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Why is sulfur dioxide more stable to UV radiation than ozone?
Ozone readily decomposes when impacted by UV radiation.
Sulfur dioxide, having the same bond structure and valence electrons in each atom as ozone, is much more stable.
Would someone please help me understand what makes sulfur dioxide a more stable compound?

I'm going to attempt to answer this from three points of view.
Qualitative considerations and MO theory:
First, why isn't ozone linear? You can imagine $\ce{O3}$ like a $\ce{CO2}$ molecule that lacks two electrons. This lack of electrons will result in less $s$-character, and less bonding "attitude". That's why ozone is bent.
Still, the repulsion in its electrons ($\ce{O}$ is a small central atom, very small) will want it to be free. Oxygen's van der Waals radius is 1.5 $\mathrm{\overset{\circ}A}$. Thus, $\ce{O3}$ has a bond length of $\approx 1.2\ \mathrm{\overset{\circ}A}$. That's really cramped. Electrons aren't happy with this and will try to get free as soon as $E_a$ and other conditions are provided.
Now compare that with sulfur's 1.8 $\mathrm{\overset{\circ}A}$ radius; which, results in a $\approx 1.43\ \mathrm{\overset{\circ}A}$ bond length. Less cramped, the electrons aren't as much complaining as they do in $\ce{O3}$'s case. We can't get really further as these two molecules aren't isoelectronic.
_____________________________________________
Approach using classic high school knowledge:
It's because oxygen has an arbitrarily positive formal charge. Take a look at $\ce{O3}$'s electron-dot diagram:
$\hspace{18ex}$
$\hspace{45ex}$ $\small{Source}$
The oxygen atom, which has an electronegativity of $3.6$ in Allen scale, has formed $sp^2$ orbitals. This, a very electronegative atom with a positive formal charge, is highly unfavorable for the molecule, and acts very "non-bonding".
The sulfur, on the other hand, has an electronegativity of $\approx 2.5$. It will also get a positive formal charge. But since it's less electronegative, it bears it.
_____________________________________________
Thermodynamics and kinetics:
Ozone decays into diatomic oxygen under certain conditions:
$$\ce{2O3 -> 3O2}$$
Three moles of oxygen will be produced per two moles of ozone. The measured enthalpy of formation for ozone is $142~ \mathrm{kJ/mol}$, while it's 0 for diatomic oxygen. So
$$\mathsf{\Delta H_f~of ~the~ products - \Delta H_f ~of ~the ~reactants} = \mathsf{reaction~ enthalpy}$$
$$ 0 - 142 = \fbox{-142} ~\mathrm{kJ/mol}$$
Furthermore, this reaction will increase entropy, as two moles of a gas produce another three moles. Thus, it's thermodynamically highly favorable.
Above that, its energy of activation is a measly $\approx 32 \mathrm{~kJ/mol}$, which is a very small kinetic barrier.
Thus, you can also conclude that $\ce{O3}$ is exceptionally "inclined to decompose". Also, note that ozone is a very strong oxidant; while $\ce{SO2}$ is a weaker one.

Just wanted to say thanks for the quick response and thorough work on this!

The bonding of ozone is explained in other questions. The important point is that its bonds are weaker than SO$_2$ even in the ground state. The dissociation energy of SO$_2$ ground state is approx 5 times larger than  O$_3$. The effect of 3 vs 4 bonds to put it casually.
However, this question asks about the stability of SO$_2$ vs. O$_3$ to ultra-violet radiation. This means that the thermodynamic stability and bonding in the ground state is not important as dissociation occurs from one of the excited states.  Excited states are generally less stable than the ground state as a photon promotes an electron from a bonding to an anti-bonding orbital
As it happens ozone can be dissociated from its Chappius bands (visible absorption approx 500 to 800nm) into an oxygen atom (in term symbol notation $^3P$) and an oxygen molecule ($^3\Sigma ^-_g$) with a yield of >90 %. (The products also form for absorption up to 320 nm. At shorter wavelengths O($^1D$) and O$_2$ ($^1\Delta_g$)are produced). Dissociation happens from the lowest excited state because this excite state is repulsive, i.e. has no minimum as the dissociating bond stretches. The ground state does have such a minimum which is why it is stable.
Dissociation for other excited states either occurs by pre-dissociation, where one excited state crosses another, or when the uv light is of sufficient energy that it exceeds the dissociation energy of the excite state. Clearly this is a complex subject but I hope you get the idea.
In SO$_2$ the first excite state absorbs in the 340-400 nm region (far blue) and is the result of interaction between two excited state. It has a large dissociation energy compared to ozone and fluorescence can be observed from this state. However, SO$_2$ will dissociate from this state if excited at a short enough wavelength.
The picture shows a schematic only of several potential energy profiles, energy vs bond extension (e.g. O$_2$ - O or SO-O). The units are arbitrary but typical if in nanometres and energy in wavenumbers but only for molecule with low energy excited states. The shapes of the curves are typical (Morse potential) but do not represent any particular molecule)

The black curve is the ground state, the grey a repulsive excited state (no minimum on energy) that causes the red curve to pre-dissociate where it crosses it. The blue curve is another higher energy excited state. You can see that excited states generally have a smaller dissociation energy than the ground state as an electron is promoted into an anti-bonding orbital


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Alloys/metallurgy : promising elements for for Sn-based solders
I am trying to research potential modifications to lead-free solders (intended to be used in electronics). Base alloy is eutectic $\ce{Bi_{58}Sn_{42}}$, and I am going to add various combinations of a bit of copper, silver, gold, platinum, gallium, lithium, and indium.
Is there something there which is known to not work? I am wondering this because I see that gold, platinum, gallium, and lithium are not typically used as dopants for electronic soldering alloys. Should I expect a lot of problems attributed to intermetallic compounds? (I know there are issues due to a Sn-Cu reaction, but nothing about the rest).
Is it theoretically possible to find a combination with significantly high copper content (> 10-20%) while still having low melting temperature (< 250 $^{\circ}\mathrm{C}$)?

Lithium might be tricky to alloy in an oxidizing atmosphere (e.g. air), and Pt might take some effort to form an alloy, because of its high m.p., which, however, is still less than than the boiling points of Bi or Sn. Other than that, there should be little trouble forming alloys.
You can search on line for "phase diagram alloy components" to find the melting point for at least some of the alloys. E.G. an online search for
phase diagram "copper-indium" alloy

turned up this link at Springer-Verlag GmbH, with the diagram below.

