text:
Which of the following are true based on this comparison?
In the following chart, the horizontal line is time and vertical line is temperature. $K$ and $L$ are solid matters that are placed in two identical pots within the same room and are being heated up with identical heaters.

(Translation: Sıcaklık = temperature, Zaman = time, dak (dakika) = minute, maddesi = substance)
Which ones of the following are true?
A-) If $K$ and $L$ are different matters, they have the same mass.
B-) If masses of $K$ and $L$ are different, they can be the same matter.
C-) If masses of $K$ and $L$ are equal, they are different matters.
I chose "only $B$" which turned out to be wrong while the correct answer is "$B$ and $C$" and in the explanation it said "You picked the wrong answer because you think different matters can't have the same boiling/melting point." which is true because they can't, right?

In B) and C), I assume that one of the "matter"s should be "mass". Different matters can have the same boiling and melting points. It's not a unique characteristic and was used for identification in the past only because of its simple determination.

Your book refers to: substance A always has boiling point P, otherwise, it is not A. Substance B always has boiling point Q. P can equal Q, however.

A is not true - no such connection exists.
B is true (and given the same temperature indicating the phase transition it is in fact likely that K and L are the same matter).
C is true and can be explained as follows: if K and L have the same mass, they must have different specific heat capacities, defined as:
$$
c = \frac{\text{the amount of energy to be added to raise the temperature by 1°C}}{\text{mass}}
$$
This is a property that is unique for each matter - same as the melting and boiling point. It can be said to be a function of the matter. Note that the $T(t)$ graphs appear to start at the same temperature, but it takes longer to heat L than K to 85°C although the same heating is applied.

Maybe it is badly translated, I dunno. It can be water - true. If boiling point is not 100°C, it cannot be water - true.


text:
Why is an alkenyl hydrogen more acidic than the alkyl analog?
For example, in ethene the sigma bonds are stronger overall than in ethane.
Ethene (pKa = 44)
C=C 108 kcal mol-1 (+18)
C-H 111 kcal mol-1 (+10)

Ethane (pKa = 50)
C-C 90 kcal mol-1
C-H 101 kcal mol-1

This is because the degree of s character is higher in the sp2 hybridized bonds, giving more compact and stronger bonds and thus reducing acidity.
This greater s character also makes the carbon more electron withdrawing, polarizing the bond and making the alkenyl hydrogen more acidic.
It seems counter intuitive that the stronger bonded ethene H is more acidic than the weaker bonded ethane H. I thought the increased bond strength would outweigh the increased bond polarity. It is like saying: "this bond is stronger, but it is also easier to break."

Duplicate of Relative acidities of alkanes, alkenes, and alkynes?

You need to be careful here. "Stronger bond" here means higher homolytic dissociation energy, which is the energy needed to produce two radicals. Acidity is based on heterolytic dissociation.

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Confusing definition of the term precision
According to the textbook, we say a measurement is precise when series of measures are close to each other.
But in real use, I feel like the term is used in a different meaning.
For example, scale A weighs to cube of lead and displays 10.5g while scale B displays 10.4977g.
In this case, one say the measurement made by scale B is more precise than that of A. That is, scale B can distinguish more fine weights than A.
Isn't this usage of the word precise different from textbook definition? Please explain clearly the real usage of the term precise.

Just because a scale shows more digits than another, it does not need to be more precise.

"In science, the number of digits to which you report data is important because it indicates
explicitly the precision with which the data were gathered." this is a quote from http://minerva.union.edu/rodbelld/courses/geomorph/sigfigs.pdf. It says like the number of digits represents the precision of a measure. But the textbook definition refers to the set of measures and their closeness. So... I'm confused again.

I would say that this refers to a scientist reporting a figure, not a piece of lab equipment.

In colloquial English, precision is synonymous with accuracy. If you look it up on Google, the first definition is:

the quality, condition, or fact of being exact and accurate.

In the field of science, however, it takes on another meaning:

the degree to which repeated measurements under unchanged conditions show the same results

There is also measurement resolution:

the smallest change [an instrument] can detect in the quantity that it is measuring

Which is similar to the definition precision takes on in numerical analysis:

the resolution of the representation, typically defined by the number of decimal or binary digit

In short, precision can take on different meanings depending on how you choose to apply it. If you talk about the precision in a set of measured data, you will be using the scientific definition; if you talk about the precision of an instrument, you will be using the numerical analysis definition.

Thanks. But I'm still confused with this sentence(from textbook) "We should not gain or lose precision during mathematical operations. Follow these rules when carrying significant figures through calculations..." Does this sentence using the numercial analysis definition or scientific definition?

The numerical analysis definition. It's saying to have the same number of significant digits in your answer as in your raw data.

So the sentence means... the usage of significant figure rules is to preserve the resolution of initial data. Is this right?

Yes! Exactly right.

@benhwang I think it helps to clarify "precision" used for a single measurement/value, versus "precision" of a group of measurements. When used for a group of measurements, "precision" is how close those measurements are to each other. When there's only a single value, there's no "each other" you can compare among, so it's typically measurement resolution. Or to think about it another way, what's the (group) precision of all possible measurements which round to that resolution? (e.g. "1.5" could really be 1.451, 1.476, 1.5321, etc., so what's the precision of that group?)

Basic principles and the corresponding vocabulary used in metrology are standardized by the Joint Committee for Guides in Metrology (JCGM). The JCGM is made up of representatives from the International Bureau of Weights and Measures (BIPM), the International Electrotechnical Commission (IEC), the International Federation of Clinical Chemistry and Laboratory Medicine (IFCC), the International Organization for Standardization (ISO), the International Union of Pure and Applied Chemistry (IUPAC), the International Union of Pure and Applied Physics (IUPAP), the International Organization of Legal Metrology (OIML), and the International Laboratory Accreditation Cooperation (ILAC).
The definition of measurement precision (or short precision) can be found in the guide JCGM 200:2012 International Vocabulary of Metrology – Basic and General Concepts and Associated Terms (VIM 3rd edition) as well as in the equivalent international standard ISO/IEC Guide 99:2007 International vocabulary of metrology – Basic and general concepts and associated terms (VIM):

2.15
measurement precision
  precision
  closeness of agreement between indications or measured quantity values obtained by replicate measurements on the same or similar objects under specified conditions

Therefore, the definition given in your textbook is correct.
Furthermore, the JCGM notes that sometimes measurement precision is erroneously used to mean measurement accuracy, which is defined as follows:

2.13 (3.5)
measurement accuracy
  accuracy of measurement
  accuracy
  closeness of agreement between a measured quantity value and a true quantity value of a measurand

When shooting in a competition, if someone has a tight cluster of hits on a target that is in the corner of the target, you would say they had high precision.  They missed, but they were precise.  If they were in a loose cluster all around the target, but their center was the bullseye, you would say they were accurate.  But if there was a tight cluster centered in the bullseye, you would say they were precise AND accurate.
Does that help?  I'm better with analogies than with definitions.

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Solubility of oxygen in water with temperature, why does the change in solubility decrease around 40 °C?
Better set in context using a graph:

The solubility seems to fall quite persistently from 0 °C to 30 °C, then gradually slowing down in fall, then dropping faster and faster from 80 °C. Why does that change in drop happen?

The relatively abrupt changes in slope (do a derivative plot to get an even more distinct picture of what is happening here) are due to changes in the solubility function dominating the solubility behavior at a given temperature and $\ce{O2}$ concentration.  For example, Henry's Law will describe the behavior at low $\ce{O2}$ concentrations.  As Henry's Law applies to the theoretical "infinite dilution" condition, this assumption falls apart at higher concentrations.
The greater the concentration of $\ce{O2}$ the more it departs from the ideal infinite solution assumed for Henry's Law behavior.  At higher concentrations, one or more of the different gas laws then controls the solubility.  Based on your graph, it seems that there are three distinct regions, and thus three different gas laws defining the solubility of $\ce{O2}$ as a function of temperature.
One can also take an empirical approach to predicting the solubility based on experimental measurements.  This website has a live "$\ce{O2}$ solubility in water" calculator based on empirical data.  The authors use two different equations for two different temperature regimes: $\ce{0 ^oC - 30 ^oC and 30 ^oC to 50 ^oC}$.  So their $\ce{30 ^oC}$ cutoff seems to agree with the feature at that temperature in your plot.  I don't know the maximum temperatures / concentrations they used for their second equation.


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How can I calculate overlap integral (J) between fluorescence and UV-Vis spectra in FRET mechanism
How can I calculate overlap integral (J) between fluorescence and UV-Vis spectra in FRET mechanism?
I want to use this formula: J=(∫ϵ(⋋) F_D (⋋)  ⋋^4  d⋋)/(∫F_D (⋋)d⋋)
How can I find  ϵ(⋋) for every lambda in suitable range, I only have fluorescence spectrum of donor and uv-vis spectrum of acceptor. How can I calculate this integral.

You can format mathematical and chemical expressions on chem.SE using MathJax; this post contains further details.

First you have to normalise the fluorescence spectrum to unit area. Then measure/obtain the absorption spectrum in terms of extinction coefficient vs wavelength. The overlap integration is done numerically; usually the fluorimeter and spectrophotometer produce so many points so that summing values is usually ok. (don't forget to multiply by $\delta \lambda$). You must have data at the same wavelength points, otherwise you will have to interpolate one spectrum to produce this. You can obtain $\epsilon$ from a plot of optical density vs concentration. Check your calc with known compounds.


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estimation of copper iodometrically using sodium thiosulphate
What is the role of iodine here? And why only sodium thiosulphate is used for titration? Why do we add $\ce{Na2CO3}$ here. Also why  do we add acetic acid here?

There is a pretty good explanation of this procedure here.
It is stated that:

Iodometric determination of copper is based on the oxidation of
  iodides to iodine by copper(II) ions, which get reduced to $\ce{Cu+}$.

The initial reaction, not the titration, is given as:

$\displaystyle \ce{2Cu^2+ + 4I- → 2CuI(s) + I2}$

Then the titration reaction is:

$\displaystyle \ce{2S2O3^2- + I2 → S4O6^2- + 2I-}$

Sodium thiosulphate makes for an inexpensive, safe and effective reducing agent to react with $\ce{I2}$.
Since starch turns to a dark blue in the presence of $\ce{I2}$, then back to white when the $\ce{I2}$ is depleted, starch makes a good indicator for the endpoint of this reaction.
The oxidation reaction of iodides to iodine by copper(II) ions is most effective under slightly acidic conditions, with a pH of around 4-5.  This is the reason for the addition of acetic acid and sodium carbonate.  Whenever a procedure calls for the addition of a weak acid and a weak base, there is a pretty good chance that the purpose is to form a buffer solution at some desired pH.  At least in this case, that is exactly the purpose.  You will notice in the reference above that they use ammonia and acetic acid instead.  Sodium carbonate is preferred in some situations for safety reasons.

