then use the energy in ways that I don't really know about.
Nonetheless, any ion which is present in chlorophyll must have an absorbance spectrum which closely matches the radiation spectrum of the sun because that's the only light the plant has to work with.
First, below is the blackbody radiation curve of the sun, i.e. the distribution light colors that are reaching us on earth.

(source: gsu.edu)
You'll notice that this has a peak near the blue-ish portion of the spectrum, but really we're just getting a lot of visible light from the sun. Ok, now let's see the absorbance spectrum of chlorophyll. There are actually several types of chlorophyll which all differ slightly in structure, but the most common are chlorophyll a and chlorophyll b.

This is a spectrum in vitro which tightens the peaks a bit compared to in vivo, but the maxima are the same.
As we can see, there are two absorbance maxima for both forms of chlorophyll. One in the blue region of light and another in the red. This is advantageous because it means for a single structure, we absorb a large portion of the light being sent by the sun because we're sampling from two parts of the blackbody distribution.
The two structure also do a good job of not overlapping but still absorbing visible light.
This gives another reason why magnesium is present instead of zinc. Because zinc has a full $\ce{3d}$ shell, it tends to form colorless complexes which would be pretty useless for transferring energy to the photosystems.
Also, this is why people who grow plants indoors often use blue light or red light only when growing the plant.

‘Zinc is fairly inert because it has a full 3d shell and thus likely would not bind very strongly or at least as strongly as $\ce{Mg2+}$, thus I would expect magnesium to out-compete zinc for the binding site.’ — magnesium(II) has entirely empty d-orbitals; I do not expect either to out-compete the other a priori. Also, zinc complexes are far from inert; they are very labile.

The fundamental concept that you seem to be missing is an understanding of chelation. In chelation a molecule has multiple sites which bind to some atom, usually a metal cation  to form a coordination complex. If the chelate can forma 2D of 3D hole for the ion, then it may be very specific for a particular ion since the charge, ionic size, and coordination symmetry are defined. So the gist is that the "hole" in the Chlorophyll molecule is "tuned" for $\ce{Mg^2+}$.

$$\begin{array}{cc}
\ce{M} & \text{Ionic radius of }\ce{M^2+}\text{ / pm} \\ \hline
\ce{Mg} & 86 \\
\ce{Ca} & 114 \\
\ce{Sr} & 132 \\
\ce{Zn} & 88 \\
\ce{Cd} & 109
\end{array}$$
The radii are from Wikipedia where you can sort the table on the +2 cations.
Note that in the chlorophyll molecule the chlorin group participates in an oxidation-reduction reaction to function. So the functionality chlorophyll of depends on more than just binding.

Zn is quite close to Mg. Can you add some info and reference on specific selectivity of the tetrapyrolle ring towards Mg?


text:
How to determine pH of food in order to avoid unsuitable storage containers
In an acidic (lemon-containing) sauce I had placed a metallic spoon. A few days later it had a metallic taste.
Stainless steal might be a significant source of nickel, chromium and iron [1]. Also, using aluminum utensils and packaging when cooking or storing food can contribute to the daily aluminum intake [2].
I want to avoid storing food in metallic containers which could increase my intake of metals.
The pH of some foods can be found here, but in some cases it's not of much use (when the food is a varying mixture of ingredients).
Is there an easy way to determine the pH of food? (whether the food is colored like a tomato sauce or transparent)

Well, foods with acidic pH would be acidic to the taste.

Why not to use indicator paper? http://www.kruuse.com/en/ecom/Forbrugsartikler/Forbrug_lab/Indikatorpapir_forbr/prod_290590.aspx

@JaroslavKotowski Because many foods are already colored. I imagine it would be hard to distinguish color changes caused by pH from changes caused by food color.

@IvanNeretin I don't think our tongue is that accurate. As an example honey has 3.70 - 4.20 which to me is quite unexpected.

mix food, add a bit water, filter a few milliliters and measure pH od the permeate

You can try wetting the tip of an ordinary pH paper, then put it into a sample of your food with that wet side. The capillary forces will suck up the liquid of your food and you should see a colour change in the formerly wetted area and above.
PH paper is not a very reliable method, but it should give you a general idea and it is easily performed at home.

I would suggest to use indicator paper. If you are worried that food coloration will make color determination impossible, you can analyze only sample.
you can take a sample of the food and mix it in a blender. Add a bit of water and filter the mixed sample using filter paper or activated carbon. Then analyze the permeate using indicator paper.


text:
How to balance the reaction equation of potassium permanganate, calcium oxalate, and sulfuric acid step by step?
I have been desperately trying to balance the following equation, and finally (ultima ratio) used an online program to get it done (posted the same question there as well).
$$\ce{KMnO4 + CaC2O4 + H2SO4 -> MnSO4 + K2SO4 + CaSO4 + CO2 + H2O}\tag{I}$$
No worries, balancing the K, Mn, Ca, C and S - but by then the H and O got out of my control. So this is how far I got myself:
$$\begin{multline}\ce{2KMnO4 + CaC2O4 + 4H2SO4 ->\\
 2MnSO4 + K2SO4 + CaSO4 + 2CO2 + 4H2O}\tag{II}\end{multline}$$
So the only thing left to be balanced, was the O, having 28 on the reagents' side and 24 on the product side. I started off, getting up to 36:33 O (reagents:products), and I felt that I was getting nowhere.
Anyone who would guide me through the steps following the balancing until equation II above?

You should balance this like any other redox: identify the oxidant, look how many electrons it requires, identify the reducing agent, look...

This is a redox reaction. While it can be balanced by the method you describe, it is much more reliable to balance it by the half-reaction method.