That said, I would expect the "10-20 % copper" requirement precludes melting point less than 300 C, even with low-melting In, Hg, Ga etc.
BTW, I would not use "dopant" for a component of an alloy that will not appreciably change its electronic properties; it is just another component, be it minor, of a ternary, quaternary etc. alloy.

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Keeping one component concentration constant and varying another in a solution
This is a simple question but it has been so long since I have done any chemistry. Basically I would like to keep the concentration of A constant and the concentration of B variable and vice versa.
Assuming we want to make a 50 ml solution with constant concentration of A and variable concentration of B.From my knowledge I would use C1V1=C2V2 in order to dilute A to its fixed concentration lets say from 4% to 3%. But how do i go about keeping it constant and B concentration variable for a final 50 ml solution.
EDIT:
Ok. So lets say A had an initial concentration of 4%. I want a constant A concentration of 3% in a volume of 50 ml. So I calculate my starting volume according to C1V1=C2V2 and get V1 to be (3*50)/4 and a starting volume of 37.5 ml. My remaining volume is 50-37.5 and get 12.5 ml. I want to make 5 solutions with 50 ml end volume. I put 12,5 ml of A in each flask giving me a constant A concentration of 3%. How do i finish up my 50 ml with different B concentrations?
2nd EDIT:
Apologies, this is my first time using this site and was told to edit my question and add my comment. I didn't put values as I believed it was clear enough but I guess not.
@ y chung that is what I have carried out in order to create the solutions, so I'm guessing I did it correctly then.
@ bon thanks for the comment. I will tell you exactly what Im trying to accomplish. I have two solutions one is A with a concentration of 9% and one is B with a concentration of 7%. I want to create 5 solutions each 50 ml with a constant 5% concentration of A and varying concentration of B from 1-5 %.
What I have done is as follows:
I reduced my A concentration to the required 5% by C1V1=C2V2 for a 50 ml solution getting 28 ml. I put 28 ml in each volumetric flask of 50 ml capacity.
I then calculated how much volume of B with the 7% concentration I would need to get it to say the 15 concentration based on my remaining volume of 22 (50-28). For the 1 % concentration this yielded a volume of B at 7% of 3.14 ml. I added that to my flask contain the 28 ml of A and made up the solution to 50ml.
The question is, is this correct? from what I gathered from y chung ,it is.
Thanks

Could you please write the main question in full? You've only vaguely shaped your thoughts around it but since we haven't seen it we can't really tell what's being asked here.

Ok. So lets say A had an initial concentration of 4%. I want a constant A concentration of 3% in a volume of 50 ml. So I calculate my starting volume according to C1V1=C2V2 and get V1 to be (3*50)/4 and a starting volume of 37.5 ml. My remaining volume is 50-37.5 and get 12.5 ml. I want to make 5 solutions with 50 ml end volume. I put 12,5 ml of A in each flask giving me a constant A concentration of 3%. How do i finish up my 50 ml with different B concentrations?

@John - If you can edit your question to include your comments, that will be a big help in getting a useful response.

Diluting A by a factor of four i.e. a 3:1 ratio,  will give you a 1% solution. If you can use another diluent that is not A or B, this is a trivial exercise. Just add five different volumes of B (that are less than 12.5ml) to your five lots of 37.5ml  A solution and top up to 50 ml. Unless I'm missing the point?

@John You might have accidentally created two accounts here, which can be merged, so that there is only one account that you have to take care of.

Add varying amount of water and a standard B solution on top of the constant amount of A you have added. Then you will finish with several solutions with different B concentration.

Perhaps you could be more specific about exactly what you would do. At the moment this would probably be considered 'low quality'.

He did not ask for what different concentrations of B he wanted. Also, He has done the keeping constant A part in his "EDIT" part himself. I don't think a very long and detail answer is required

Personally I have difficulty understanding what your answer means. I'm not asking for a long answer, just some additional explanation as to exactly what you are trying to say.

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Plasma creation and condensation
Every chemical has a decomposition temperature.  My understanding is that above that temperature molecular bonds are broken.  And if we raise the heat high enough then all molecular bonds will break and we will be left with a "plasma" of ionized atoms, right?
Can the decomposition temperature of bonds be predicted by models, and if so on what data do those models depend?
Now suppose we heat a chemical in an inert (noble) atmosphere, or in a vacuum, to the point of total decomposition.  Now we cool the plasma back to STP.  Are there models that can predict the products we would expect to find – i.e., the right hand side of the reaction equation?
I'm guessing that thermodynamics stipulates we should find the lowest-"energy" compounds that can be created from the constituent elements on the right hand side.
In practice I imagine that achieving that result could require that cooling be done at some "sufficiently low rate."  So, to increase the complexity of the question: Are there any models that can predict what products will be found as the cooling rate is increased?  I believe that in the limit (a virtually instantaneous quench) one ends up with a "glass" (or combination of glass, condensate, and gas) but even then one can't be left with free ions so can we predict the composition of the quenched plasma?

I'm afraid that asking sth like that without being specific is much too broad

Also edited tile - it was very vague

@Mithoron: I disagree that the title was vague, but we can try your suggestion for a bit.  It is a general question.  The answer may simply be, "No, there are no general models of decomposition or molecular reconstruction."  If there are, then  I expect them to be well known by theorists.  Would it help if I provided an example of a left-hand-side and then a bunch of right-hand sides that "balance" without any reason to believe they will actually occur?

Two factors that are going to make it difficult to predict the outcome of this type of experiment (except perhaps for very simple systems like hydrogen gas or chlorine plus bromine) are (1) the complex combinatorics lead to a huge number of intermediates and final products, many of whose energies/reactivity are not known; (2) kinetic barriers are harder to predict that ground state energies. As is often the case, you may have to go into the laboratory and do some experiments. Possible approach: aim the plasma at a supercooled probe in a vacuum and spectrally characterize the resulting matrix.

@iad22agp: That sounds like the summary of a good answer.  Could you expand points 1 and 2 a little bit and post as answer?  The whole question is whether a general theoretical solution is possible.  Obviously one can decompose a particular chemical and analyze the condensate in a lab, but I was wondering if theoretical chemistry offers models of recombinetrics of decomposed molecules.