The presence of iodine is detected because of the equilibrium $$\ce{I- + $n$ I2 <=> [I_{$2n+1$}]^-}$$ which forms a complex with starch; just to be more precise. Otherwise, this is of course correct.


text:
What does it mean "A oxidizes on B"
I have just managed to understand the difference between oxidation and reduction. So basically if A oxidises B, then B loses electrons, right? But what does this formulation mean: "A is oxidized on B"? What is this "on" suppose to mean? That A is losing electrons to B or vice-versa?
Sorry if this sounds like a very stupid question, but I'm really confused by presence of "on something"
EDIT:
If you search for this

"Carbonate-based electrolytes commonly used in Li-ion batteries are
  reduced on negative electrodes"

on Google, you will find the first paper and this specific formulation.
And here you can find the paper: http://pubs.acs.org/doi/abs/10.1021/acs.jpclett.5b01727?journalCode=jpclcd

I've never heard such a phrase before. I would assume it is a typo. Any more context? What is A and B? || And if A oxidises B, then B is the one that is being oxidised; B is the one that undergoes oxidation, or gains electrons. So you have it the wrong way round.

I have added the source. Thank you for the answer!

Wait, what? B undergoes oxidation means it gains electrons? Isn't oxidation the process where a compound loses electrons? Did I get also this one wrong? I'm looking at this: http://www.ict4us.com/r.kuijt/en_oxidation_reduction.htm

Google does not give the same search results for the same search terms for everyone.  They return results based on your location (guessed from your ip address) and other things they know about you based on their tracking cookies.  For example, my search did not turn up the reference that you sited.  I would just use the actual link rather than giving instructions for a google search.  Just my 2 cents on that.

Where you say "if A oxidizes", do you mean that A oxidizes B or that A gets oxidized?

I have added the link (I hope you have access to this article) and 2. I'm interested in the "on" formulation. If I see A oxidizes B, I understand that B loses electrons, as it gets oxidized, right? Please correct me if I'm wrong!

Yes, if A oxidises B, then B loses electrons. However in your question body you wrote that "A loses electrons to B" which is the wrong way round. In this case, the preposition "on" indicates the place at which the electrolyte is being reduced, which is the negative electrode.

So let me see if I got it right. This formulation "Carbonate-based electrolytes commonly used in Li-ion batteries are reduced on negative electrodes" means that the negative electrode loses electrons, while the electrolyte gains them?

This answer is a summary of the string of very good comments to the OP's original set of short questions.  This is both my take on the questions and an attempt to produce a more readable "net" answer than those contained within the set of individual comments.
Regardless of the intended context, "A is oxidized on B" means that A looses electrons.  It's seems this context likely implies that B is an electrode surface "on" which A is being oxidized.
Regardless of context, if A is oxidized by B, then B is reduced.
If B is an electrode surface, then the electrons from A needn't reduce B itself, but could flow through that part of the cell (the anode).
I hope this (correctly!) answers each of the OP's questions.

Thank you for you answer. Just to make it clear. This formulation: Carbonate-based electrolytes commonly used in Li-ion batteries are reduced on negative electrodes means that the electrolyte gains electrons, right?

Correct as stated.  All that maters is ...electrolytes...are reduced... And that makes your statement correct.


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Why in molybdenum disulfide the oxidation state of Mo is IV?
On one hand, stoichiometrically, it makes sense that in $\ce{MoS2}$ the $\ce{Mo}$ would have +4 and each sulfur would have -2.
(source)
However, in reality, every $\ce{Mo}$ atom is bonded to 6 sulfur atoms and every $\ce{S}$ atom is bonded to 3 $\ce{Mo}$ atoms. Why in such a complex configuration is it still correct to assign such oxidation numbers? $\ce{Mo}$ participates in 6 bonds, so how are they translated into an oxidation state of 4?

In sodium chloride each sodium ion has six nearest chloride neighbours. However, sodium is unambiguously +1 oxidation state. If you were to count every one of the six chloride neighbours as belonging to one sodium atom, you would double-count the chloride ions. In fact, you wouldn't even double-count; you would count each chloride ion six times. Likewise the fact that each Mo has 6 nearest S neighbours doesn't mean anything. It could have 4 or 6 or 8 or 12 (it just happens to have 6) and the oxidation state would not be any different as long as the material has the bulk formula MoS2.

and whether or not Mo in MoS2 has anywhere near a +4 charge (probably not since sulfide is quite polarisable) is another matter altogether. Oxidation states aren't necessarily supposed to accurately reflect charges though, so there's no real problem there.

@orthocresol But each chloride also has 6 nearest neighbors. So theoretically you could "pair up" every Na and Cl in "donor-acceptor pairs".

Yes, you could. In the same vein you could pair up every Mo with two S.

@orthocresol but in NaCl those "pairs" are immediately next to each other, while in MoS2 that is not the case.

That doesn't matter. Solid NaCl doesn't divide itself up into pairs before reacting. All that is required is for the ratio of Mo to S to be 1:2.

@orthocresol  yes I see your points now. thanks.

@orthocresol I think you can digest that into an answer ;)


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Can liquids be made a lot purer than by fractional distilation?
Suppose you have a fused quartz tube with one closed end coiled into a spiral with a diameter of more than 10 m and you submerge it into a mixture of water and ethanol in a way that gets rid of all air then reposition it in such a way that it's partly submerged with the axis of the coil going nearly horizontally and slightly up in the direction from open end to closed end. Furthermore, suppose that the coil then starts moving and spinning along its length in the direction from open end to closed end and once the closed end passes the height where cavitation occurs, the closed end is cooled to recondense liquid there and after cavitation occurrs again on the next loop, liquid is condensed again in the same way? Is it true that if the concentration of water is above a critical concentration, it will become more concentrated after it cavitates and is recondensed because a smaller fraction of ethanol vapour will condense than water and if the concentration of water is below that amount, the water will become less concentrated after it cavitates and then is recondensed because a smaller fraction of water will evaporate than ethanol? Could really pure water be gotten in that way because fused quartz is extremely insoluble in water?

Units: 1 picture = 1,000 words ;)  Also, do really mean a 10 meter diameter tube (or coil)?

Since ethanol and the water-ethanol azetrope boil at a lower temperature than water, if you start with a lot of water with a smaller amount of ethanol  then you can "distill" (however you do it...) all the ethanol out and be left with "pure" water.
But water and ethanol form an azeotrope which is 95.5% ethanol by weight and boils at 78.1 °C. For the water-ethanol azeotrope the wt. % ethanol in the liquid is the same as the wt % of ethanol in the vapor. Thus there is no way to exploit liquid to gas phase transition to enrich the ethanol more than that.

How do you know a mixture of water and ethanol with 95.5% ethanol by weight can't be made into purer ethanol in that way? You haven't proven to me why that should be the case. Wouldn't very little water go from the liquid phase into the gas phase?

Proof? LOL - Let's start with 1+1=2... // See phase diagram for water-ethanol here, not exact scale, but gives the idea. http://www.chemguide.co.uk/physical/phaseeqia/bpcompn1.gif // webpage http://www.chemguide.co.uk/physical/phaseeqia/nonideal.html


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Why do rare earth metal oxides vary in color so much?
I only know a little about what causes different color, but I know it has to do with the electron orbitals and how they reflect or absorb/emit incoming light. The rare earth metals are in the lanthanide series so they all have an identical valence shell.
I've worked with lots of chemicals, but there's not much that is quite the same as the bright and exotic colors produced from the rare earth metal oxides. I worked with the oxides in powdered form in ceramic glazes and they each vary greatly in color, with colors including pink, light blue, black, creamy yellow, and white. (After firing, lime green and grey can also be introduced). Why do they each produce such different and vibrant visual colors?

Very related albeit considering d-orbital splitting rather than f-orbitals: http://chemistry.stackexchange.com/q/62715

The very general answer is because they have orbitals that are close together in energy (corresponding to photons in the 100-1000 nanometer range) and allows for efficient non-radiative dissipation of exitations. According to wikipedia, f-f transitions are probably not the culprits, since in contrast to transition metals they are non-bonding  - there is no vibronic coupling between them to "allow" for forbidden transitions. However: Europium, with half filled f orbitals - is colorless... There is something about the f-shells.

Are they completely nonbonding?  There may be enough vibronic interaction to get those pale colors.

TL;DR f-f transitions in lanthanide ions give rise to the colours of the ions. The electronic states can be derived from quantum mechanics, and their energies vary from ion to ion, which give rise to different colours.. The colours are pale because these transitions are Laporte-forbidden and therefore weak.

These absorption bands do indeed arise from electronic transitions, but the theory required to describe them is more sophisticated than simple crystal field theory where one speaks of d-d* transitions. Instead one would have to construct the symmetries of the electronic states via direct products. In the case of the transition metals, there is some description here. For example, a $\mathrm{d^2}$ octahedral transition metal complex would have a $\mathrm{^3T_{1g}}$ ground state and there would be three spin-allowed transitions to excited states ($\mathrm{^3T_{2g}}$, $\mathrm{^3T_{1g}}$, $\mathrm{^3A_{2g}}$). Not all of them may fall within the visible region, but often, more than one of them do.
As a concrete example, here's the spectrum of $\ce{[V(H2O)6]^2+}$.1 In wavenumber units the visible spectrum is roughly $14\,000\mathrm{-}25\,000~\mathrm{cm^{-1}}$:

You can hopefully already appreciate that it is not as simple as "d-d* transitions" which would imply that the complex only absorbs at one wavelength.
Now, for the lanthanides, exactly the same principles apply. The lanthanides characteristically do not interact strongly with any bound ligands, and therefore the effect of the ligand field is almost negligible. The relevant electronic terms are therefore the atomic term symbols, which are obtained by Russell-Saunders coupling, which is more fully described in numerous physical chemistry textbooks. For example, consider $\ce{Pr^3+}$ ($\mathrm{4f^2}$). The allowed term symbols are
$$\mathrm{{}^3H, {}^3F, {}^3P, {}^1I, {}^1G, {}^1D, {}^1S}$$
and following spin-orbit coupling (which is quite a significant effect in the lanthanides) one gets the electronic terms
$$\mathrm{{}^3H_4, {}^3H_5, {}^3H_6, {}^3F_2, {}^3F_3, {}^3F_4, {}^3P_0, {}^3P_1, {}^3P_2, {}^1I_6, {}^1G_4, {}^1D_2, {}^1S_0}$$
The ground state as predicted by Hund's rules is $\mathrm{^3H_4}$. There are therefore up to 12 different transitions that must be considered. In some other lanthanide ions, there are even more possible states.
The bottom line is that each lanthanide ion, with its unique electronic configuration, will have different excited states and different excitation energies. Therefore their absorption spectra differ and they display different colours. $\ce{Pr^3+}$ itself has the following spectrum.2 The lower excited states don't fall within the visible region.