Step 1 - Identify your spectator ions (so we do not spend time trying to balance them)

Three ions appear unchanged on both sides of the reaction: $\ce{K+},\ \ce{Ca^2+},\ \& \ \ce{SO4^2-}$

Step 2 - Identify your redox pairs.

You have one manganese species on each side. They must constitute one half-reaction:
$$\ce{MnO4- -> Mn^2+}$$
Likewise, you have on carbon species on each side. They must constitute the other half reaction:
$$\ce{C2O4^2- -> CO2}$$
For each half reaction:

Step 3 - Balance O by adding $\ce{H2O}$
Step 4 - Balance H by adding $\ce{H+}$
Step 5 - Balance charge by adding $\ce{e-}$
Step 6 - Combine the half reactions

Can you take it from here?


text:
Why can't you use m-CPBA to synthesise isophorone oxide from isophorone?
Apparently you must use sodium hydroxide and hydrogen peroxide because using m-CPBA gives an undesired product

Why doesn't m-CPBA work and what does it do instead? Why does sodium hydroxide and hydrogen peroxide work?

When using MCPBA, the undesired by-product is probably formed due to Baeyer–Villiger oxidation of the ketone to the corresponding ester (here: lactone).

and why doesn't this happen with the peroxide?

Iirc the Dakin oxidation of a ketone generally only works with benzaldehydes or acetophenones with EDGs o- or p- to the carbonyl


text:
What does solubility limit mean in practical terms?
I am not really a chemist guy so please bear with me. Glycerol monolaurate has an upper limit of solubility in water of 20 μg/mL, which is enough for my experiment. Does this mean that no matter how much glycerol monolaurate I add to water the concentration will always stay at 20 μg/mL? Glycerol monolaurate is in powder form so does this mean that the powder would flow on top of water and the concentration of 20 μg/mL glycerol monolaurate would be in the water below?
Thanks.

Q1) If 20 micrograms / ml is the upper solubility limit and you added less than 20 micrograms the solution could be less concentrated but a concentration higher than 20 micrograms / ml H2O can't be achieved.
Q2) No more than 20 micrograms of glycerol monolaurate will dissolve in 1 ml of water so if more than 20 micrograms was added the remaining undissolved powder would stay separated in some way (eg. precipitate) which could be filtered (as @MaxW has stated), or a colloidal suspension could also form. If more than 20 micrograms / ml were to be used the supernatant (GML/H2O layer) could be estimated to be 20 micrograms/ml) after filtration.

It seems that the density is just below 1.00 for glycerol monolaurate , so the undissolved portion will float. If it had a very fine particle size you might be able to get some sort of colloidal suspension. // There real notion here is two phases if you add more than 20 micrograms per mL. A liquid phase that contains water and the dissolved glycerol monolaurate, and a solid phase that is the undissolved solid glycerol monolaurate.

@MaxW yes I agree, I haven't made this solution before so I don't know if it separates cleanly or not. Ill edit the emulsion part (:

You can't really "emulsify" a solid. An emulsion is for two liquids. So the excess solid either floats, sinks, or forms a colloidal suspension - or possibly some combination due to particle size variations, wettability, and surface tension.

@MaxW yes the colloidal suspension was what I meant ..

Thanks MaxW and Joel. The below poster stated that the concentration will increase.  I assume this is wrong?

Must be referring to the overall concentration of the GML in the container not dissolved in solution , maybe the colloidal suspension?


text:
How is knowing about the symmetry of a crystal useful
I am studying the space groups and know that we use them to talk about crystal symmetry, I would like to know what material properties may be produced or predicted from knowing the symmetry or lack thereof in a crystal system of a material?

How is this primarily opinion based? He's asking for examples, not opinions.

I concur, this post should be edited to be more semantically poignant, but as someone in MSE I find no cause to believe that proper answers to this question will be based largely on opinion.

Thirded. Symmetry is one of the fundamental tools in physics and chemistry, many specific examples can be provided. And I have to say I don't really see how rewording will markedly improve "I would like to know what material properties may be produced or predicted from knowing the symmetry of lack thereof in a crystal system of a material?"

Thank you, to be fair my original question didn't specifically ask for examples. Is there a way to get an answer on this?

@chilliBeanDream The question currently sits with three re-open votes. Once it reaches five, the question will be unlocked for answering.

I think you may have better luck in Physics SE. Nevertheless, let me chip in. The periodicity of the crystal (lattice)  influences the form of its wavefunction. This is due to Bloch's theorem. This doesn't mean that the wavefunction is completely determined just by its periodicity. Other things such as the types of atoms, bonds, surfaces, defects, etc influence the wavefunction. Nevertheless, many properties (e.g bandstructure) are dependent on the wavefunction. Ergo they are influenced by the symmetry.

Properties that are directly dependant on crystal symmetry are

birefringence
piezoelectricity
(obviously) x-ray and electron diffraction pattern
pyroelectricity
nonlinear optical properties
...


text:
Correct term/word for bias in calculations?
I ran an experiment recently and made several approximations in the calculations that led to the results being around 2 orders of magnitude less then the literature values.  I will be re-doing the calculations but in the meantime I need a word to describe what is wrong with this data set (ie what the effect was on the data by making the approximations).
I thought saying the approximations resulted in a "statistical bias", or "analytical bias" might be right, but this is a little different that regular bias because the data is all fine and unbiased, its the approximations/calculations that are making the final results off by ~100x literature values.
I'm not sure if this is the right place to ask this, so if you think it belongs on the statistics stackexchange, please let me know.
If you're interest (or if it matters), the approximation was to ignore epithermal neutron capture in NAA analysis.

Two orders of magnitude might best be termed "calculation error". More seriously, it is helpful to know where making approximations will have the largest impact on the answer to guard you against cutting too many corners in your analysis. This may suggest that some approximations are safe but others are catastrophic.