To begin with, dissociating molecules to a plasma at the atomic level will not only create ions but also neutral atoms. These will indeed reassociate on cooling, and the question is how to predict what the products will be. In the extreme case of a simple system such as pure hydrogen gas (dissociating, then recombining to H2) or, say chlorine plus bromine dissociating/recombining to give BrCl, it should be easy to predict the products formed and perhaps more about the behavior of the system as a function of temperature.
If you have a more complex system, including (for the sake of argument) atoms such as carbon that can form complex ring and chain structures, predicting the outcome will be impossible - since (1) the different ways to combine atoms may lead to a huge number of possible intermediates, many of whose energies/reactivity are not known; (2) even if you could predict these intermediate structures, the number of further possible reactions between them becomes astronomical, and (3) kinetic barriers (activation energies) for  reactions are generally harder to predict that ground state energies. As is often the case when theory comes up short, you may have to go into the laboratory and do some experiments. Possible approach: aim the plasma (along with a stream of inert gas such as argon) at a supercooled probe in a vacuum and spectrally characterize the species in the resulting matrix. I imagine that you could end up with some pretty amazing chemical species that way.
I would urge those among my colleagues who are familiar with the current state of theoretical chemistry to add to this discussion. I love a question that makes you go back to basics and rethink everything from the ground up.

Rethinking theoretical chemistry from the basics? There is something they call the world formula. But that's as much a theoretical physics approach as it is chemistry. And it is very hard to get much grip on that. And I think they are not even close. And if/when it exists I would assume it is almost impossible to treat with or current hardware, even the most simplest systems, but that might change in a couple hundred years


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Why do we not see silicon in a structure like graphite?
Silicon exists in a structure similar to diamond, with 4 silicon atoms bonded to each other in a tetrahedron.
Why do we not see it exist in a structure like graphite?
Is it even possible to get silicon to exist like this? If so could we make graphene but with silicon atoms?
Not sure why you would, since I understand carbon graphene will still be cheaper. But I would just like to know.

An answer is this: http://www.madsci.org/posts/archives/1997-08/866552381.Ch.r.html.

It's possible, and it's already been done! The compound is called Silicene.

However, silicene generally isn't perfectly flat. (Then again, as  Karl Ratzsch points out in the comments, graphene isn't perfectly flat either). The Wikipedia article notes that it usually takes a puckered shape.
I don't know the reasons for the puckered shape, but I recall being quite surprised that silicene could be synthesized at all. The reason why is that I was taught in high school that silicon very rarely forms double bonds due to its large size. For a visualization, please refer to the absolutely awful image below (if anyone could provide an actual graphic I'd be grateful).

The black atoms on the left are carbon. The $p$ orbitals overlap to a significant degree, which means that pi-bonds can easily form. The green atoms to the right are silicon. You can see that, because of the larger atomic radius, the p-orbitals don't overlap as much, meaning that double bonds are generally not favorable in silicon. This explains why compounds like silenes are relatively unstable.
Pi systems are crucial to the chemistry of carbon--they allow carbon to form multiple bonds, flat rings (like in graphene), and generally be stable in a lot of configurations. The lack of pi-bonding in silicon is sometimes given as a rationalization for why silicon-based lifeforms can't exist, and explains why it's so hard to get silicon into the flat sheets that form silicene.

Graphene is not flat, either. Cannot be, because only 3D crystallites have perfect long-range order. Landau-Peierls instability this is called.

@Karl I've just asked Are all stable 2D Xenes and binaries buckled?


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How determine the number of significant figures of a feet/inch system?
How many significant figures are there in $5~\mathrm{ft}\ 10~\mathrm{in}$.
I know that $5~\mathrm{ft}\ 10~\mathrm{in}= 5.8\bar{3}~\mathrm{ft}=70~\mathrm{in}$, so I am not sure exactly how much significant digits are in this number.

Significant figures is a rule of thumb, to avoid giving overly precise results when the input doesn't warrant it.
In actual scientific calculations - where keeping track of the precision matters - significant figures aren't used. Instead, explicit error ranges (confidence intervals) are used. Notation of these varies, but often takes the form of a plus-minus format (14.5±0.3 m). The concept of significant figures uses implicit error ranges, conventionally set to one half of the last significant digit.
When error ranges are explicitly given, the error of a calculation result is found with formal propagation of uncertainty, rather than the rule-of-thumb significant figure rules. Sometimes this is based on a model of the distribution of error around the central value (for example, assuming that the error is normally distributed around the central value with a given standard deviation), but most commonly the error is taken as the minimum and maximum of an acceptable range of values around the main measurement. Under this scheme, the propagation of uncertainty often takes the form of the "crank three times" method, whereby you run the calculation once for the central value, once for the minimum value, and once for the maximum value.
So if you were doing a calculation like $\pu{14.5\pm0.3 m} - \pu{1.5\pm0.2 m}$, you would calculate once for the central value ($\pu{14.5 m - 1.5 m = 13 m}$), once to give you the smallest value you could get with the range ($\pu{14.2 m - 1.7 m = 12.5 m}$), and once to give the largest value your could get with the range ($\pu{14.8 m - 1.3 m = 14.5 m}$). You'd then convert this into a range, normally rounding both values such that the error value only has 1-2 significant figures: $\pu{13\pm0.5 m}$
It turns out that the rules for significant figure calculations are just a quick-and-dirty rule of thumb of doing more-or-less the same thing. I'd encourage you to try it out with a number of examples. If you assume that the error in each quantity is one half of the last significant digit, and you do the crank three times error propagation, you should come up with a result that has an error that is about one half the last significant digit of the answer by the significant figures approach - give or take a digit.
So really what you want to do is to convert the measurement of $\pu{5 ft 10 in}$. into one which has an explicit error associated with it. For this measurement, I'm guessing you'd have an error of $\pu{\pm0.5 in}$. - I say this because rounding to the nearest 10 inches seems rather silly to me. (Feel free to work the calculations below with different assumption, say with ±5 in., if you disagree.) So you have a measurement of $\pu{5 ft 10 in. \pm 0.5 in}$ in other words $\pu{5 ft} (\text{exact}) + 10\pm\pu{0.5 in}. = \pu{5 ft} (\text{exact}) \times \pu{12 in./ft }(\text{exact}) + \pu{10\pm0.5 in.} = \pu{70\pm0.5 in.}$ Note that is the same as $70$. in with two significant digits.
Since we have explicit error bars, we can use the crank three times method to convert it to feet: $70~\mathrm{in.} / (12~\mathrm{in./ft}) = 5.8\bar{3}~\mathrm{ft}$ and $69.5~\mathrm{in.} / (12~\mathrm{in./ft}) = 5.791\bar{6}~\mathrm{ft}$ and $70.5~\mathrm{in.} / (12~\mathrm{in./ft}) = 5.875~\mathrm{ft}$. This is $5.8\bar{3}±0.041\bar{6}~\mathrm{ft}$. Convention holds that we round the error to one or two significant figures, and also round the main value to the same decimal place: $5.83±0.04~\mathrm{ft}$. -- This is basically what you'd have from $\pu{5.8 ft}$ with two significant digits.
The upshot is that you'd never use $\pu{5 ft 10 in.}$ in a calculation. You'll want to convert it to a single consistent unit first. When you do that, keep in mind the actual error associated with the measurement, and use that error as a gauge at how precise the converted quantity is.
(Note that if this is a question on a test or on a homework, ask your teacher how they think you should get the "correct" answer. When dealing with rule-of-thumb things like significant figures, intelligent people can vary in their interpretations. This means an answer valid in one context can be marked as wrong on the answer key because it doesn't match the formalism the teacher is using.)