As for why they are "bright", that is simply related to the selection rules. For transitions between atomic states, the requirements are
$$\Delta S = 0; \Delta l = \pm 1; \Delta L = 0, \pm 1; \Delta J = 0, \pm 1 \text{ except }J = 0 \not\leftrightarrow 0$$
and for centrosymmetric systems like atoms there is an additional constraint that
$$\mathrm{g} \not\leftrightarrow \mathrm{g}; \mathrm{u} \not\leftrightarrow \mathrm{u}; \mathrm{g} \leftrightarrow \mathrm{u}$$
which is the Laporte selection rule. The bottom line is that these "f-f transitions" (I am using this term to loosely refer to the collective set of electronic transitions) are parity-conserving and therefore Laporte-forbidden. Vibronic coupling, which is the pathway which allows this selection rule can be relaxed, is not of consequence in the lanthanide ions because of the lack of covalency; the colours are not intense, and therefore typically a pale colour is observed. (Compare $\ce{KMnO4}$ which has a fully parity-allowed LMCT transition; its dark purple colour is exactly the opposite of pale.)
Much greater discussion of the electronic spectra of lanthanides can be found in ref 3.
References

Oxford lecture notes.
Aspinall, H. C. Chemistry of the f-Block Elements; p 35.
Cotton, S. Lanthanide and Actinide Chemistry; pp 61 onwards.

Hmmm, I don’t agree with the bright part. The colour of $\ce{[FeCl6]^3-}$ is also due to (parity allowed, iirc?) LMCT transitions and I perceive its yellow as very bright … d-d transitions are spin-forbidden ($\mathrm{d^5}$).

@Jan $\ce{[FeCl6]^3-}$ has a LMCT transition at at $24,900~\mathrm{cm^{-1}}$ so you're right on that. I guess it comes down to what one perceives as bright, but I didn't really explore that very much.

as to 'bright' I suppose you mean intense, in which case all of these are very feeble compared to fully allowed transitions as in aromatics such as dyes where extinction coefficients are many many tens of thousands in $\pu{dm^3mol^{-1}cm^{-1}}$

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text:
Removal of calcium from calcium-ammonium nitrate?
I am interested in model rocketry, and need to produce small quantities of pottasium nitrate. For this process I need ammonium nitrate, but I only have calcium ammonium nitrate. Is there a simple process for removing the calcium?

decompose it at 210 C

You mean by electrolysis?

No I meant by heating.

Ah, so I would heat the compound (dissolved in water) to 210 C and then I should get calcium precipitate that I can filter out?

Yes, that would remove the calcium and give ammonium nitrate.

Thanks for the help. The rest of the internet wasn't very helpful, because most people associate ammonium nitrate with bombs, not rockets. I really appreciate it :)

If you add ammonium sulfate then the calcium will precipitate out and not waste as much of your nitrate content.

Why not just get potassium nitrate ?

Soak it in water and decant off the aqueous solution freeze and the crystals should form its just calcium carbonate and ammonium nitrate
dont recommend using heat.

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text:
Chemical reaction and shift in magnetic susceptibility sign
I'm trying to find an easy experiment involving a chemical reaction in which two paramagnetic reactants A and B react to give a product which is diamagnetic (or the reverse). The idea is to show that A is paramagnetic, B is paramagnetic, but A+B --> C is diamagnetic.
Or, another experiment: an experiment in which a change in temperature changes paramagnetic to diamagnetic (or the reverse). I've seen such an experiment but cannot remember the details. All I can say is that there was a color change and, at the same moment, a paramagnetic/diamagnetic shift

Thermal decomposition reactions which produce oxygen would satisfy your requirement since oxygen is diamagnetic, right.

Oxygen in its ground state is famously paramagnetic! For OP: for the second part of your question, you may be interested in complexes which can undergo spin transitions, in particular the $\mathrm{d^6}$ Fe(II) complexes. High-spin Fe(II) is paramagnetic, but low-spin Fe(II) is diamagnetic. https://en.wikipedia.org/wiki/Spin_transition Also how about Ni(II) complexes $\ce{Ni(PPh3)2X2}$ which can flip between square planar (diamagnetic) and tetrahedral (paramagnetic) geometry? I don't know much about those so just putting them out there.

Seems to me any reaction involving the recombination of two radicals leads to a closed-shell diamagnetic species (which oxygen is definitely not!). I can't think of any other examples immediately. You could make singlet oxygen, which is diamagnetic, from paramagnetic triplet oxygen by irradiation, but then there is only one reactant.
If you heat a paramagnetic substance enough so that the thermal energy can overcome the energy that is gained by the alignment of your magnetic moments, I believe the substance will not exhibit paramagnetism in a magnetic field above that temperature. So that seems doable.
Edit: Actually, I found an example: ozone is diamagnetic and can be made from paramagnetic oxygen and an oxygen radical. So that could be a possible reaction where you have two paramagnetic reagents combining into a diamagnetic product.


text:
Is Dalton's Atomic Theory a theory or a postulate?
It is always mentioned that Dalton proposed an Atomic Theory. But when you read it, it consists of the following:

The main points of Dalton’s atomic theory, an explanation of the structure of matter in terms of different combinations of very small particles, are given by the following postulates:

All matter is composed of indivisible atoms. An atom is an extremely small particle of matter that retains its identity during chemical reactions.
An element is a type of matter composed of only one kind of atom, each atom of a given kind having the same properties. Mass is one such property. Thus, the atoms of a given element have a characteristic mass.
A compound is a type of matter composed of atoms of two or more elements chemically combined in fixed proportions. The relative numbers of any two kinds of atoms in a compound occur in simple ratios. Water, for example, a compound of the elements hydrogen and oxygen, consists of hydrogen and oxygen atoms in >the ratio of 2 to 1.
A chemical reaction consists of the rearrangement of the atoms present in the reacting substances to give new chemical combinations present in the substances formed by the reaction. Atoms are not created, destroyed, or broken into smaller particles by any chemical reaction.

(From the book: General Chemistry by Ebbing and Gammon)
I'm sure many of you might be knowing that postulates and theories are quite different. Can someone explain me whether Dalton proposed a postulate or a theory?

It is most definitely a theory. The 4 points you mention could be construed as hypothesis by themselves. But Dalton went further, he put them to the test.
The results of his exeriments formed yet another component of his atomic theory called the Law of Multiple Proportions. This law served as proof that the other components of his theory were true. I used to spend most of a class period explaining this law to students, it's best you search the web for a complete explaination but I'l give you a brief summary via example.
Say an experiment shows it takes exactly 16 g of oxygen to combine with 12 g of carbon to make carbon monoxide. Now suppose we make carbon dioxide. We would find that it now takes 32 g of oxygen to combine with that same amount, 12 g, of carbon. So, for a fixed amount of carbon, the amount of oxygen required to make the two different compounds can be expressed as a small, whole number ratio. What would explain this better than to surmise that elements come as particles and thes particles combine in small, whole number ratios to form compounds.
When a hypothesis have been proven by repeatable experimentation, we call it a theory.


text:
What are the physical properties of nitrosomethane?
I've searched all over the web and I can't find any information about this compound. Can anyone help?

https://en.m.wikipedia.org/wiki/Nitromethane

I'm closing this question as off-topic because requests for specific data points do not belong on SE.

It seems to be a rare one and quite unstable under standard conditions.
According to Chemspider:

Chemical formula: $\ce{CH3NO}$
  Also known as  NITROSO-FORMAMIDE OXIME METHANE
  It melts at $2.5~\mathrm{^\circ C}$ (measured) and has a density of $0.9~\pu{g cm^-3}$ (predicted).

According to Paul D. Adeney, Willem J. Bouma, Leo Radom, William R. Rodwell. J. Am. Chem. Soc. 1980, 102 (12), 4069–4074:

Nitrosomethane is known to be less stable than its isomer formaldoxime
  and original attempts to isolate this molecule failed owing to it's
  speedy isomerization to formaldoxime.


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length of bond, bond angle and p character in bond
My reference data: http://www.rsc.org/suppdata/ra/c4/c4ra06781a/c4ra06781a1.pdf
Consider these five chemicals:

A: $\ce{CH4}$ (methane)
B: $\ce{CH3Cl}$ (chloromethane)
C: $\ce{CH2Cl2}$ (dichloromethane)
D: $\ce{CHCl3}$ (trichloromethane/chloroform)
E: $\ce{CCl4}$ (tetrachloromethane/carbon tetrachloride)

From what I know, an orbital is longer and has higher energy as its p character increases.
From A to E, both $\ce{C-H}$ bond and $\ce{C-Cl}$ bonds get shorter, and $\ce{Cl-C-Cl}$ bond angle in D is smaller than that in C.
Nothing makes sense, first one implies more p character in $\ce{C-Cl}$ bond i.e. D < C < B and second order implies the opposite. I assumed equal x for every C's orbital taking part in $\ce{C-Cl}$ bond by symmetry.
The angle between two orbitals with $sp^x$ hybridisation is $\cos^{-1} \left( - \frac 1 x \right)$.
Also, from A to E, the electronegativity of central atom C increases, so C-Cl bond should become longer as the difference between EN decreases.
So my question is the p character and hybridization logic only working for 2nd-period elements exclusively?
Maybe relevant: It is a widespread opinion that bond energy in sigma bond p-p>s-p>s-s but sigma bond energies of $\mathrm{sp>sp^2>sp^3}$. The rationalization for the first order is more overlapping is possible due to directional and long nature of p orbital and rationalization for 2nd order is that energy of p is more than s, so the order of energies of molecular orbitals is more for more p character.
EDIT: In compounds containing $\ce{H,\ C,\ N,\ O,\ F}$ only I can predict bond lengths and bond angles almost every time using relative properties of hybrid orbitals and their relation with p character in a respective bond otherwise known as Bent's rule. But, once $\ce{Cl}$ enters the picture, there is no simple trend in any scenario, bond angle could be rationalized due to steric reasons but the cos and x relation should be valid if it follows simple hybridisation which should allow me to compare bond lengths by comparing lengths of participating orbitals (more p in carbon's hybrid orbital towards $\ce{Cl}$, longer orbital, and $\ce{Cl}$ has 3p in every case), but it seems it doesn't follow simple hybridisation. As an example, as bond angle increases, bond length must decrease, which is not the case for $\ce{CCl3}$, $\ce{CH2Cl2}$ etc

The electronegativity of any element is constant.