I believe the fundamental notion that you're looking for is a "systematic error in the data analysis".
A "statistical bias" would imply some sort of sampling problem which doesn't seem to be the case from your description.
An "analytical bias" would imply a different/better laboratory technique which again doesn't seem to be the case. So although the overall results are biased, you don't need to redo the experiment itself but rather reanalyze the data using the proper calculations.

The individual approximations are indeed "biased", if they add up to systematically underestimate the result. It is definitely necessary to stress that the approximations are independent(ly biased), because a single approximation that leads to a result two magnitudes off would in most cases be regarded as just invalid, stupid, wrong.


text:
Lubricants and rubber
I've been reading a lot of conflicting articles/threads pertaining to which lubricant is compatible with rubber/plastics and I'm stumped.
From what I've come across, the consensus is that silicone is safe to use on rubber and that petroleum based eats at it over time. But then I heard the opposite.
What I'm wondering is, if rubber is composed of some petroleum then why would a petroleum lubricant or grease have a negative effect on it? Also what about plastics and vinyl, do they get damaged as well?

You need to know what sort of rubber. "Natural" is made from plant extracts not petroleum and is severely degraded by many petroleum-based solvents. But there are artificial "rubbers" of many varieties that are totally different and react very differently with solvents. And there are a lot of different types of solvents. So: no easy way to know the general answer. Bottom line research the material you are working with specifically or risk problems.

The simple version:
Petroleum-based rubbers can be dissolved by petroleum jelly eating away at the mechanical integrity.
The slightly longer version:
Petroleum jelly incorporates its way into the rubber, changing the plasticity (ie how rubbery or stiff) the rubber is.  Changes in plasticity change the mechanical strength and toughness such that the rubber is then outside the material property specifications of the product it is with/on/in, and hence fails.
Even longer version:
Most rubbers have small molecules inside them that help the molecular chain is the rubber move.  These molecules are called plasticizers (and occasionally tackifiers).  They are small molecules usually 200-500 MW which is less than the several hundred thousand of the rubber itself.  Petrleum jelly is basically hydrocarbon of roughly the size of the tackifiers.  The petroleum jelly molecules can 'wiggle' their way into the rubber, moving and adjusting how the existing plasticizers interact with the rubber.  In some cases the plasticizers may be extracted out, also changing the interaction, fundamentally changing the energies involves in intermolecular motion which is then related to toughness and strength (many steps in that derivation omitted).  In extreme cases in poor quality rubbers, some of the rubber molecules themselves can be dissolved by the petroleum jelly.  Changes in toughness, plasticity and mechanical integrity are common when petroleum jelly interacts with rubbers, and hence it should be avoided as a lubricant.
As you have pointed out silicone oil is a costly replacement, but not unbearable more expensive for most jobs.

I see. I figured if a rubber had a petroleum base then greases with petroleum would be safe but in fact it causes the composition to change since it's made of the same base which allows it to change the characteristics?

The same 'base' allows it to solveate/dissolve, a chemical phenomena.  The change in characteristics noticed is normally a physical phenomona, based on breakdown in mechanical qualities (see my more extended answer posted after your comment)

Thanks for that. So now I'm curious about the rubber boots used in some automotive parts that hold in petroleum based greases like chassis grease. Are these rubbers made without a petroleum base?

Probably.  But I can't know for sure without a pair in my hands.

What about stuff like armor al or tire shine. Are they safe to use? The bottles don't list the chemical make up and now I'm wondering if some have petroleum additives that could degrade the surface over time.

I am sorry I do not feel comfortable validating every product to you, particularly when I cannot speak to their contents  A wikipedia search for Armor All indicates its likely composition.  .You'll have to study some basic organic chemistry or hire a consultant.


text:
Spreading of Chemicals (chemistry lab)
Is it possible to spread a chemical or the danger of a chemical if the chemical has dried up on a piece of paper? For example if in a lab I have gotten a certain chemical on my lab paper but the chemical has dried up does that lab paper now pose a threat to my health and the health of others? Also if after touching the lab paper I touch my jacket and then after touching my jacket I eat something later on would I be in danger?
The chemical I am referring to is copper sulfate.
I have not experienced any irritation or burns (maybe some weird stomach or digestive problems).
Thank you

Yes, depending on the chemical of course.

You need to specify, there could be lots of different chemicals there you know

No offence, but you should not work in a laboratory if you ask questions like this, and your supervisor ought to be hanged for letting you. You are a danger to the health and safety of everyone around you, including yourself. Unless all the chemicals around you are indeed totally harmless, which I can't know, but doubt. Get your supervisor to give you a thorough safety briefing, and try to learn some basic chemistry.

I second Karl's comment. Think about it. This is your only life.

@Karl, I understand you sentiment completely, but it could have been phrased less harshly.

Always remember: the sign of a true chemist is that they wash their hands before going to the toilet.

There is no sensible single answer to this: it depends on the chemical (and you really should read the safety notes before using anything or you will cause severe trouble.)
Consider these scenarios.
You are working with a solution of common salt and you spill it on your lab coat. No problem at all; you could eat your dinner off the lab coat as it is the same slat you might sprinkle on food. But you spill a solution containing some mercury salts on your coat, get it cleaned and decontaminated immediately as it will eventually poison you even if you don't use it to wrap your dinner.
Or you spill some ether on your lab coat. Don't stand near naked flames until it has evaporated, but once it has gone, it has gone with no lingering effects. Spill some nitric acid/sulphuric acid mixtures on it and even if you wash it off before wearing it again you might still have the problem that it has created some nitrocellulose, which can spontaneously combust, taking you with it. Equally badly, you spill some tellurium salts on your coat. They might not kill you but you will be shunned for weeks as you will smell strongly of something like garlic, but more offensive.
Bottom line: read the safety instructions and act accordingly: there is no general rule.