"The ± format should be avoided whenever possible". Source: Guide to the expression of uncertainty in measurement by ISO, IUPAC, IUPAP etc., see also https://chemistry.stackexchange.com/a/58667/24052

All non-zero digits are significant(62 has 2 sig fig)
In a number without a decimal point, only zeros between non-zero digits are significant(206's 0 is significant).
In a number with a decimal point, all zeros to the right of the first non-zero digits are significant(eg. 15.30 the zero is significant but 015.3 the zero is not significant. 15.30 has 4 sig fig but 015.3 only has 3).

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help].

This is true but doesn't address the OP's problem which is specifically about significant figures in non-SI unit systems.

Express the specified value in inches, as you have done (since that is a more precise measure of length than feet) and then count the significant figures.


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What is the geometry of an alkyl radical?
I encountered this question which asks how many possible radicals are formed when $\ce{CH3CH2C(CH3)3}$ is monosubstituted by $\ce{Br2}$. The answer given is 3 while I think it  should be 4, reasoning that there is a "chiral" carbon if the radical is formed on carbon 3. While arguing with my tutor two questions came to my mind:

What is the geometry of the $\ce{CH3\dot{\ce{C}}HC(CH3)3}$ radical?
If the radical is trigonal pyramidal, does it transition quickly between the two possible states? If so are they considered two radicals?

My findings so far:
This answer on SE hints that the exact shape depends on the substituents on the alkyl radical.
Modern Physical Organic Chemistry states that all other (non-methyl) localized radicals are not planar.
Organic Reaction Mechanisms states that the geometry of free radicals is still controversial.

First off, it was very astute of you to recognize that if the radical formed at carbon 3 (the methylene carbon in the starting compound) is pyramidal, then the radical would be chiral ( the lone radical electron serving as the 4th different substituent on that carbon) and 4 different radicals would be possible.
In a typical carbocation, the carbon bearing the positive charge is $\ce{sp^2}$ hybridized, meaning that the p-orbital at the cationic center is empty and the carbocation is planar.  On the other hand, in a typical carbanion, the negatively charged carbon is pyramidal and approaches $\ce{sp^3}$ hybridization. This $\ce{sp^3}$-like orbital contains 2 electrons.  Therefore, it would be reasonable to expect that a typical hydrocarbon radical, where the orbital holds 1 electron would be somewhere in between these two extremes and be slightly pyramidal.

This is indeed the case and the degree of non-planarity is very sensitive to the substituents attached to the radical center.  In the case of hydrocarbon radicals, such as those created from your starting material, the barrier to inversion from one pyramidal form to the other is extremely low, on the order of a few kcal/mole or so.  Therefore, interconversion between the two enantiomeric forms of the radical would be extremely rapid at room temperature.
So yes, the $\ce{CH_3\dot{C}HC(CH3)3}$ radical is slightly pyramidal and it does transition rapidly between the two pyramidal enantiomers.  Your answer of 4 radicals is correct, as long as you understand that 2 of them rapidly interconvert at room temperature.

Indeed it should be less pyramidal than the carbanion. Chiral hydrogen-bearing carbon atoms adjacent to C=O groups will racemise slowly in basic solution through the carbanion intermediate (how much time they actually spend in the carbanion form will depend a lot on pH). One would expect racemization of the radical to be considerably more rapid than that.

Be careful! It is not reasonable at all to assume that the alkyle radical is "something in between a carbanion and a carbocation". A quick look into the OC textbooks tells you that the methyl radical is in fact planar. The tBu radical on the other hand is pyramidal, but for steric(!) reasons, with an inversion barrier of a mere 1.2 kcal/mol. For the given example,the barrier is probably less, IF the thing is not in fact planar. And unless you have evidence for the contrary, i would assume the latter, and say "4" is wrong.

@KarlRatzsch  Why is it not reasonable to assume that the alkyl radical is something in between a carbanion and a carbocation?  I would agree that the barrier to planarity is very low in these cases.  So low in fact that experiment (or theoretical calculations) cannot distinguish between planarity and a very small deviation towards pyramidal.  OC textbooks are of little help in this case.  Bottom line, there is an electron that can be stabilized by mixing in some s-character which will cause (slight) pyramidalization...

@KarlRatzsch  ...The 3 $\ce{C-H}$ bonds will give give up some s-character in order that the lone electron can exist in a lower energy orbital containing some s-character.

120° and 109° are the two possible geometric extremes, saying that the radical is something in between does not explain or predict anything.  And nobody has been able to provide synthetic proof for a pyramidal C radical (not talking about ones with hetero atoms or cycles). You are right, it is reasonable to believe that secondary orbital interactions will lead to a slight deviation from planarity, but as long as you cannot quantify or even prove it, I would always put a question mark behind the "4". And anyway a single number is never an answer. Right? Least that's what I tell my students.

@KarlRatzsch  You make some good points. However,  I don't need to measure the 4 $\ce{C-H}$ bonds in methane to "know" that they will all be equivalent, symmetry tells me that is so.  In the case at hand, when I compare a radical with 1 electron in an orbital to a cation with no electrons in an orbital and to a carbanion with 2 electrons in an orbital, I "know" that the radical is in between.  It will not be as flat as the carbocation, nor as pyramidal as the carbanion.  If an experiment suggests that the radical is planar, I would say that you need tomeasure to the next decimal place.

I would also take some care with the categorising the radical in between a cation and an anion - the geometry certainly will be different from both cases. The radical has ungerade spin, while the others have gerade spin, which can often lead to significant deviations. However in computational chemistry, as a good practise kind of thing, it is often advised to calculate the singlet cation of the radical as a guess for the radical itself, since in most cases the behaviour is quite similar (no rule though).