Maybe you need to note that the active orbital in chlorine is 3p. Comparing 3p with 1s isn't exactly fair.

Well, I did not compare 3p with 3s for bond lengths I compared C-Cl bonds in different compounds with different atoms attached to C. Also, Electronegativity of (CHCl2)-X>(CH2Cl)-X, more Cl attached to C make the C slightly positive and increase "electronegativity". I suppose I'm gonna have to scrap every hybridisation concept for every element other than 2nd period, attaching any number of Cl breaks away from every trend.

Your question is a bit confusing, maybe you could clear it up, write down the numbers you mean and why you think they are wrong and so on. What do you mean by first and second order?  If I understand it correctly, you want to argue with this hybridization concept on why there are marginal differences in bond lengths? I'm starting to think they should stop teaching that anywhere.
If you want to stay in the primitive world of hybridization, you should look at calculations that only consider s and p-orbitals. The basis sets used in your reference have d, f and g-orbitals in them.

edited with an explanation

You sure ask a lot of hybridization questions. What exactly is it that you want to do? Is this for a publication? Or just for your personal amusement? You might be ready to leave those concepts behind that as you noticed really only work for the handful of examples they were made for and learn some actual quantum chemistry


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Confusion in naming a carboxylic acid with a cyclohexyl group
For the below compound  .

According to me the answer should be 1-carboxymethylcyclohexane.
But the answer given as 2-cyclohexylethan-1-oic acid
Why the name given by me is wrong .
Whereas the name given to

Is cyclohexane-1-carboxylic acid
Why it is not 1-cyclohexylmethan-1-oic acid

Related: What is more important, to have the parent chain be the cycle or the chain that has an hydroxyl group

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).
(…)

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).

The compound given in the question contains a characteristic group $(\ce{-COOH})$. Since there is only one characteristic group, the seniority order of classes is not relevant in this case; thus, the $\ce{-COOH}$ substituent corresponds to the principal characteristic group.
Since the principal characteristic group is a substituent of the chain, and the ring does not have any substituents corresponding to the principal characteristic group, the senior parent structure of the compound given in the question is the chain and not the ring. The rules P-44.1.2.2 and P-52.2.8 for the selection between a ring and a chain as senior parent structure are not relevant in this case.
Simple carboxylic acids are named substitutively using the suffix ‘oic acid’ as follows.

P-65.1.2.1 Carboxylic acid groups, $\ce{-COOH}$, that conceptually replace a $\ce{-CH3}$ group of methane or terminate an unbranched hydrocarbon chain are named by replacing the final ‘e’ of the name of the corresponding hydrocarbon by the suffix ‘oic acid’. No locants are necessary to denote the positions of the carboxylic acid groups in a hydrocarbon chain; locants are used when hydrocarbon chains are modified by skeletal replacement, as shown in P-15.4.3.2.3. Except for formic acid, acetic acid, oxalic acid (see P-65.1.1.1), and oxamic acid (see P-65.1.1.1), systematically formed names are preferred IUPAC names; the names given in P-65 .1.1.2 are retained names for use in general nomenclature.

Therefore, the systematic name for the unsubstituted carboxylic acid structure that is given in the question is ethanoic acid (not ‘ethan-1-oic acid’ since no locants are used to denote the positions of the carboxylic acid groups in a hydrocarbon chain). However, the preferred IUPAC name (PIN) of ethanoic acid is the retained name acetic acid according to rule P-65.1.1.1.
The acetic acid chain is substituted using the usual principles of substitutive nomenclature.
However, according to rule P-14.3.4.6, locants are omitted where there is only one kind of substitutable hydrogen.

P-14.3.4.6 All locants are omitted for parent compounds when all substitutable hydrogen atoms have the same locant.

Therefore, the preferred IUPAC name name for the complete compound given in the question is cyclohexylacetic acid since the locant ‘2’ of ‘2-cyclohexylacetic acid’ is omitted.

So is it cyclohexylmethanoic acid or cyclohexanecarboxylic acid?

A good explanation of IUPAC rules for naming organic compounds may be useful.
For carboxylic acids (which it is due to the $\ce{-COOH}$ group), the webpage says:

Carboxylic acids are named by counting the number of carbons in the longest continuous chain including the carboxyl group and by replacing the suffix -ane of the corresponding alkane with -anoic acid.

The reason it should be named as a carboxylic acid (instead of using carboxyl as a mere group) is due to precedence (and this is related to the comment by Loong):

The carboxyl group takes precedence over alkyl groups and halogen substituents, as well as double bonds, in the numbering of the parent chain.

The correct name is thus 2-cyclohexylethan-1-oic acid, as you said.


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Changing iron(III) acetate to iron(II) acetate
Is there a way simple way to turn iron(III) acetate into iron(II) acetate or vice versa?
Edit (as per answer below): an oxidiser like hydrogen peroxide will turn iron(II) acetate into iron(III) acetate. A reducer (not sure what household product) will do the reverse.
Some background: my partner soaks scrap cast iron in white vinegar for several weeks and then uses the resulting liquid as a mordant for dyeing. According to wikipedia's page on iron acetate, soaking iron in vinegar creates a mix of iron(II) acetate and iron(III) acetate. As I understand it, the oxidation state of the iron used as a mordant effects the colour of the dye on the fabric. As such, it would be nice to be able to experiment with just iron(II) or just iron(III). Is this possible to achieve using household chemicals?
Even more background: The dyeing process is soak fabric in mordant water. Dry fabric. Wrap fabric with leaves and place in boiling water. Unwrap and admire beautiful prints :)

Did you use hydrogen peroxide in the synthesis or just iron + vinegar?

Just iron and vinegar. I think that iron(III) predominates because the water is a red-brown. Would h2o2 force the mixture to iron (III) and a reducing agent (not sure what) force it to Iron(II)? Or am I misunderstanding the chemistry?

No, you're not misunderstanding anything, that is exactly my line of thought. I think introducing hydrogen peroxide at the right time in the right conditions will push the oxidation to iron(III). It's probably something you can just experiment with, carefully. I'd say boil the solution in peroxide but I don't know if you have a fume hood, etc. The tougher one is the reduction. Anything that burns in air is being oxidized, and thus is a reductant to some degree. I don't know if sawdust or something like that would be a strong enough reductant, I seriously doubt it.  I'll think on it some more.

You need a reducing agent that will reduce iron(III) back to iron(II). Actually finely distrubuted iron could be the best reagent to do this, as it will not introduce any contaminants. The reaction taking place is 2 Fe(III) + Fe = 3 Fe(II).
Of course the iron used should not be rusty. I am not sure that you can easily get hold of fine iron powder, but I would give a try boiling the vinegary solution with fresh iron shavings or steel wool and as little contact with air as possible. The color change should be visible to greenish from the red-brown of the iron(III) acetate.
Actually, during dissolution of Fe in acetic acid you should get iron(II) acetate predominantly until the Fe runs out, but usually scrap iron is covered by iron(III) oxides/hydroxides, and these also dissolve in the vinegar. When the iron(II) solution is in contact with oxygen in air, it oxidises rather quickly. It is a good idea to start with a batch that was not left standing for weeks.
Other reducing agents work quicker than iron (e.g. tin(II) salts or zinc powder), but these introduce other products that would interfere with your dyeing.


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anode and cathode in lithium ion battery?
By definition, the anode is the electrode where oxidation takes place, the cathode is the electrode where reduction takes place.
When I am charging the Li-ion battery, positive electrode undergoes oxidation and hence I may call it the anode. When I am discharging the Li-ion battery, negative electrode undergoes oxidation and I may call it the anode. Is this correct?
Do the anode and cathode interchange for charging and discharging?

This is indeed correct, since, as you said, the cathode is per definition the site for reduction and the anode for oxidation, both names have to be "switched" when viewing electrolysis (charging of the Li-ion battery) and galvanic cells (discharging of the Li-ion battery). What does not change is the location of the + and - pole.
A good way to remember the site of oxidation and reduction is the word pair an-ox /cat-red.


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What gives milk its characteristic white color?
The other day, my kid brother asked me this:

What pigment gives milk a white color?

Now, I found that particular question quite interesting since, as a student of Biology, I already had a rough idea as to what constitutes (cow's) milk: Water, sugars, fats, and a whole bunch of vitamins and minerals. Yet as far as I know, none of them play a "pigment"-like role anywhere else in nature... which has left me deeply perplexed.
So what pigment is responsible for milk's characteristic white color?

related http://chemistry.stackexchange.com/questions/66060/what-gives-fresh-milk-a-yellow-tinge

There is no "pigment" that makes milk white.
According to Wikipedia:

A pigment is a material that changes the color of reflected or transmitted light as the result of wavelength-selective absorption.

Going by that definition of a "pigment", then there is no (white) pigment in milk. In fact (as @Ivan points out in the comments) there is no such thing as a "white pigment", because that would require the "pigment" to absorb no visible radiation (if it absorbed all visible radiation, it'd appear black)
That's because the real reason for the white color of milk is a physical phenomenon called Scattering. (Source: Optics, 4th edition, E. Hecht)

(Diagram's from Wikipedia)
Milk is an emulsion of fat dispersed as these really tiny droplets in water. These fat globules (and some milk proteins) scatter light of all wavelengths in the visible spectrum. This is why milk appears white
This is the same reason (albeit different cause) as to why clouds often appear white; except here, the scattering is due to suspended water droplets and dust particles.

I'd rather stress there is no pigment that makes anything white.

Isn't the chief reason for the white color the fact that fats/proteins are imbued with calcium? (homogeneously). Roughly every calcium compound in water renders a white solution.
Note heterogenous milk is a little yellowish, which you would expect as the color of typical fats/oils.

Nope. Solutions of calcium compounds are mostly colorless. Arguably, there is no such thing as white solution.

@khaverim http://chemistry.stackexchange.com/questions/66060/what-gives-fresh-milk-a-yellow-tinge

Precipitates in water*, I should have said.

Scattering by fat doesn't completely explain things by itself, as that would mean that skim (nonfat) milk should be clear, not white. (It's definitely clearer, but not completely clear, which implies there's still something else in the milk which is scattering light.)


text:
What is the use of chloroform in the procedure of determining peroxide value?
In this procedure of determining peroxide value, the oil sample is dissolved in acetic acid chloroform solution. I understand that the acid is added so that the following reaction occurs
$\ce{2ROOH + 2H^+ + 2I^- -> 2ROH + I2 + H2O}$.
But what is the point of the added chloroform?