I disagree, there is a general rule: Things that are done in a lab shall remain in the lab, no exceptions. Safety is number one priority!

@Martin-マーチン this is a good rule, but insufficient. Just because your lab coat never leaves the lab doesn't make it safe (a contact poison on your coat will still poison you in the lab if you keep wearing it).

What do you mean exactly with "dried up" ? Did you spill hydrochloric acid on your lab book, and you worry about the paper being corrosive by now ?
It is always best to know what you are dealing with and its reactivity.
As rule, what belongs in the lab stays in the lab. This means lab books and labcoats. And washing your hands before eating. If you suspect something has been contaminated, the best thing is to treat it like hazardous waste, dispose of it accordingly, and do not bring it outside the lab/at home, etc. These are only basic safety measures. Stuff does not "disappear". If it evaporates, it goes somewhere else. If it reacts, you have another species to deal with. If the solvent evaporates, you are left with a product in the solid state, all the more reactive...
Can the danger/toxicity propagate to the products of a reaction ? Yes, eg., if you prepare a cyanohydrin using a toxic cyanide salt, you can still get cyanide poisoning if you are exposed to the cyanohydrin.
Ask your safety officer.


text:
Simplified Chromatography - will this work?
I am interested to separate a sample matrix via a method very similar to chromatography. The reason I am asking about this is because I am planning to integrate this method with my SERS application.
This is what I have in mind:

A droplet of sample is placed at position X.
50/50 Methanol/water is dropped at position Y.
The chromatography plate is outlined as red. It is made of silica gel.
This plate is placed flat on a table.

Assuming:

The solvent does not overflow to the sample drop area.
The solvent can only travel via the direction of the arrow
The solvent will dissolve the sample while it is travelling.

Will this actually work?

What does 'work' mean in this context?  Are you asking if there will be a chromatographic separation?

Why not just do TLC, since that is pretty much what you have described the set-up for?

@ringo Because TLC requires evaporation chamber and a bunch of other stuff. I am building a special type of stripes that cost less than a dollar for my SERS application.

@Lighthart Yes. I am referring to sample separation

You can do TLC without a vapor chamber.  The separation isn't as good.


text:
Can one predict the entropy change for aqueous phase reactions without calculation?
I have seen several sources that claim aqueous solutions always have more entropy than pure liquids or solids, however, some aqueous ions have negative standard entropy.  This opens the possibility of solubility reactions with negative standard entropy change.  For example,
$$\begin{align}
\ce{NaF (s) &<=> Na+ (aq) + F- (aq)}&&\Delta S^{\circ} =-6.26 \ \mathrm{J/(mol\cdot K)}\\
\ce{AlCl3 (s) &<=> Al^3+ (aq) + 3 Cl- (aq)}&&\Delta S^{\circ} =-261.5 \ \mathrm{J/(mol\cdot K)}\\
\ce{AlF3 (s) &<=> Al^3+ (aq) + 3 F- (aq)}&&\Delta S^{\circ} =-429.58 \ \mathrm{J/(mol\cdot K)}
\end{align}$$
I have used thermodynamic data from this source.
So it seems to me that it is impossible to easily predict whether the entropy change for a reaction involving an aqueous phase is either positive or negative without calculation (unless a gas phase is involved, in which case the aqueous phase is moot)
Am I missing something, or are all the sources that claim an aqueous phase always has more entropy than a liquid phase oversimplifying the situation or just plain wrong?
I brought up the example of aluminum chloride to a chemistry instructor who then said that it is possible for molecular compounds to have negative entropy change when dissolved, but never for ionic compounds.  Yet both sodium fluoride and aluminum fluoride are ionic, and seem to have a negative entropy change when dissolved.

I understand why the standard entropies are negative, and about solvation shells. What I'm asking is whether the change in entropy can be predicted at a glance instead of looking up values and calculating as I have done.  Many textbooks claim all of these reactions should have a positive $\Delta S^{\circ}$ simply because "aqueous solutions have more entropy than solids or liquids." The instructor I talked to claimed that $\ce{AlCl3}$ will have negative $\Delta S^{\circ}$ because it is a molecular solid whereas $\ce{NaF}$ and $\ce{AlF3}$ should have positive $\Delta S^{\circ}$ because they are ionic solids. Are the textbooks and this instructor incorrect?

BTW, aluminum chloride is also pretty much ionic when dissolved.

Some ions when dissolved in water may form a solvation shell that orders the solvent molecules in a way that the entropy decrease through this ordering outweighs the entropy increase through the ions leaving the crystal structure. Then you would have an overall decrease in entropy when dissolving the salt.
Small ions with high charge have a tendency for this.

When textbooks say (and many do) that we can predict the sign of $\Delta _{solution} S^\circ$, and that it is positive, they are wrong.  Many salts have a negative solution entropy changes, as you already saw.  The problem is that they are neglecting that water is part of the system in a solution process, and that the number of microstates for the water, and hence its entropy, can decrease when ions are dissolved in it.  In pure water the molecules are constantly changing their orientation, but in an ionic solution water molecules in the hydration sphere become oriented, and so lose orientational (really rotational) entropy.

Welcome to Chemistry SE! Take the Tour to familiarize yourself with this forum.