As your books clearly indicate, nobody seems to have been able to find a chemical proof for the geometry of C-radicals. If it's not indeed planar, it equilibrates extremely quickly between the two pyramidal shapes. Unless there is some further steric constraint that keeps it in one conformation.
For the argument with your tutor: The correct answer is "two diffent primary radicals and one secondary radical". If someone asks for a number, and you have the slightest doubt, write two, three senctences explaining it. If you're lucky, the tutor might say "extra point for the one guy who noticed the question was stupid". If not, he'll still happily give you 8 out of ten. And if you're at a school where this is not the case: Get the hell out of there.

Let's rephrase the question. How many different radicals can be formed from 2,2-dimethylbutane? Answer = 3
OR
How many monobromo stereoisomers can be formed from 2,2-dimethylbutane? Answer = 4. Two primary bromides and a racemic secondary bromide.


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What makes an X-ray machine's white photographic film go black?
I read on a   bbc gcse bitesize webpage "In older X-ray machines, white photographic film" [edit- I now know that bbc webpage is wrong, and it's not a white film. I'll leave the rest of the post after the next sentence, as is, but note that from the next paragraph, the rest of the post is based on the misunderstanding from that bbc link that there is white film  (And I guess it's talking of any pre digital when it says 'older' - and anyhow prior to film they used photographic plates - glass rather than plastic/film ).]
I understand that X-ray images are black and white. You have the X-rays themselves, the white film, and an object/person in-between.
The white parts of the image are bones and/or things the X-rays couldn't get through, the white film remains white, and the black part is where the X-rays did pass through, and it blackens the white film.
I've also heard that they have a cassette/casing, and in that is fluorescent material or coating and film.  The X-rays hit the fluorescent material and (I guess by absorption and reflection, or absorption and emission) converts it into regular light, and there is a phosphorescent material/coating there, too, so the light remains for a while, and the light, I guess, blackens the film.
I understand they don't work like that anymore and now you can have other types of X-ray detection mechanisms. But I want to understand at least how the traditional ones are black and white.
I understand that X-ray machines used to use a white photographic film.
What is the white photographic film made from that distinguishes it from a black photographic film? Can I buy white photographic film?
I have seen old-fashioned camera film before.
In black and white photography, was it a black photographic film that went white when exposed to light?

Answer not worth putting as answer: https://en.wikipedia.org/wiki/Photographic_processing

@ssavec I don't see the answer there. It mentions about film that is black and goes white on exposure. But white film used in xrays is white and goes black on exposure.. and i'm asking about that white film.. how its composition is different

@barlop Normal B&W film is transparent ("white") if not exposed and becomes dark where exposed and developed, same as x-ray film. This is known as a negative image. ssavec's link mentions a reversal process for making positives (for making projection slides), but apart from a few changes to make them work better with x-rays, the film behaviour and chemistry is no different than normal silver halide B&W film. (incidentally, x-ray film is still very commonly used, especially in things like dentistry where the cost of digital imagers is a concern)

@MichaelD.M.Dryden  interesting.. it doesn't seem to be white   it seems to be like this(photo of a black and white film selling on ebay)  http://i.imgur.com/fkT2jK8.png   which is this colour  #8c866c  http://www.backgroundcolor.net/hex/8c866c  whereas xray photos seem to have proper white. Would it have been white like white paper until it was yanked out of the roll? if so then e.g. how soon after being taken out of the roll would it have changed from paper white to that hex #8c866c colour?

@barlop That's before it's developed and fixed. Once developed and fixed, if not exposed, the silver halide is dissolved and you only see the transparent plastic base of the film (they come in a few colours from grey to purple). The silver halide that has been exposed to light is converted into elemental silver by the developer and thus appears dark.

@MichaelD.M.Dryden what about colour film.. film that went in cameras that supported colour. Were they also white "once developed and fixed" and "if not exposed"?

Yes, colour print (negative) film is also transparent if not exposed, but the base is usually an amber/orange colour. Colour slide (positive) film is dark when unexposed.

Let us continue this discussion in chat.

X-ray film is very similar to standard black and white negative film. It's basically a transparent plastic base coated with an emulsion of silver halide particles (a salt of $\ce{Ag+}$ with a halide, usually $\ce{Cl-}$ or $\ce{Br-}$). As shown below, when light hits one of these particles, it reduces a small amount of the halide to elemental silver (1). When this is exposed to a photographic developer (essentially a mild reducing agent), more elemental silver is produced around the initial sites (2).$$\ce{Ag^+ +e- -> Ag}$$
Finally, a fixing agent is added which solubilizes the unexposed silver halide so it can be removed from the emulsion, preventing further light exposure from fogging the film.
Thus, the regions that are exposed to light appear dark when developed as the tiny silver particles ($\ce{Ag}$) block light passing through the film and the unexposed regions are transparent, taking on the colour of the plastic base (often blue or grey). X-ray films are typically viewed on an illuminated white background, making the unexposed areas appear white (or off-white, coloured by the plastic base).
Normal B&W film is actually slightly sensitive to x-rays but, the emulsion of dedicated x-ray film contains other components to facilitate the silver reduction from x-ray exposure.

Source

Thanks. When you write " facilitate the silver reduction from x-ray exposure."   do you mean to lessen or prevent the silver from being reduced due to xray exposure?  (I understand that what happens with xrays is they hit a fluorescent  material which lights up with normal light and hits the film - or I guess, hit the silver)

Other way around—facilitate = make easier. You want the silver to be reduced when it's hit with an x-ray. Mixing it with a fluorescent dye would be one way of doing it. I don't know if there are any other methods in use.

I see, so when the developer gets the film, the parts that got exposed to light, the black parts, have no silver, and the parts that remain clear, are the ones covered in silver.  But then what's the point of the silver in the first place? If the silver disappears when the light hits.. e.g. I presume the film is protected in an environment of blackness/darkness(with the exception of the parts that intentionally got exposed to light when the picture was taken), and the film remains in an environment of darkness until after the developer develops it.

Other way around again. The parts that are dark are where the silver halide has been reduced to elemental silver. The transparent parts were not reduced and the silver halide is washed away in the fixing step. The film does have to be protected from light except when intentionally exposed, though this is easy with x-rays since they will pass through materials that keep visible light out.

thanks.. I understand that now. So why should elemental silver on a clear film , after development, make blackness?

oh I see "Thus, the regions that are exposed to light appear dark when developed as the tiny silver particles (Ag) block light passing through the film " <-- now I understand that from your post. thanks.  So they look at the film from the other side, the side where the silver isn't.. though i'd still wonder why if it's transparent, you wouldn't see silver, and you'd see black.. but I have a better idea now. Thanks

It seems2me that the metallic silver on the negative, is black, but that's nothing to do with the density of the silver. Dense silver that appears silver coloured fine it wouldn't allow light through, but it would be reflecting light hence dense silver doesn't appear black unless something makes it black, regardless of its density.But I see that the density will have the effect that a picture taken of that negative,will produce a +ve (-ve of -ve). 'Cos the black areas are also dense on the 1st -ve,and so a photo of the negative will produce clear areas in the same locations on the second -ve.