Probably to help dissolve the sample - think vinegar/oil on salad, which is typically not really homogenous. The sample chemical would still react with the iodide due to the swirling of step 5.

Acetic acid is required for this method for exactly the reason you gave.  It also has the fortunate property of being miscible with many polar and non-polar solvents.  Because this is a method for the analysis of oil samples, it is necessary to chose a non-polar solvent capable of dissolving the sample that is also miscible with the acetic acid.  Chloroform is one solvent that fits that description nicely.
As a quick side-note, the use of chloroform in this test procedure is actually now discouraged by the Standard Methods for the Analysis of Oils, Fats and Derivatives.  Isooctane seems to be the preferred solvent, apparently for health and safety reasons.


text:
Does fumaric acid form stronger hydrogen bonds than maleic acid?
My book claims:

Fumaric acid can form intermolecular hydrogen bonds (strong) while maleic acid forms intramolecular hydrogen bonds (weak), hence fumaric acid has higher boiling point than maleic acid."

The intermolcular hydrogen bonds in fumaric acid are said to be stronger than the intramolecular hydrogen bonds in maleic acid, while I thought that intramolecular hydrogen bonds should be stronger than intermolecular hydrogen bonds. So, which one is correct?

Which book is it?

ok I see but what I did not understand was at this level "fumaric acid can form intermolecular bonds (STRONG) and maleic acid intra(WEAK) " well then before i thought intramolecular bonds were stronger than intermolecular bonds, so I am completelly lost.

"The easy way Chemistry"

Intramolecular hydrogen bonds do not have to be weak, but the strength of a hydrogen bond depends a lot on the $\ce{O-H\bond{....}O}$ angle. With intramolecular hydrogen bonds, it is often hard to achieve the perfect $180^\circ$ angle while the same is not a real exercise for intermolecular hydrogen bonds which have full degrees of freedom of movement on both sides.
In fact, I want to argue that maleic acid only forms intramolecular hydrogen bonds if there is no solvent (or crystal water) that can otherwise supply hydrogen bonds. If there is, all hydrogen-bonding sites — both donating and accepting — will be fully saturated with external hydrogen bonds instead; they are slightly stronger due to the better angle.
The question remains why fumaric acid — which is definitely unable to build up intramolecular hydrogen bonds — has a higher boiling point, because in theory maleic acid should also be able to form as many intermolecular hydrogen bonds. However, fumaric acid’s structure already dictates that it practically must form a network of coplanar molecules interacting with each other via hydrogen bond dimers from carboxylate to carboxylate. This is a very nice structure to keep. Maleic acid cannot form such an extensive network no matter how much it distorts. It will be restricted to much less favourable crystal structures meaning weaker intermolecular attraction.
In the liquid phase, fumaric acid will keep the near-sphere ordering somewhat intact, i.e. conserve a high number of hydrogen bonds to its neighbours. However maleic acid will probably stick with one fully intramolecular hydrogen bond that does not contribute to intermolecular attractions, thereby lowering its boiling point.

What your book (probably) meant is that the hydrogen bonds in fumaric acid, being intermolecular, does a better job of holding the molecules together, so it has a higher boiling point than maleic acid, which instead, displays intramolecular hydrogen bonding which doesn't hold the molecules together just as well.

I thought that intramolecular hydrogen bonds should be stronger than intermolecular hydrogen bonds. So, which one is correct?


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Why don't hydride shifts occur with hydrohalogenation of 4-methyl-cyclohexene?
My book (see bottom) implies that hydrohalogenation of 4-methyl-cyclohexene follows this mechanism:

Why not undergo consecutive hydride shifts to furnish a tertiary carbocation?
I would guess that the second step, nucleophilic attack by bromine, is much faster and prevents this from happening. But when I try to confirm this by looking up some examples I see this one at masterorganicchemistry.com:

This implies the following very similar mechanism:

The carbocation in the example is a smaller ring, meaning it should be less stable and even more reactive (less able to perform hydride shifting.) But then again the charge is closer to the tertiary carbon and easier accessible.
Could someone explain what is happening?

Vollhardt, P.; Schore, N. Molecules: Structure and Function, 6th edition; W. H. Freeman: New York, 2014, p491 exercise 12-6:

Exercise 12-6
Predict the outcome of the addition of HBr to (a) 1-hexene; (b) trans-2-pentene; (c) 2-methyl-2-butene; (d) 4-methylcyclohexene. How many isomers can be formed in each case?

It's an exercise and the book only gives the starting material as "4-methyl-cyclohexene" and in the back the two structures which I drew in my pictures (1-bromo-[4/3]-methyl-cyclohexane.)

Maybe the textbook is just wrong?

I'll settle for that for now.

The reason for not having consecutive hydride shifts is that generally, a new carbocation is formed through shift only when it is more stable than the immediately previous one.
To form a tertiary carbocation, 3 consecutive hydride shifts would have to occur consecutively (when + is at c-4 from the alkyl group, your first product) in which more than one intermediates have same stability (same number of alpha-H). Hydride shifts will not occur for your second product to form tertiary carbocation as the next intermediate is infact less stable (3 alpha-H as compared to 4 alpha-H previously).

Since the ring already has 6 members (least strain) so ring expansion/contraction will not occur either.

On second thought, there will also be (+)ve inductive effect on the c+ in my first product, increasing it's stability but I think that HyperConjugation has greater effect on stability of carbocation than inductive effect so loss of alpha-H should have a greater effect.

Please check out the mechanism of squalene cyclisation which includes four consecutive methyl and hydride Wagner-Meerwein rearrangements, neither of which is truly downhill.

For rearrangement of carbocation there should be some driving force like increase in stability of carbocation.


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Electron pair repulsion theory qualm
Is the there not repulsion between the individual electrons within the bonding pairs ? Why does the "bonding pair " have an overall repulsive effect on the other lone and bonding pairs ?

There is repulsion between the individual electrons of a bonding pair (and within a lone pair, for that matter). The reason that the electrons can be found in that "pair" nonetheless is that those "pairs" are actually orbitals - allowed "spaces" for electrons. Of these orbitals, there are many more than there are electrons in a molecule: they can thus be classified as occupied and unoccupied. They differ in energy: the "closer" an orbital is to a nucleus, for instance, the lower the energy of the electrons occupying it. Often, the electrons are lower in energy when sharing an orbital than when not sharing it - because in case of the latter, one of the electrons ends up in a higher orbital (think Aufbau principle). One might also say that the electron-nucleus attraction overcomes the electron-electron repulsion of the electrons within a pair.
The natural question to ask next is: why do we not have more than 2 electrons per orbital? The answer is that you cannot have 2 electrons with the same spin (a fundamental relativistic property of an electron) in one orbital (There are reasons for this, but elaborating them goes too far at this point). Electrons have either +1/2 or -1/2 spin - thus you get one of each within a VSEPR pair.
There is repulsion between electron pairs, simply as a result of the identical charges of the electrons.


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Does gold always become gold chloride when dissolved with HCl and an oxidizer and can it always be precipitated with sodium metabisulfate?
I tried to extract gold from computer parts by STRIPPING the gold using $\ce{HCl}$ and $\ce{H2O2}$ by dissolving the copper & nickel underneath but i added FAR too much $\ce{H2O2}$ by a stupid mistake and I ended up dissolving the gold... I didn't know that could happen? I am correct in understanding the $\ce{HCl}$ dissolves the gold, but it needs both energy and oxygen to do so? Nitric acid (commonly added to $\ce{HCl}$ to form aqua regia which dissolves gold) is $\ce{HNO3}$ which has lots of oxygen and Hydrogen Peroxide is $\ce{H2O2}$ which has oxygen also, albeit less than Nitric Acid.....
My mixture is VERY dark red... It's fuming very lightly, and makes you choke if you get near it. It's not see-through but rather very dark and a bit murky, but could be transparent if you added 200% of water.
I am afraid to try anything because if I screw it up I will lose $200 of gold but am I correct in understanding that adding sodium metabisulfate will drop the gold? What if it is too acidic? Will I have to add Urea or Sodium Bicarbonate to neutralise the pH??? HELP!!

You may find to your dismay that getting dissolved gold separate from dissolved copper and nickel is dicey (to give you some idea, I still have a dissolved solution of gold and copper sitting in my garage ca. 2 years ago because I hesitate to mess with it; also it looks kind of cool).
You should use urea to neutralize first, just going on experience. I've done the same procedure you describe here to try to recover gold from RAM cards etc. Add urea slowly and check pH regularly. I wouldn't try sodium bicarbonate only because I've never done that and it seems you ought to go with the most well-used method if you're risking a big loss.
Sodium metabisulfite is indeed a common reagent for precipitating the gold. It smells pretty terrible before and after adding it (because of sulfur content). I think $\ce{SO2}$ gas is released when you react with gold so you should have a well-ventilated area for this. You should get a brownish precipitate. I would keep adding it to the neutralized solution until no more precipitate is formed, then let it settle.
This point is probably your best hope for separation of the gold. Recover the precipitate by filtering and leave off as much of the fluid as possible (wash with DI $\ce{H2O}$, and melt the brown salt to remove metabisulfate and get the pure metal.
In the future I would not use any oxidizer when stripping the gold from $\ce{Cu}$/$\ce{Ni}$. I found that a much better method is to stir/air-bubble the material in $\ce{CuCl2}$ which can be made from pure copper, $\ce{HCl}$, and careful additions of $\ce{H2O2}$. It will take longer but it is much safer. You'll recover solid bits of gold instead of dissolving it undesirably. i.e.
$\ce{CuCl2}$ > $\ce{HCl} + \ce{H2O2}$ > aqua regia
A lot of this procedure is difficult to improve because (even academic) literature is nebulous on the topic. I once bought a ~$250 book on the topic. "The Properties of Electrodeposited Metals and Alloys: A Handbook" by Safranek. It's hard to get a hold of.
P.S. it does not form gold chloride when dissolved in aqua regia ($\ce{HCl}$ alone won't effectively dissolve it). It forms a complex with $\ce{Cl-}$, chloroauric acid $\ce{HAuCl4}$.

Bisulfite HSO3-  not Bisulfate HSO4- . you want to reduce the gold ions to gold metal. Since when is a copper chloride solution aqua regia?

That is not jow I read it. Rathetlr the > symbol is used for "better than"; CuCl2 is better than HCl +H2O2 and that is better than aqua regia (for selective extraction of the gold).


text:
Is following graph is of a dibasic or monobasic acid?