When a source says that the entropy change is positive when a solid/liquid is dissolved in water, what they're trying to say is that the system becomes more disordered relative to the solid or liquid alone. This intuitively makes sense, on the basis that the now-dissolved ion/molecule is "freed" from the bonds that once held it in place. It can roam freely through the entire volume of the solution now, rather than when it was all clumped together as a solid or liquid.
Now, why are the calculations seemingly showing otherwise? The $\Delta S$ you have calculated above is the change in terms of hydration rather than dissolution. A negative $\Delta S_{hydration}$ is sensible because some ions do in fact impose a local order (relative to the proton) whereas others do not. However, if you want to look at the entropy of solution (which is then comparable to enthalpy of solution), you have to consider the case of infinite dilution.
Infinite dilution, by definition, is such that no more change in concentration is achieved despite adding more solvent. Assuming that the substance dissolves in the first place, at infinite dilution the individual solute molecules/ions are probably substantially far apart from each other such that they no longer interact with each other - they are totally free from each other but not the solvent. So relative to their 'bound state' of solid and liquid, dissolving the substance will then result in a $\Delta S$ that must be positive.

Note that the reference state is not one with infinitely dilute concentration (which would result in a nonsensical value of the entropy), but rather the interactions between ions are as in an infinitely dilute solution.

The order of decreasing entropy from the state with highest to lowest entropy is- gas>aqueous>liquid>solid. Hence, just know that the entropy is positive when state changes from either liquid or solid to aqueous and it is negative when there is change of state from gas to aqueous.


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Selectivity in Suzuki coupling of 2,4-dichloropyrimidine
I've seen many papers (see e.g. Synthesis 2010, 16, 2721–2724) in which the 4-chloro group of 2,4-dichloropyrimidine is selectively reacted under Suzuki coupling conditions. Is there any way where this regioselectivity of the coupling can be predicted?

There are the Handy rules for predicting regioselectivity of cross-coupling for polyhalogenated heterocycles, based on the 1H NMR shifts, but for pyrimidine there is a general preference for cross-coupling at C-4 and C-6 which runs contrary to the Handy predictions. See also Baran group meeting on heterocycle haloselectivity


text:
What reactions does this steel cold-bluing solution undergo?
Bluing steel generally refers to the process of forming a protective layer of $\ce{Fe_3O_4}$ ("black iron oxide") that prevents normal rust from forming. The chemistry behind the traditional industrial process for bluing steel involving multiple baths in hot alkalis and acids ("hot bluing") is reasonably well documented (and is not the focus of this question).
My question is about a different procedure referred to as "cold bluing" in which steel parts are treated with a solution at room temperature and blacken instantly without heat. The ingredients from the MSDS of a popular cold bluing product are:

Phosphoric Acid
Copper (II) Sulfate
Nickel (II) Sulfate
Selenous Acid

(The compounds are listed in descending order by percentage.)
Phosphoric acid seems like a reasonable ingredient. It's well known as a rust remover and iron phosphate protects some steels from oxidation. Unfortunately, from there the ingredients seem to become increasingly perplexing.
Copper sulfate isn't itself an unusual compound, but I'm not sure what it's doing here. Nickel sulfate is even more confusing, but nickel salts are toxic so it must be there for a pretty good reason.
The relevance of selenous acid escapes me. All I really know about it is that it's an oxidizing acid and is basically selenium dioxide dissolved in water.
Treatment of steel with this solution causes a black coating to form almost instantly. I've been starting to wonder if cold-bluing is actually "plating" the similarly colored copper (II) oxide onto the steel rather than forming black iron oxide.
If this is the case, I would expect the reaction to be something vaguely like:
$\ce{Fe + CuSO_4 +\ ?-> FeSO_4 + Cu +\ ?}$ (copper is more noble)
$\ce{Cu + H_2SeO_3\ +\ ?\ -> CuO + Se +\ ?}$ (is selenous acid really strong enough to oxidize copper?)
...but this doesn't involve the nickel sulfate at all and just generally feels wrong.
So, overall my question is:
What is the black compound that forms on the outside of steel when treating it with the solution described above, and how does it form?
edit: Additional thoughts:
While there is very little information on this compound online, copper selenide is described by Sigma-Aldrich as being a black powder. I can't find any discussion of its synthesis outside of some articles about nanoparticles, but I wonder if it's a more reasonable outcome than copper oxide.

Nickel(II) is often used as a catalyst in metal redox reactions. Selenous acid is apparently doesn't do redox quickly, so maybe the nickel speeds that up?

Here's this, but it doesn't mention the nickel. Apparently copper selenide is the final product.

I believe the nickel makes another compound (possibly nickel selenide, but maybe not) that gives the product a desirable color. I base this on this patent for coating metal in copper telluride in which the author speaks of using nickel sulfate to get a desired color. Obviously copper telluride isn't copper selenide, but there are parallels, obviously.

So based on everything here so far, the end products are presumably copper selenide and nickel selenide then. That still leaves me wondering where the iron comes into play. The solution remains pristine until iron is introduced, which would lead me to believe that the selenous acid can only attack elemental copper/nickel and not the salts. It is widely reported in the industry that uses these solutions that they do not work on aluminum, but if iron starts this reaction just by being a reducing agent, I see no reason why raw aluminum (oxide scraped off) wouldn't work just as well.

Yeah, it's definitely more complex than I would have expected going into it. I know iron and sulfur have a particular fondness for each other. Maybe it has something to do with the iron-selenium interaction. Or maybe the selenious acid can only reoxidize the aluminum metal back to aluminum oxide, reprotecting it and halting the reaction?

I'm somewhat sure the phosphoric acid alone would do the trick. The rest is probably a lot of vodoo. ;-)

@Karl how would that work? The iron phosphates I'm aware of are all tan-colored. I've used phosphoric acid to remove rust before and it only seemed to react with existing iron oxide (and didn't turn anything black). The cold bluing solution turns the surface of clean steel black instantly (and the black is fairly durable). And when I say instantly, I mean like magic trick instantly, like watching the iodine clock reaction. Phosphoric acid alone definitely doesn't do that.