Michael's answer is very good, i'll just simplify

I understand that X-ray images are black and white.

No, it's clear, then black and clear. Not white. When they are on a white background e.g. white table or white wall, they show as white and black.
If they were on a purple background then they'd show up as black and purple (purple bones). Because the clear part of the xray takes the colour of the background the xray image is on. But it's clear and black.
And re what makes the film go black, the film itself doesn't, but the exposed silver on it does.

You have the
  X-rays themselves, the white film, and an object/person in-between.

The X-rays themselves, the clear film, and an object/person in-between
Apparently   clear/transparent film may come in a few background colours can be grey, can be purple, can be amber.. i've seen amber transparent colour film  (that's the case with colour films used by photographers, and may or may not be the case with a photographers films for black and white photos, and the films used in xrays). The ones for black and white photos, and for xrays, may be grey
And yeah..
The CLEAR parts of the image are bones and/or things the X-rays couldn't get through (e.g. tumours), the clear parts of film remain clear,

and the black parts [on the film] is where the xrays hit

The film is covered in a kind of silver particles, silver halide.
The silver can be blackened by xrays or light..
It seems that technically, most of the rays hitting the silver are light, because in order to require less dosage of radiation, the scientists thought of using a fluorescent material, the xrays pass through that and the fluorescent material converts the xrays into visible light, it absorbs the xrays and emits visible light.
Fluorescent material does that(material with fluorescent phosphors). For example some "invisible ink" is florescent, it lights up when UV light is shone on it. It takes the high energy photons/short wavelength waves and absorbs it and emits lower energy photons/longer wavelength waves/visible light. Wikipedia says most fluorescent material converts that way, high energy to low energy.
In addition to flurescent material, there is also phosphorescent material, it has phosphorescent phosphurs,which keeps the light stored for some time.
The parts of the silver metal on the film that will be darkened, will be darkened mostly by light, and partly by xrays, since there's more light than xrays hitting it.

The clear parts of the image are bones and/or things the X-rays
  couldn't get through, the clear film remains clear, and the black part
  is where the X-rays did pass through, and it blackens the clear film.

It blackens the silver on the film, not the film itself. The film itself is a clear film with silver on it.
The film though is predevelopment.. And predevelopment, the entire film is covered in silver.
The parts of the silver that got hit with xrays or light, will go black. And change in form from silver halides to metalling silver, and they won't get washed away in development.    The parts of the silver that didn't get hit, will get washed away during the development.

I've also heard that they have a cassette/casing, and in that is
  fluorescent material or coating and film. The X-rays hit the
  fluorescent material and (I guess by absorption and reflection, or
  absorption and emission) converts it into regular light, and there is
  a phosphorescent material/coating there, too, so the light remains for
  a while, and the light, I guess, blackens the film.

yeah I suppose so. And the light blackens the silver on the film not the film itself.
You ask about the materials used for xray film and regular photographic film.. Apparently the differences are minor.. And in both cases the film is clear.
He showed a colour photographic film that was amber/orange, and transparent.
The black and white one maybe is clear like regular glass is clear..I don't know if it has any slight colour at all but even if it did, it's transparent no question about it.
I think maybe the term "negative" doesn't apply with xrays. Because in regular photography the rays bounce off the subject rather than through the subject, and with a normal camera and film,  what is light comes out dark and what is dark comes out light, hence it's called a negative.
Another thing that happens, (at least with regular photopgraphy, perhaps not with xrays), is the negative is turned into a positive.. And that's done by taking a picture of the negative (A), making a negative of the negative, which is a positive (B). I believe that's done with an enlarger that also enlarges it.  The silver is dense and blocks the light. The fact that silver on A is black is irrelevant, since it's the denseness that blocks the light to produce B.
Michael mentions about the denseness of the silver, but I don't think it's relevant to the xray..in making a new image, but I think he means you'd get little clear areas / specks that aren't black in the xray if the silver wasn't dense, but it is dense.. So the denseness and the blackness helps there in the black parts of the xray image appearing black.

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text:
Comparing bond lengths of a molecule using valence bond theory?
My textbook says that all the bonds of $\ce{SF4} $ are not equal while in the case of $\ce{XeF_2}$, $ \ce{BF4-} $ or $\ce{SiF4} $ all the bonds are equal. No reason is stated for this observation.
So is there a general way or formula to check if all the bond lengths are equal or not for a given molecule?

Please use the mhchem package for $\LaTeX$ formatting. Chemical formulas and equations mustn't be slanted.

You have to check geometry of the molecule

So the geometry will be trigonal bypyramidal, trigonal bypyramidal, square planar and square planar for $\ce{SF4} $, $\ce{XeF_2}$, $ \ce{BF4-} $ and $\ce{SiF4} $, respectively. What should I do next?

Where in this question comes the valence bond theory into play? Currently it reads more like a question about the simple predictions of the VSEPR model, geometries of certain molecules. It does not seem to be at all about VBT. Maybe some other tags and a more descriptive title would help this question.

4 times wrong, sorry but without knowing VSEPR case is bad. http://chemistry.stackexchange.com/questions/18427/why-f-replaces-axial-bond-in-pcl%e2%82%85 may help

Did you mean the the shape of the molecules or the geometry of the molecules?   The VSEPR of the molecules will be 5(4 bond pair, 1 lone pair), 6 (4 bond pair, 2 lone pair), 4(4 bond pair, 0 lone pair) and 4 ( 4 bond pair, 0 lone pair) for $\ce{SF4}$ , $\ce{XeF_2}$, $ \ce{BF4-} $ and $\ce{SiF4}$ ,respectively. I asked my teacher and he said that the molecule with VSEPR 3 or 5 has to be the answer. So the answer has to be $\ce{SF4} $