Is the above graph is of a dibasic or monobasic acid?
There is slight hint of it being of dibasic around 15. But is it enough to conclude ? I am confused.

Looks like dibasic.

As I understand it, if the data is absolutely correct, the curve should not change direction of curvature in that manner if it is monibasic. Each time it's double differentiation changes signs 2 times, I think it passes an equivalence point no matter how shallow it is. It should be a dibasic acid. You can get a function of [H+] using equilibrium constant (assume it to be monobasic) and taking x as moles of naoh added in other words, the "initial" moles of salt x is and "initial" moles of acid is a-x, a is the moles of acid we started with.By "initial" I mean the moles before equilibrium. You would get a quadratic equation, see at how many times sign of double differentiation of -log([H+]) wrt x changes for any value of x. Here is oxalic acid's curve.


text:
Has a carbon analogue of poly(dimethylsiloxane) been synthesized?
Consider poly(dimethylsiloxane) (PDMS), a member of the family of silicone polymers:

Now, imagine its carbon analogue:

I've not been able to find any evidence that this carbon analogue has been experimentally prepared. Searching for "poly(dimethylalkoxane)" got me nowhere.  I eventually searched for "dimethyl polyoxymethylene," which turned up the unsubstituted variant of the above, polyoxymethylene (POM):

as well as methyl-capped POM oligomers, polyoxymethylene dimethyl ethers:

But, no "poly(dimethyloxymethylene)" was forthcoming.
Has the carbon analogue of poly(dimethylsiloxane) been synthesized? If so, what name has it been given?

What you are looking for is the polymer of acetone, not "poly(dimethylalkoxane)".

@DHMO Sure! Whatever it would be called, I haven't found the name yet. If you find it and post an answer with a link & description... winner!

I have not been able to find anything... acetone does have dimers and trimers but those are not what you would expect...

http://www.sciencedirect.com/science/article/pii/0032386170900807 http://onlinelibrary.wiley.com/doi/10.1002/apmc.1996.052360101/pdf

So, if I read those links correctly, @Mithoron, the acetone polymerization apparently goes to products other than the polymer of interest?

Yeah, condensation seems prevalent.

http://chemistry.stackexchange.com/questions/25115/acetone-and-sodium-hydroxide/

So, @CurtF., this is just supporting the argument that polymerization of acetone does not yield the polymer of interest?

Yes, that's right Brian.

The dimer exists, at least in chemical databases, and it is called 2,2'-oxydipropan-2-ol.  However, according to ChemDoodle the proper IUPAC name is 2-(1-hydroxy-1-methylethoxy)-2-propanol.  I didn't find any evidence for the existence of the trimer or polymer in any databases, though.

The comments to the original question do well to demonstrate that acetone polyacetal (or polyacetone; names in the literature vary, but these are the most common) is extremely difficult to produce by typical polymerization methods; this is mainly due to thermodynamic reasons, though acetal dimers and trimers of acetone have been produced at very low temperatures.1
Polyoxymethylene is the polymeric form of paraformaldehyde (formaldehyde oligomers), which forms spontaneously in aqueous solutions of formaldehyde. The methyl-capped paraformaldehyde and polyoxymethylene would form if methanol is added to the solution; indeed, this is actually typically done to limit the length of paraformaldehyde oligomers. Polyoxymethylene, as a homopolymer, is quite stable and machinable, provided end-capping is performed; this is typically done with acetic anhydride.2

Burnop, V. C. E. Polymer 1965, 6 (8), 411-417. DOI: 10.1016/0032-3861(65)90025-X
Boyer, Lyndon D.; Sorenson, Wayne R.. Polyoxymethylene process. U.S. Patent 3355428, issued November 28, 1967.


text:
Difference between quick lime, slaked lime, lime water and soda lime
The chemicals "quick lime", "slaked lime", "lime water" and "soda lime" all have "lime" in common. What is the difference between them?

Aha, I wondered if this would be a self-answer -- most or all of this is on Wikipedia.

'Milk of lime' is somewhat similar to limewater. It is the suspension of slaked lime in water whereas limewater is a true solution.

Quick lime is calcium oxide, $\ce{CaO}$, in the solid state.
Slaked lime is calcium hydroxide, $\ce{Ca(OH)2}$, in the solid state.
Lime water is a dilute solution of calcium hydroxide, $\ce{Ca(OH)2}$, in water.
Soda lime is a mixture of sodium hydroxide, $\ce{NaOH}$, (also known as caustic soda) and calcium oxide, $\ce{CaO}$, (also known as quicklime).

Thanks for posting this! Can you add links to Wikipedia or elsewhere? Most of these are on Wik.

@hBy2Py, I did not refer to wikipedia. I knew this.

You may know these things, but including links to outside sources increases the usefulness of your answer to others.

Slaked lime comes from the idea of slaking the "thirst" of quick ("live" or "active") lime with water. Sort of an alchemical etymology.

To add to the family of materials in the 'lime' category: The sedimentary rock limestone is composed of calcium carbonate, $\ce{CaCO3}$.

According to Wikipedia:

Quick lime is $\ce{CaO}$ (calcium oxide)
Slaked lime is $\ce{Ca(OH)2}$ (calcium hydroxide)
Lime water is just a dilute solution of calcium hydroxide.
Soda lime is a mixture of about 75% calcium hydroxide, a few percent each of sodium hydroxide and potassium hydroxide, and about 20% water.  It is commonly used to adsorb $\ce{CO2}$ from the air.

Soda lime works great for adsorbing other acidic gases, too: $\ce{NO2}$, $\ce{SO3}$, etc. I used it in my grad work for $\ce{NO2}$.

I completely understand your frustration as all those names are trivial names (non-systematic.)
The "lime" is derived from the use of Limestone
https://www.etymonline.com/word/lime
I would have to mention that even those names are associated with chemical names that are not pure and contain some impurities.
Lime = quicklime ($\ce{CaO}$ - Calcium oxide) is made from limestone ($\ce{CaCO3}$ - Calcium carbonate) by decarbonisation at high temperatures.
To get slaked lime = hydrated lime ($\ce{Ca(OH)2}$ - Calcium hydroxide) you mix Lime with water, resulting in a white powdery substance.
In another process, you can make lime water - where you add lime into a greater amount of water resulting in water solution/suspension of hydrated lime
$\ce{Ca(OH)2}$  in water - this is also called milk of lime.


text:
Volume in decimetres cubed dimensional analysis
In chemistry class, we were learning about concentration, which is the number of moles of a substance per decimeter cubed of water, expressed as $\mathrm{mol} \cdot \mathrm{dm}^{-3}$.
However, I was confused when my teacher wrote decimeters cubed as $\mathrm{dm}^{-3}$ instead of what I presumed would be $\mathrm{dm}^3$. I asked why and he didn't know. Why do chemists write it like this?

This touches a very powerful analytical technique called unit analysis (aka dimensional analysis).
Moles per decimetre cubed is $\dfrac{\text{moles}}{\text{dm}^3} = \text{moles}\cdot \text{dm}^{-3} \ne \text{moles}\cdot \text{dm}^3$
So the -3 indicates that the unit is in the denominator not the numerator.
PS - I learned chemistry when there were only four elements - earth, wind, water and air. I think of liters not $\text{dm}^3$.

A useful addition may be that $\frac{1}{L} = L^{-1}$ and $(L^{-1})^3 = L^{-3}$.
And +1 for litres :)

The unit of volume is $\mathrm{dm}^{3}$. When you write moles per cubic decimeter (concentration), it can be expressed as
$${\mathrm{mol}\over\mathrm{dm}^{3}}$$
which is equivalent to
$$\mathrm{mol}\cdot{1\over\mathrm{dm}^{3}}$$
and because
$${1\over\mathrm{dm}^{3}} = \mathrm{dm}^{-3}$$
it can therefore be expressed as
$$\mathrm{mol}\cdot\mathrm{dm}^{-3}$$
In summary, it's simply because the volume term is in the denominator and the sign of the exponent changes when the product of it and $\mathrm{mol}$ is written instead of the fraction.


text:
Meaning of m2 in the symbol for an isotope of an element
What does the notation $^{197\mathrm{m}2}\ce{Pb}$ mean? Specifically, the '$\mathrm{m}$2' part.
I've found this and it appears to have something to do with charge distribution. The original notation was found here.

Related: How do I correctly typeset metastable radionuclide symbols?

As per this Wikipedia page on isotopes:

The letter $m$ is sometimes appended after the mass number to indicate a nuclear isomer, a metastable or energetically-excited nuclear state (as opposed to the lowest-energy ground state), for example $\ce{^{180\mathrm{m}}_73Ta}$ (tantalum-180m).

Note that the lower subscript to the left of the chemical symbol is the atomic number of the element (in the example above, 73 is the atomic number for tantalum) and is omitted in your example.
However, your question specifically is about the "m2" part. I'm quoting the Wikipedia page on nuclear isomers here:

For isotopes with more than one metastable isomer, "indices" are placed after the designation, and the labeling becomes m1, m2, m3, and so on.

In your case, 197 is the mass number for this lead isotope, and m2 is the second excited metastable state, with a half-life of 1.15 $\mu$s.


text:
What is the mechanism of addition of glycine into 5-chloro-N-methylisatoic anhydride?
The following transformation is involved in a step in the synthesis of diazepam:
(†)
What is the mechanism of the transformation from the second to the third intermediate (N-methylisatoic anhydride to 1 via glycine in triethylamine)?
There are a few possible electrophiles and nucleophiles that could be involved here. Either the aniline could attack the glycine carboxyllic acid, or the glycine could attack either of the anhydride carbonyls. The paper, however, says that the glycinate is formed, so I don't think the former option is viable. Based on the pKa's involved, what might be the most reasonable proposal?
(†) J. Org. Chem., Vol. 45, No. 9, 1980, 1679

I wonder if they mischaracterized the intermediate 1. The addition of acetic anhydride would still lead to 2 if we had the aniline-acid before right closure. I am a bit suspicious that the aniline would be able to add to a carboxylic acid with just triethylamine. Of course, I could be missing some kind of activation pathway, but I don't see it at the moment.

well in the paper they say that they use triethylamine to generate the primary amine. I was thinking that the primary amine attacked the anhydride carbonyl, not the aniline attacked the glycinate carbonyl.
Would you mind drawing the aniline acid you had in mind?

Welcome to chem.SE. Please see our policy on homework, and amend your question to include your thoughts on the answer. Otherwise, it will likely be closed by the community.

@ToddMinehardt This doesn't apply to self-questions so strictly right?

@DHMO - Perhaps, perhaps not. See this meta thread for clarification. My interpretation is that the OP must show some effort towards an answer, be it an attempted solution, research, or a general report on what they have found so far.