By "voodoo" i meant that even the inventor of this mixture probably doesn't know why it works. I would guess that the actual black plating you find is an iron oxide/phosphate mixture. Selenium is somewhat toxic, Nickel highly allergenic, and copper would cause instant electrochemical corrosion at every scratch. Just guessing, though.


text:
Does exothermic solvation mean solute is more soluble at low temp?
(I'm surprised this question or a very similar one hasn't been asked, so I wouldn't be surprised if I just failed to find a question with my search that does answer this question.)
I dissolved some $\ce{K3PO4}$ in water a while back, and the solution became very hot. I commented on this to my labmate, and he said "So that means it's more soluble at low temperature, right?" I evaded response because I disagreed but didn't have a thought-out response yet.
I think he thought of the reaction as:
$$\ce{K3PO4 -> 3K^+ + PO4^{3-} + heat}$$
But I think that's missing an important step. I think it's more like this:
$$\ce{K3PO4 -> 3K^+ + PO4^{3-}}$$
$$\ce{PO4^{3-} + H2O <=> HPO4 + OH- + heat}$$
Phosphoric acid's third pKa is pretty high, so the second reaction is pretty significant. This, I would say, is why this particular reaction is so exothermic. So, in this case, I don't really think the heat plays into the solubility of potassium phosphate.
Sometimes, though, solvation comes with a positive change in temperature. For instance, the enthalpy of solvation of sodium iodide is $\mathrm{-7.53\ kJ/mol}$. Sodium iodide, though, is still more soluble at high temperatures. I could be missing something here, though.
In the cases where it is specifically the solvation reaction that causes a temperature change, is it the true that lower temperatures always increase solubility of salts with exothermic solvation reactions (and vice-versa)?
I'd say that the answer is no because the solvation should be dependent on $\Delta H - T\Delta S$ and, since the enthalpy change half of that formula is temperature independent, as $T$ goes up, solvation should be favored by any solvation that has a positive change in entropy. However, I can't really think of any solvation that has a negative change in entropy.

I don't think it matters what goes on with the base forming an acid or all that. What matters is that the overall process--whatever happens--is exothermic. Thus, if you increase the temperature, you are favoring the side that "contains" more heat, the "reactants." Thus, the equilibrium constant with an exothermic forward reaction decreases as temperature increases, and in this case, solubility decreases; it is more soluble at low temperatures.

@YunfeiMa Well then why is potassium phosphate more soluble at high temperatures if it gets too hot for me to touch as it dissolves?

Consider Soda Pop.

@Lighthart Done. Chose some Code Red. Thanks.

CO2 has a negative solubility curve in water (Lower solubility with higher temperature). (And an exothermic heat of reaction).  But feel free to have a pop anyways.

@Lighthart Well, that would be a case where it agrees with the idea that exothermic solvation = higher solubility at low temp. But I think that also has a negative $\mathrm{\Delta S}$.

Think of this in terms of equilibrium constant.
$$K=e^{-{\Delta G\over RT}}=e^{-{\Delta H-T\Delta S\over RT}}=e^{-{\Delta H\over RT}}\cdot e^{{\Delta S\over R}}$$
Now it is pretty clear that exothermic reaction means negative $\Delta H$, which in turn means positive coefficient in $-{\Delta H\over RT}$, which consequently is a decreasing function, as is the whole exponential expression. The entropy change (or indeed the sign of it) does not matter.
The example of $\ce{CaCl2}$ is irrelevant, for anhydrous $\ce{CaCl2}$ is never the phase in equilibrium with the solution. Roughly speaking, one simply cannot just up and dissolve $\ce{CaCl2}$. In effect, there would be two reactions: first the hydration, which produces $\ce{CaCl2.6H2O}$ (that's the exothermic part), and then the dissolution of that.
The implicit approximation we relied upon is that our $\Delta H$ and $\Delta S$ won't change much throughout the dissolution process. This fails when it comes to insanely well soluble things (think of $\ce{NaOH},\;\ce{KOH}$) where by the time we reach equilibrium, we might even have more solute than water. Naturally, the properties of such solution would be quite different from those of pure water.

So, based on this, the case of $\ce{K3PO4}$ must be exothermic due to the acid-base reaction or something else other than the actual solvation process because its solubility goes up with temperature, right?

I'm not sure. This might fall in the last category (where the assumption of constant $\Delta H$ fails).

Good point. It is very soluble.

No. Here is the data for the solubility of $\ce{CaCl2}$ from Wkipedia which dissolves exothermically.
     Temp    g/100 g Water
       0          59.5
      10          64.7
      20          74.5
      30         100
      40         128
      60         137
      80         147
      90         154
     100         159

So the solubility of $\ce{CaCl2}$ increases as the temperature increases even though the dissolution is exothermic.

Doesn't answer the question.

Unless of course you were purposely curt, in which case: kthxbi. I get the feeling, however, that you just didn't read the question fully.

Huh? You asked "... is it the true that lower temperatures always increase solubility of salts with exothermic solvation reactions?... " I gave a counter example which disproves the premise. So the answer is no.

And used exactly the reaction I mentioned where I said that I don't think it's the solvation that causes the exotherm in that case, but rather the acid-base reaction that occurs afterwards. So the answer is unaffected by your response. Unless you care to describe how it isn't due to the acid-base reaction in your answer.

Ok, you're right I didn't read and process all you question. I found a different black swan - $\ce{CaCL2}$. So it can't be "always."


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How to deterime the number of equivalents of Grignard reagent used for this reaction?
I know that the amount of substance of Grignard reagent used is 5 moles. But the answer given is 2. Is this anything to do with 'equivalents'?

What is the acetaldehyde even doing in this reaction?