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Freeze distillation: Does efficiency depend on solubility?
I was surprised to learn that freeze distillation has the same inefficiencies as vapor distillation: Namely, that it is impossible to get two miscible fluids to fully separate in one "pass" of subjecting the mix to a temperature between the melting point of its constituents.
I can understand why this is the case in vapor distillation, because vaporization of a liquid still occurs below its boiling point.
But AFAIK the same is not generally true of freezing: assuming it has a nucleus, a single liquid will either crystallize completely at a given temperature or it won't at all.
My guess is that the inefficiency of freeze distilling has something to do with the chemistry of miscible fluids: I.e., whatever phenomenon allows them to mix also drags molecules of the lower-melting chemical out of the mix above its melting point when the higher-melting chemical crystallizes.  If so then the efficiency of freeze distilling should have some relationship to a measurable quantity (like a miscibility coefficient?).  What theories and models of miscible fluids are applicable to the question of efficiency of freeze distillation?
(If there is no attraction between the two fluid chemicals then presumably "single-pass freeze distillation" could be perfect.  It would probably require agitation to help the more volatile of the two stay out of the crystals forming from the first chemical to freeze.  For example, agitation during freezing is how ice is formed without visible bubbles from dissolved gasses.  But if "there is no attraction" then presumably the fluid chemicals are not miscible, right?)

The whole thing gets weirder when you toss in the fact that recrystallization from solution is a common technique to purify a chemical...

Solids have solubility, just as liquids do. This is easily seen in metal alloys, where different amounts of different compounds can cause different crystal structure, low-level phase separation, and large-scale phase separation (if the solid solution is not stable). The changing of these nanostructures of the material can even occur with a material fully in the solid state, such as tempering of steel.
This relates to your question in that liquids freezing will behave very similarly with respect to phase composition as vapors condensing. The composition of the solid will be proportional to the composition of the liquid and affected by differing freezing points, just like you would expect from a liquid-vapor Txy diagram.
There are some differences, however. Molecules in solids can, as I said above, move around within the solid. However, this diffusion is substantially slower in solids than in liquids, and in non-metal solids, slow enough to be largely insignificant within the process of freezing. This means that the freezing process will create regions of differing composition within the solid. Further and related to this phenomena, the speed of freezing will impact the concentration of the solid, as faster freezing will push it even further from equilibrium.
The slow diffusion in solids also means that once trapped in the crystal structure, compounds with higher freezing points will not as easily exit the solid as they might from a liquid. This means that melting can behave very differently from vaporization, and solids might become a slush in the process of melting.
An explanation as to why this all occurs could go as follows: solubility is a factor of the strength of the interaction between different types of molecules. Those interactions must be stronger than interactions within the individual types to avoid phase separation. The process of freezing tends to not fundamentally shift the strength of these interactions. It is possible for materials partially soluble in liquid to be mostly insoluble in solid, but pairs of miscible liquids have such intense cross-compound interactions that freezing typically can only impact concentration of the separate phases.
The really important chemistry in this case takes place at the phase boundary between the solid and liquid, which behaves almost identically to that between a liquid and a gas. Molecules with weaker interactions with the solid phase (weaker interactions with other molecules in that phase) will tend to leave the solid phase more often - but there is an equilibrium where molecules will enter their less-favored phase. Essentially, everything you would have learned about liquids and vapors applies here, except diffusion is too slow to allow good mixing within the more condensed phase.
Of course, there are, in specific cases, more complicated explanations. One is of interstitial compounds, where a second compound might occupy spaces within a solid. While this happens to a degree in all solids, interstitials that are small enough to fit fully in a crystal structure without deforming it might exhibit unusual solid solution behavior, such as a compound with a lower freezing point showing a greater affinity for solid phase than the other compound.
To answer the title question, yes, separation depends on solubility. Unlike vapor distillation, though, it can also depend on temperature difference (thus speed) and mixing.

If I understand this correctly: Then in all cases the most efficient freeze distillation (i.e., maximum separation) will occur at the slowest rate of freezing combined with maximum agitation? And factors that determine how much of the lower-freezing chemical get stuck in the solid include: Interstitial compounds, and the strength of solution forces? So pairs of miscible liquids that would most efficiently separate are those with the weakest interactions (which are measured/characterized how?), and where higher-freeze-point molecule forms the tightest crystals (minimizing interstitial voids)?

@feetwet Pretty much. Those do to an extent impact vapor distillation, too, though. That's part of why distillation columns maximize surface area. Miscible are miscible because they have strong interactions, which is why it's harder to separate those compounds (via any kind of distillation). My post doesn't fully address interstitials (and I do not feel qualified to fully discuss them), just that voids can cause unusual behavior in general. Like slow diffusion, the existence of voids is one of the ways solids differ from liquids/vapors.

Got it; thanks!  Are you aware of general methods other than distillation that are commonly used to separate miscible fluids?

@feetwet I suggest you look here: https://en.wikipedia.org/wiki/Separation_process . With regards to liquids, the best options are usually centerfugation (which can separate basically any compounds) and absorption/adsorption. In a lab context, liquid-liquid extraction (introducing a non-miscible liquid) could allow some separation, as could chromatography. Reactions could also be used to enhance separation by producing a separate, more easily separated compound.


text:
Geometries of methyl and silyl radicals
The $\ce{.CH3}$ radical is planar but the $\ce{.SiH3}$ radical is a trigonal pyramid. Why are they different? After all, $\ce{C}$ and $\ce{Si}$ are in the same group.

As is currently being discussed in this question, I'm not convinced that the methyl radical is planar, but even if not planar, it is certainly very close to planarity.  It is also certainly less pyramidal than the $\ce{.SiH3}$ radical as you state.
The $\ce{.SiH3}$ radical is more pyramidal than the $\ce{.CH3}$ radical due to differences in electronegativity between $\ce{Si}$ (1.9) and $\ce{C}$ (2.5) and hydrogen (2.1) and how these differences play into Bent's rule.
In a nutshell, Bent's rule states that, "atomic s character concentrates in orbitals directed toward electropositive substituents" (ref).
Since hydrogen is more electronegative than silicon it will draw more of the electron denisty in a $\ce{Si-H}$ bond towards its end of the bond (the reverse of what happens in the case of a $\ce{C-H}$ bond where the carbon is more electronegative than hydrogen) decreasing the electron density in the silicon atomic orbitals involved in the bond.
If there is less electron density in the silicon atomic orbital used in the $\ce{Si-H}$ bond, then according to Bent's rule, the silicon atom will hybridize so that less s-character is used in these atomic orbitals that now have a lower electron density, and will instead use that s-character in the orbital containing the single electron and stabilize it.  Increasing the p-character in the $\ce{Si-H}$ bonds and increasing the s-character in the orbital holding the single electron will tend to push the molecule towards an $\ce{sp^3}$-like, pyramidal geometry.
The reverse is true in the case of carbon.  Due to electronegativities, there will be more electron density towards the carbon side of the $\ce{C-H}$ bond (compared to the $\ce{Si-H}$ bond).  Due to this increased electron density around carbon, it is energetically less favorable to remove s-character from the carbon atomic orbitals involved in the $\ce{C-H}$ bonds (and destabilize this increased electron density) and transfer it to the orbital containing the single electron.  As a result the carbon atomic orbitals involved in the $\ce{C-H}$ bonds will have higher s-character compared to the analogous silicon case and $\ce{.CH3}$ will be less pyramidal.


text:
How are chemicals prilled?
What does the process of prilling something like, say, calcium nitrate, involve? Do you need other chemicals to prill the substance?