Generally, homework assignments don't constitute references to papers from JOC...

Follow up to my previous comment. Do they have supporting information for any intermediates? NMR, perhaps?

@Zhe No, it's a purely practical synthesis paper, but given that we know for sure that we have the glycinate anion, it doesn't seem too reasonable to me to think that the methylaniline moeity would attack the carboxylate. Really, all I wanted was to understand what was going on, and DHMO's answer satisfies that requirement.

A mixture of finely ground 5-chloro-N-methylisatoic anhydride (5.19 g), 2.25g of glycine, 4.15 ml of triethylamine. and 30 ml of water was stirred at room temp for 5 hours. All solid material had disappeared after 3.5-4 h. Voilatile material was removed as completely as possible on a rotavap and the residue treated with 60 ml glacial acetic acid and heated to reflux for 4.5 hours. After the mixture cooled, as much acetic acid as possible was removed on the rotary evaporator, and the tan oily residue was treated with 30 ml of ether. On brief swirling of the mixture, crystallization set in, and the colorless crystalline materiual was collecting after standing overnight and was washed with ether (4.60g, mp 176.5-178°C). The etheral filtrate (two phases) was diluted with enough ethyl acetate to render it homogenous, washed twice with dilute sodium carbonate, then with water, filtered through anhydrous sodium sulfate, and concentrated. Recrystallization of the crystalline residue (0.53g) gave 0.43 g of product, mp 177-179°C. Total yield 5.03g (91.8%).

This implies that we  have an initial basic aqueous environment, and we have an acidic aqueous environment afterwards.

The glyine is deprotonated by the triethylamine to form $(1)$.
The nitrogen in the primary amine carboxylate attacks the carbon of one of the carboxylates in the 5-chloro-N-methylisatoic anhydride to form the complex $(2)$.
Some rearrangements of the electrons give $(3)$.
Decarboxylation gives $(4)$.
Then, our environment becomes acidic. All the carboxylate anions are protonated to become carboxylic acids. The ketone is not protonated. We have a neutral complex $(5)$.
It proceeds through addition/elimination to give us our desired product.
Problems

In $(4)$, we have a negative charge on the nitrogen, in a basic environment, which shouldn't be favoured.
The nitrogen in $(5)$ isn't exactly nucleophilic enough.

An amide nitrogen (N-phenylated too) acting as a nucleophile in a 7-membered transition state - unusual.

The nitrogen in compound (4) is not very nucleophilic.

Indeed, on reading the procedure more carefully, refluxing glacial acetic acid  (!) is what's required to close the ring. In this case, you might well form the mixed anhydride as an intermediate to promote the ring closure.

The nucleophilic nitrogen is part of a carbamate and has an electron-withdrawing aryl group on it; the electrophile is an unactivated carboxylic acid (!). I'd go one step further than ron and say it's not legitimate. I agree with Zhe, there probably has to be some sort of activation of the carboxylic acids. Hmm, under acidic conditions, the carbamate should spontaneously lose CO2 anyway. That may pave the way to a cyclisation of aniline N onto an anhydride. I'm not sure, but it's definitely not what's presented above. The alternative is cyclisation of an enol onto the carbamate.

Here's a possibility 1) cyclize [3] to the expected uracil analogue, 2) then cyclize to generate a 5-membered lactone appended to the uracil, 3) eliminate $\ce{CO2}$.  I don't know if the resultant nitrogen ylide is a TS or intermediate, but it would readily ring expand to the desired product.

Compound 3 is very likely not stable and will decarboxylate immediately.

@orthocresol et al. (I can't ping multiple users) is it alright now?

I propose the following mechanism. Contrary to the answer provided by DHMO, I do not believe that the hemicarbamate intermediate is stable. Instead, I assume that it breaks down into carbon dioxide and an amido anion immediately upon formation or very shortly afterwards. This gives us a nucleophilic amido residue which can attack glycine’s carboxylate function. If need be, we can supply catalytic protons to the carboxylic group before the amido nitrogen’s attack but I have omitted that in the scheme.

Scheme 1: Proposed reaction pathway.
I consider the amino group of glycine to be the first attacker since it is most nucleophilic, and I consider it attacking the ester group since that should be more susceptible than a substituted hemicarbamate. Upon breakdown of the tetrahedric intermediate to give the first amide group, the free hemicarbamate anion is liberated which should immediately liberate carbon dioxide as stated above. The formation of the final amide linkage is thermodynamically favourable, even in spite of the unfavourable ring size.

That would necessitate the nucleophilic attack before protonation, though, since there's no other way the aniline gets deprotonated with mild base.

@orthocresol But even if the methylaniline is protonated, its lone pair is still sufficiently nucleophilic to attack a carboxy group. It’s just less nucleophilic than a free amine.

Yeah I agree with your mechanism in general. I think this (or some slight variation of it) makes the most sense.

I think this is like what I did in "combining some steps...", I think. Good point about the anionic hemicarbamate intermediate. You are saying that DHMO's intemediate (2) would not exist right?  It makes sense that CO2 would be liberated immediately, but for me this is a finer point. The essential points were that the ester is more electrophilic than the substituted hemicarbamate and that the glycine amino group is the nucleophile. Thanks, @Jan !

I came up with this, but DHMO proposed the same mechanism first.


text:
Why can't I balance this reaction algebraically?

$$\ce{aCu + bHNO3   ->  cCu(NO3)2 + dNO + eH2O}$$
  (Source)

I calculated:
\begin{align}
a &= c &(\ce{Cu})\\
b &= 2e &(\ce{H})\\
b &= 2c + d &(\ce{N})\\
3b &= 6c + d + e &(\ce{O})
\end{align}
And I tried substituting $a= 1$, but it couldn't solve the whole equation. How can I solve this algebraically?

$3 \ce{Cu} + 8 \ce{HNO_3} \rightarrow  3 \ce{Cu(NO_3)_2} + 2 \ce{NO} + 4 \ce{H_2O}$ is the balanced equation, found by accounting for electron transfer.

@TAR86 I want to know how to find that using equations.

You'd never solve it without some assumptions. See, you have 4 equations and 5 variables. Essentially, this means you have infinitely many solutions.

@IvanNeretin I tried to assume numbers, but every time one of the equations gives error.

Wolfram alpha finds: $b = \frac{8a}{3}, d = \frac{2a}{3},  e = \frac{4a}{3}$. Plug in $a=3$ to get whole numbers.

@Ivan there are not really infinitely many solutions if we regard the coefficients as molar ratios.  All the mathematical solutions have a unique set of ratios, e.g. we could have $a=3, b=8$ or $a=42, b=112$ but always $b/a=8/3$.

Given the reaction $\ce{a Cu + b HNO3 -> c Cu(NO3)2 + d NO + e H2O}$, we can set up four equations:
$$\begin{align}
a &= c & &\text{(for Cu)} \tag{1} \\
b &= 2e & &\text{(for H)} \tag{2} \\
b &= 2c+d & &\text{(for N)} \tag{3} \\
3b &= 6c+d+e & &\text{(for O)} \tag{4}
\end{align}$$
Substituting equation $(1)$ into the rest, we get
$$\begin{align}
b &= 2e & &\text{(for H)} \tag{2*} \\
b &= 2a+d & &\text{(for N)} \tag{3*} \\
3b &= 6a+d+e & &\text{(for O)} \tag{4*}
\end{align}$$
Substituting equation $(2^*)$ into equations $(3^*)$ and $(4^*)$, we have
$$\begin{align}
2e &= 2a+d & &\text{(for N)} \tag{3**} \\
6e &= 6a+d+e & &\text{(for O)} \tag{4**}
\end{align}$$
We can now multiply equation $(3^{**})$ by $3$ and subtract that from equation $(4^{**})$, which leaves us with $0 = -2d+e$. Therefore,
$$d = \frac{1}{2}e \tag{5}$$
Now, we try to express the constants $a,b,c,d$ in terms of $e$. Substituting equation $(5)$ back into equation $(3^{**})$, we find that
$$\begin{align}
2e &= 2a + \frac{1}{2}e \\
a &= \frac{3}{4}e \tag{6}
\end{align}$$
and since $a = c$ (equation $(1)$), we also find that
$$c = \frac{3}{4}e \tag{7}$$
So, using equations $(6)$, $(2)$, $(7)$, and $(5)$, we can now rewrite the reaction as:
$$\ce{$\frac{3}{4}e$ Cu + $2e$ HNO3 -> $\frac{3}{4}e$ Cu(NO3)2 + $\frac{1}{2}e$ NO + e H2O} \tag{8}$$
Clear the fractional exponents by multiplying by 4:
$$\ce{$3e$ Cu + $8e$ HNO3 -> $3e$ Cu(NO3)2 + $2e$ NO + $4e$ H2O} \tag{9}$$
Now $e$ can be any positive rational number ( such as an integer $1,2,3,4,\cdots,n$ or a fraction such as 1/2 or 3/4) and for a "balanced" chemical equation, we divide by the greatest common factor of all the integer factors, thus 1 is the correct value to use. Thus the balanced equation is:
$$\ce{3 Cu + 8 HNO3 -> 3 Cu(NO3)2 + 2 NO +  4 H2O} \tag{10}$$

Note that if the equation had been something like
    $$\ce{$6e$ Cu + $16e$ HNO3 -> $6e$ Cu(NO3)2 + $4e$ NO + $8e$ H2O} \tag{11}$$
then $e$ would have been 1/2 since all the species in the equation have a factor of 2 in common. For a "balanced" chemical equation integer values where the gcd is 1 are desired.

As Paul Harvey used to say here is "the rest of the story..."
The actual chemical reaction is:
$$\ce{a Cu   +   b HNO3   ->  c Cu(NO3)2    +   dNO   +    eNO2 + fH2O}$$
and this reaction doesn't have any unique solution.
In weak nitric acid the favored reaction is:
$$\ce{3 Cu   +   8 HNO3   ->  3 Cu(NO3)2    +  2 NO   +    4H2O}$$
but in strong nitric acid the favored reaction is:
$$\ce{ Cu   +   4 HNO3   ->   Cu(NO3)2    +  2 NO2   +    2H2O}$$

This is a very good step-by-step walkthrough on how to solve a system of linear equations, but I must complain about one line. There is no constraint that $e$ is a positive integer, it can be any real number. If $e$ were to be $\pi^2$, the equation is still as balanced as it would be if $e$ were $1$. The choice of setting $e = 1$ is simply one of simplicity - nothing more than that.