The trouble here is that the number used is ambiguous as the grignard is doing two different things: three equivalents deprotonate active alcohols (why not protect them with something else?) and two react to appear in the product. Maybe the question his only interested in the equivalents that get into the product?

@matt_black If there are no side reactions, the reason not to protect them is because PhMgBr is a relatively cheap, commercially available reagent. Protection could waste precious material.

@SendersReagent as finally alcoholic and amino groups are as it is (i.e. not reacted with grignard as base) it must be ensured that finally no PhMgBr is left. So first acetaldehyde uses all left grignard, and then H+ protonates acidand amino group. {Therefore , this information gives us a hint that grignard reagent is in excess.}

Acetaldehyde seems like a much harder way to go than acetone. Maybe they were worried about making an acetylide hemiaminal ether if they use acetone.

No you are right it should be five. there are three dissociative protons thus one from the secondary amine, and two from the hydroxy groups.  These must be reacted first then you need two to react with the ester. The question was made by a human and the book answer of 2 is probably just an error from such.

You only use two equivalents of the Grignard reagent. The other three go to waste. I think it's a question of semantics rather than an error of a human.

@Martin-マーチン I see how you might think that, but if you break down the question it asks for the value of $x$.  If $x$ is not greater than or equal to 5, there reaction will not proceed to completion.

It's bad question as deprotonation of those groups would lead to side reactions


text:
Is it possible to write a redox reaction with two oxidations and one reduction together?
Let's take the Redox reaction between ${MnO_4}^-, {C_2O_4}^{2-}$ and  ${SO_3}^{2-} $
Is it possible to write the complete reaction as below or do I need to consider the two reactions separately?
$\ce{4MnO4- + 5C2O4^{2-} + 5SO3^{2-} -> 4Mn^2+ + 10CO2 + 5SO4^{2-}}$

I wouldn't try to combine these reactions. You would want to consider both reactions and maybe look up the standard reduction potentials of all reactants and products involved to give yourself an idea of what is oxidizing what and what is reducing what. In general, you should let the SRP's be a guide for trying to figure out what actually happens in these reactions.
The reaction you wrote doesn't really describe anything chemically. We have to make sure not only the math works out, but also the chemistry. Also, your charge balance is wrong, so even if the chemistry did work out, the reaction would be impossible. You seem to have 24 (-) on the reactants side and net 2 (+) on the products side.
Remember to always check BOTH your mass (MB) and charge balance (CB) work out when you're doing REDOX problems.
The following two reactions are the acidic balanced REDOX reactions for the the species you have written. Of course, these may not actually be the real reactions that would occur. For that, you would have to look things up in tables in your book.
$\ce{6 H+ + 2MnO4- + 5SO3^2- -> 5SO4^2- + 2Mn^2+ + 3H2O} \quad\text{(CB: -6 : -6)}$
$\ce{16 H+ + 2 MnO4- + 5C2O4^2- -> 10 CO2 + 2Mn^2+ + 8 H2O} \quad\text{(CB +4 : +4)}$

With two species being oxidized the overall reaction is complex. Given some starting concentrations it is unlikely (more like impossible...) that both reactions proceed at some constant relative rate. So the net overall reaction would depend on what the absolute and relative concentrations of the two species being oxidized. So trying to think of one overall reaction is almost certainly likely to lead to problems.


text:
What is the mechanism for the conversion of thiourea to diazomethane?
I really want to know how this reaction works:
\begin{align}
\ce{SC(NH2)2 + 2 NH3 &-> CH2N2 + 2 NH4+ + S^2-}\\
(\ce{S=C(NH2)2 + 2 NH3 &-> H2C=N=N + S(NH4)2 })
\end{align}
This seems like it ought to be a pretty simple mechanism, but I don't know a lot about organosulfur chemistry and I am having difficulty explaining this reaction to myself. This reaction will also occur with thioacetamide, forming N-mathanimine ($\ce{CH3-N=CH2}$) if that helps. Also, thiourea does thione-thiol tautomerism.
MY friend and I scribbled some arrows. I don't think it's a good mechanism, but is this the general idea?

Does this make any sense?

Have you a reference or source for the reaction?

I am doing the chemical bath cadmium sulfide precipitation, so my real mechanism may involve coordination around cadmium (II), but the J. Chem. Educ. paper says "it is well-known that (NH2)2CS yields S2-ions in basic media", so I assume the reaction between thiourea and ammonia works even without the cadmium (Ibanez, J. G. et al., J. Chem. Educ. 68(10), 872.).
This paper has it reacting with hydroxide to form water (rather than ammonia to ammonium), but it shows that CH2N2 is indeed the byproduct: http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=1022931


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Absorption of Diazepam/Valium
Why diazepam absorbs quickly when taken with metoclopramide? Does it because diazepam is absorbed from intestine like acetaminophen? If it does, why presence of food increases the absorption of diazepam. Meals can slow down gastric emptying and delay diazepam reaching small intestine. So if diazepam is absorbed from intestine, presence of food tends to slow down diazepam absorption, correct?

There are a lot of questions here.  I think this is better suited for [Biology.SE], but I fear they will close it as being too broad.  A more general question about absorption might be a better route to go initially.


text:
Difference between aldehydes and ketones
Why are ketones and aldehydes two different categories? To me it seems like aldehydes should be a subset of ketones. Since the general formula for ketones is CnH2n+1COCmH2m+1 and for aldehydes it is CnH2n+1COH which is just the formula for ketones but with m=0.
I am not looking for an opinion, but rather an explanation behind the reasons to have two different groups. Are there any inherent differences between them in how they react with other compounds or how they are produced?

Ketones are to aldehydes as ethers are to alcohols. You could extend this to water. Is water formalcohol? Once you begin to explore the periodic table deeply enough, you will see that there is no distinctions really exist. Only trends. The same is true of organic structure. Organic chemistry is a continuum and the groups we create are constructs -- but they are not arbitrary. Especially in organic chemistry, we group things into these categories because they help us to make assumptions based on consistent trends in reactivity. Ketones and aldehydes have significantly different chemistry. There is only a minor difference between the chemistry of acetone and butanone, but there is a pretty big difference between the chemistry of acetone and acetaldehyde.
To make a long story short, aldehydes are categorized differently because their chemistry varies significantly enough that different sorts of reagents are needed for the same types of reactions.
Aldehydes are much more electrophilic. This is because they don't have the electron donating effect of an alkyl group. Aldehydes have much more acidic alpha-protons. Aldehydes are also more easily oxidized. Aldehydes pose problems different problems in synthesis. For Their non-acidic proton, for example can be hard to remove for substitution.
An interesting example of a useful tool for dealing with aldehyde protons is dithiane chemistry. You can use a dithiane protecting group to make the aldehyde proton acidic, substitute it, then put the oxygen back on in a deprotection step.

I agree, but I do think that the analogy between ether/alcohol and ketone/aldehyde is a stretch. Ethers do (or don't do) vastly different chemistry from alcohols. Same could be said for aldehydes/ketones, but the difference is much less extreme, and the overlaps is far greater.

good point. I guess I'm just trying to say we split them up because their chemistry is different enough that the category is valuable.


text:
What is the meaning of "charge separation" in resonance?
I've read that the resonance structures in which there is lesser 'charge separation' contribute more to the resonance hybrid than the ones with greater charge separation.
But I am unable to understand why it is so.
I mean isn't it correct that the presence of two charges on adjacent atoms leads to "less stability" due to bond stretching or/and compression?
But apparently, the actual observation is contrary to this argument.
I guess I'm actually confused about the meaning of 'charge separation'? Does it refer to the distance between the location of the charges or the difference between their amounts?

Resonance structures have nothing to do with any observation, because they exist only in our mind.

@IvanNeretin  I understand that, but the contribution of the structures can be observed right? What I am asking is why the contribution of structures with less charge seperation more

You might want to clarify your question. I think there are two concepts being mixed up here.
If I interpret properly:
In a simple argument for a neutral molecule, such as hexatriene, the Couloumbic interaction between the charges would not favor charge separation.

Simple classic electrostatic should help. I would say that if there must be formal charge separation, then the less unstable limiting form is that whit minimal distance between them, ie minimal "formal electrical dipole". So you have to look for what minimise separation, both for a formal physical distance as well as the overall amount of separated formal charges.

There are two ways to think of charge separation and both make resonance structures less stable:

Moving electrons from a bond to a single atom to generate a cation/anion pair. For example:
$$\ce{R2C=CH-CH=CR2 <-> R2C=CH-\overset{+}{C}H-\overset{-}{C}R2}$$

Moving double bonds to separate opposite charges. For example:
$$\ce{R2C=CH-\overset{+}{C}H-\overset{-}{C}R2 <-> R2\overset{+}{C}-CH=CH-\overset{-}{C}R2}$$

Usually, however, the first way is meant. The general idea behind required a minimal charge separation is that in general (as long as we do not form super-octet structures) double bonds are lower in energy than separate charges on adjacent atoms as you correctly mention. Therefore, if we minimise the number of formal charges the idea is that we are concurrently maximising the number of double bonds, thereby generating a better resonance structure which contributes more to the actual composition.
On the other hand, even in the second case it is better to keep the charges together (in the absence of other effects such as strongly electronegative or electropositive atoms). This is because the charges attract each other and they do this better if they are close.

I feel it's kinda missing point. Mesomeric structures are only lame representation of real bonding. If there's no separation of charge (or ionic character) in reality, then what would be the point of writing such structure?

@Mithoron I have no argument against that. With the exception that I think that is how it is taught in school (wrongly)

Exactly, they tell us Resonant structures contribute to the Resonance Hybrid based on their energy levels


text:
Basicity of Acetic acid
Acetic acid is a monobasic acid as it disassociates into CH3COO$^-$ and H$^+$.
Why don't the hydrogen atoms of CH3COO$^-$ disassociate too?

The $\mathrm{p}K_\mathrm{a}$ of the acidic hydrogen in $\mathrm{CH_3COOH}$ is 4.76. The $\mathrm{p}K_\mathrm{a}$ of the remaining hydrogens would be of the order of 50. These atoms are not labile.
When discussing strengths of acids, we consider the conjugate base. The more stable it is, the stronger the acid is. $\ce{CH3COOH}$ is an acid because is conjugate base $\ce{CH3COO-}$ is resonance-stabilized. This means that the negative charge is delocalized on the whole $\ce{CO2-}$ group, making the anion more stable than, say, ethanolate $\ce{CH3CH2O-}$. If one of the remaining hydrogen is dissociated, this would give a dianion: $\ce{H2CCO2^2-}$, which would be extremely reactive (unstable). So such a species is simply not observed.
Perhaps you had this question because you were thinking of the acidity of the alpha protons (the one you are thinking of) in esters (like $\ce{CH3COOR}$), that have a $\mathrm{p}K_\mathrm{a}$ around 25. Malonic diesters can even be deprotonated in aqueous medium.
The alpha protons are "acidic" because the ester function has electron-withdrawing properties. So are ketones, aldehydes, nitriles, sulfones, etc. The carboxylate ($\ce{COO-}$) moiety already has excess electrons (negatively charged), so it is not electron-withdrawing.

50?! I would have thought lower than that. I've seen sources where LDA has been used to form the acetic acid dianion.

@SendersReagent Indeed

\ce{} is a better choice for chemistry formatting than \mathrm{}: http://meta.chemistry.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here