@orthocresol - I guess we'll have to agree to disagree. Having $\pi^2$ atoms of Cu makes no sense in what is traditionally thought of "balancing" equations. The mineral olivine on the other hand with the formula formula $\ce{(Mg^{2+}, Fe^{2+})2SiO4}$ can have any ratio of  formula  $\ce{Mg^{2+}}$ to  $\ce{Fe^{2+}}$ and almost certainly would not have definite integer values for  $\ce{Mg^{2+}}$, $\ce{Fe^{2+}}$ and $\ce{SiO_4^{2-}}$.

Ok, edited the answer a bit to make the part about $e$ better. $e$ could also be a rational fraction like 1/2 or 4/3. $\pi^2$ however makes no sense to me in a "balanced chemical equation."

Here's how to solve the system in python:
>>> import numpy as np
>>> stoich_mat = np.array([[-1, 0, 1, 0, 0],  # copper balance
                           [0, -1, 0, 0, 2],  # hydrogen
                           [0, -1, 2, 1, 0],  # nitrogen
                           [0, -3, 6, 1, 1],  # oxygen
                           [1, 0, 0, 0, 0]])  # impose constraint that a == val
>>> rhs = np.array([0, 0, 0, 0, 1])  # last number is assumed val of a
>>> np.linalg.solve(stoich_mat, rhs)
array([ 1.        ,  2.66666667,  1.        ,  0.66666667,  1.33333333])

Thus this solution says:
$$\ce{Cu   +   2.67 HNO3   ->  Cu(NO3)2    +   0.67 NO   +    1.33 H2O}$$
Multiplying through by 3:
>>> rhs_times_3 = np.array([0, 0, 0, 0, 3])
>>> np.linalg.solve(stoich_mat, rhs_times_3)
array([ 3.,  8.,  3.,  2.,  4.])

$$\ce{3 Cu   +   8 HNO3   ->  3 Cu(NO3)2    +   2 NO   +    4 H2O}$$
If your question is how computers (or people) can solve systems of linear equations, you can read up on Gaussian elimination and related techniques.

Well, I'm not sure about using python to solve stoichiometry problems. Could/Would anyone do that for a chemistry test now using a smartphone or some other computer device? (Being an old coot I still have a flip-phone.) I just can't imagine reverting to programming for a chemistry test.

@MaxW FWIW You could use python or a number of other languages on a smart phone using an online service like https://repl.it, but I don't think its very practical. More pragmatically, pulling up Wolfram Alpha on your phone is a pretty easy thing to do, and I personally have a TI-89 app on mine that will make quick work of a system of linear equations

The point isn't Python per SE, but to show how to set up problems like this for solution using linear algebra and matrices.  These days, I'm not sure its even fair to call the act of using a computer to solve a matrix equation "programming".

Using SymPy instead of NumPy, from sympy import *; Matrix([[-1,0,1,0,0,0],[0,-1,0,0,2,0],[0,-1,2,1,0,0],[0,-3,6,1,1,0],[1,0,0,0,0,1]]).rref() produces (Matrix([ [1, 0, 0, 0, 0,   1], [0, 1, 0, 0, 0, 8/3], [0, 0, 1, 0, 0,   1], [0, 0, 0, 1, 0, 2/3], [0, 0, 0, 0, 1, 4/3]]), [0, 1, 2, 3, 4]).

You seem to have had the right idea of fixing the scale by arbitrarily choosing the value of one coefficient, and then solving for the rest.  Apparently, you just got stuck at some point, presumably either because you couldn't solve for $b$ just with simple substitutions, or because your initial choice of $a = 1$ gave you fractional values for the other coefficients that you didn't know how to deal with.
If you plug $a = 1$ into your original equations, you'll immediately get $c = 1$, and by substituting that into the other equations, you're left with:
\begin{aligned}
b & = 2e, \\
b & = 2+d, \text{ and} \\
3b & = 6+d+e.
\end{aligned}
One way to solve that remaining set of equations is to first subtract the second one from the last one, to get: $$3b - b = (6+d+e)-(2+d) \implies 2b = 4+e,$$ and then substitute in the first one to get: $$2(2e) = 4+e \implies 4e = 4 + e.$$  Subtracting $e$ from both sides then gives you $3e = 4$, which you can divide by 3 to get $e = \frac43$.
Once you have a numerical value for $e$, even if it's a fraction, you can then substitute that back into the first equation above to get $b = 2\cdot\frac43 = \frac83$, which you can then plug into the second one to get $\frac83 = 2 + d$, and then subtract 2 from both sides to get $d = \frac83-2 = \frac83 - \frac63 = \frac23$.
Now we have a solution $(a=1, b=\frac83, c=1, d=\frac23, e=\frac43)$, but it still contains fractional values that we'd like to get rid of.  The way to fix that, however, is simple: just multiply all the values by their least common denominator, 3, to get $(a=3, b=8, c=3, d=2, e=4)$.
The reason that works is because your original system of linear equations was, by construction, homogeneous, i.e. every term in every equation contained exactly one of the coefficients $a$, $b$, $c$, $d$ and $e$.  Thus, multiplying all the coefficients by the same scaling factor multiplies every term in the equations by the same amount, and thus turns any valid solution into another equally valid one.
Such rescaling does, in fact, have a reasonable physical interpretation: if we have $n$ instances of the reaction going on at the same time, then the combined reaction will obviously consume $n$ times as many of each reactant and produce $n$ times as many of each product.  Allowing $n$ to take on fractional values does require a bit more of a mathematical leap of imagination, but we may e.g. interpret the fractionally scaled reaction as describing just a fraction of the original equation — which may or may not be chemically meaningful, depending on whether all the scaled coefficients work out to whole numbers of molecules, but which nonetheless correctly describes the proportions of reactants consumed and products yielded.
Of course, you could've also arrived at the appropriately scaled solution directly, if you had happened to start with the initial guess $a = 3$ instead of $a = 1$.
Indeed, the fact that you didn't need to make any further arbitrary choices during the solution, after this initial choice of scale, proves that all the possible solutions to this homogeneous system of linear equations are simply scalar multiples of each other.  On the other hand, if you'd added a $+\ce{f NO2}$ term to the products of your reaction (as MaxW suggests in their answer, for extra realism), then you would've had to make an arbitrary choice about the proportion of $\ce{NO}$ and $\ce{NO2}$ products at some point during the solution (or, alternatively, leave it unspecified, leaving you with some non-numeric factors in your result), reflecting the fact that the solution space of this extended system of equations is multidimensional, i.e. has more than one degree of freedom, and that this extended reaction can thus yield varying proportions of its products depending on the conditions under which it occurs.

You have five variables but only four equations.
Algebra tells you that this is not a fully defined set of equations and that there are infinitely many possible solutions. For five variables you would need five equations.
However, a fifth constraint can be included by aiming for the smallest possible integer-only solution.

The missing equation is the scale, which, for any homogeneous linear system with non-trivial solutions, may be chosen arbitrarily. After all, if $(a,b,c,d,e)$ is a solution, then so is $(ka,kb,kc,kd,ke)$ for any $k$. Generally, we'd like to pick the smallest scaling factor that makes all the coefficients into positive integers (or, at least, into non-negative integers that are not all zero). The OP tried to do this by fixing $a=1$; unfortunately, that choice fails to yield integer values for the other coefficients.

5 variables / 4 equations - that simply isn't true. The reaction: $$\ce{3 Cu   +   8 HNO3   ->  3 Cu(NO3)2    +  2 NO   +    4H2O}$$ does have a unique solution. A fifth condition is that the gcd of all the coefficients is removed to yield the smallest set of integers possible.

@MaxW it absolutely is true.  There are 4 elements in the equation that are balanced.  I'm not sure what you mean by GCD (greatest common denominator?), but the solution that you give is not necessarily unique.  My answer gives a solution with non-integer coefficients, for example.

@IlmariKaronen Yes, that was what I was implying for infinitely many solutions.

@MaxW I was speaking from a pure algebraic point of view. Including the constraint of ‘lowest possible integer coefficients’ does indeed generate a fifth equation that can be used.

Just for fun suppose we take the element balance equations and multiply each of them by the coefficient indicated below:
Copper: $a=c$, $×0$
Hydrogen: $b=2e$, $×1$
Nitrogen: $b=2c+d$, $×5$
Oxygen: $3b=6c+d+e$, $×(-2)$
Add up that linear combination:
$0=-2c+3d$
So $c/d=3/2$. Assign values of the copper nitrate and nitric oxide coefficients accordingly and the rest of the system easily yields.
Hey wait, have we not seen that ratio before?  It's the ratio we get for those two coefficients based on the standard redox method.  When you use the redox method, you are using oxidation states as a tool to find a linear combination that simplifies the solution.


text:
hypo residual chlorine testing using a 10ml reagent pack
Why when testing a sample for chlorine residual do I get the same results for 5 ml sample of water as I do with a 10 ml sample, and a 15 ml sample of water using a reagent pack that asks for a 10 ml. The reagent container does not give the weight of the reagent in the packet (it is powder), but I think it might weigh 0.1 grams. I am just using a Hach DR 900 to test the free chlorine residual.
My thinking is if the reagent packet is asking for 10 ml sample of water shouldn't my results differ when using different sample sizes. For the three different volumes of samples, I got a free chlorine residual of 1.00 mg/L for all three sample sizes.
Sorry, the packet is a DPD free chlorine reagent for 10 ml sample. The packet contains Sodium Phosphate, Dibasic, EDTA Disodium Salt, DPD Salt, and Carboxylate Salt.

First of all welcome!  Please take note of the changes made to the formatting of your original question.  Also, before this question can reasonably be answered, you will need to provide units (grams, gallons, pounds, etc.) for every number given in your question, and a description of how much of what is in your reagent pack.  Also, where you say you get "the same results" you should say what the results are.  I think that once your question is thoroughly and properly expressed, there is a good chance that someone will be able to help you find a good explanation for your problem.  Good luck!

This is sort of an extended comment, not necessarily an answer.
Overall, yes I'd expect different numbers at least. How they would vary is a matter of some conjecture.
(1) The Hach DR 900 is a colorimeter. I'm not sure how large of a sample you need to fill active volume of the colorimeter cell.
(2) The ratio of water volume relative to the amount of solid reagent is changing. So the chemical equilibrium on which the calibration curve depends might shift. If you dilute 5 and 10 ml samples to 15 ml with "pure" water and take 10 ml of each of the three samples (2 dilutions plus "full strength"), then you be comparing better.
(3) Even given (2), the results seems odd. Normal drinking water should have between 0.2 and 2.0 mg/L.

Is pH of the water ok?
Does water have some unusual buffer capacity?
Reagent pack old?
Do you have some standards which you can test?