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

@Mithoron I saw the Wikipedia article already, although it's nice to link for reference. I didn't see any mention of whether other chemicals are required to make the process possible, however (for instance, anything else besides calcium nitrate).

I don't know the answer for sure, but will take a guess based on what I read at the link provided by Mithoron.
Non-chemists outside the laboratory, and persons who have to handle and work with those chemicals would certainly benefit from prilling to reduce dusting which could be detrimental to their health in breathing and on the mucous membranes.
Gardeners use Ammonium, potassium and calcium nitrates in fertilizers and stump removers, and do not always wear eye protection or dust masks. Thus prilling helps to mitigate exposure to these chemicals.


text:
Can carbon have quadruple bonds?
Carbon has 4 valence electrons, so it means that it can have a total of four (covalent) bonds, correct?
If that's the case, why can't carbon share four bonds, such as with another carbon atom (to form $\ce{C2}$, for example)?

The question about fullerenes would probably be best asked as a separate question, with a link back to this one if you really want.

Have you heard of Valence Bond Theory (orbital hybridization), VSEPR theory, or MO theory?

I have removed the extra fluff outta your question. We really don't need to know about how you learned about the uncertainty principle. The questions should be free-of-noise and stick to the point. We have [chat] if you want casual conversation with users of the site.

s orbitals have spherical symmetry, and p orbitals have a dumb-bell shape.  When these orbitals overlap, then they form the bonding orbitals (and corresponding antibonding orbitals).  The p-orbitals are aligned along Cartesian (xyz) axes.  If all three p-orbitals overlap, then the s-orbitals can't get 'close enough' to overlap and form a fourth bond.  If we start mixing s and p orbitals, the geometry always puts electron density away from the other bonding atom.
Concisely: there are no arrangements which allow overlap of atomic orbitals that create four bonding molecular orbitals with the proper orientation.
C2 as a molecule can exist, it will simply only have 3 bonds plus two unpaired electrons confined to each carbon.
If you want a more elaborate answer we have to unpack the heavy quantum mechanics, and you won't be seeing that until your junior year.

Side note: $\ce{C2}$ molecule is more interesting than one might think.


text:
Why are the plastic balls used to reduce evaporation in this reservoir black instead of white?
Plastic balls have been added to a reservoir in drought stricken California in order to reduce evaporation.  (Link to news story here.)
Why are the balls colored black instead of white?  I would think that white would decrease heating and therefore evaporation.
Edit:  Apparently the balls are there to block sunlight from converting bromide to carcinogenic bromate.
Youtube video explaining the purpose of the balls

Cynical me thinks the reason might not be scientific, but perhaps someone here can dissuade me from such a thought.

Pure speculation: the plastic is naturally black and they couldn't find a paint that stuck to it in time.

It's all about blocking the UV wavelengths.  Those photons carry more energy than visible wavelength photons and will consequently cause more rapid evaporation.  Black or white plastic balls, we can't really know what their UV absorption\reflection  properties are without an actual spectrum of the colorant.  I'm guessing that the black colorant is cheap and permits less UV to pass through (ideally reflects UV better) than other colorants.

I'm thinking the balls may have been sold by the lowest bidder or, like Ron says, it's the UV issue.

The primary means of lowering evaporation is just lowering air-exposed surface area.  I think the black color may have to do with not wanting to change the thermal properties of the water body, but I would love to know the real reason.

Stumbled across a quote which suggests that the balls were not originally made with the purpose of reducing evaporation: "The reason they were originally referred to as shade balls is it was an issue of mitigating bromide in algae bloom..."

This piece in Bloomberg News is about Sydney Chase, the founder of XavierC, the company which makes the "shade balls" in question (there are two other suppliers, Orange Products, Inc. and Artisan Screen Printing).
From the section of the XavierC site on how they work:

Two main causes of water evaporation in reservoirs are temperature and surface area. Our hollow plastic conservation balls address these two issues effectively and inexpensively.
  The black conservation balls provide a floating cover on the reservoir taking up approximately 90% of the surface area. Simultaneously the balls prevent UV rays reaching the water and therefore reducing algae.

From the above-mentioned Bloomberg piece:

The Los Angeles Department of Water and Power has now dumped 96 million balls into local reservoirs to reduce evaporation and block sunlight from encouraging algae growth and toxic chemical reactions. The balls are coated with a chemical that blocks ultraviolet light and helps the spheres last as long as 25 years. Las Virgenes, north of L.A., now uses shade balls, too.

From the Los Angeles Times story on August 12, 2015:
What are they made of?

The 4-inch-diameter balls are made from high-density polyethylene, which is the same material you would find in a one-gallon milk jug. This plastic is approved to come into contact with drinking water.

Why are they black?

Carbon black was added to the plastic to stabilize the balls in UV light from the sun.

and, from another LA Times piece about the "shade balls" written in 2008:

The balls, which cost 40 cents each, are made of polyethylene. The coating contains carbon. Black is the only color strong enough to deflect ultraviolet rays, said Paul Sachdev, president of Orange Products.

So the carbon black protects the ball itself from the UV, right?

@DmitryGrigoryev - It appears that way. And since the balls cover the surface of the reservoir, that means that the UV radiation making it to the underlying water is attenuated, too. If I had to guess, I'd say the design of these was pretty well thought-out to optimize both the lifespan of the item as well as its function.

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text:
Is sodium percarbonate artificial?
If I add sodium percarbonate to my natural laundry soap recipe, can I still claim "No Artificial Anything"?
I know it's used in many "natural" products, but is it considered an "artificial ingredient"?
I don't want to make any false claims with my product.
thank you!

I think this is more a question of what you're allowed to do under your local regulatory agencies than it is a question of the chemistry. If it were left up to chemists, "organic" products would be completely different from what they are now.

Yeah, maybe a better fit for Law SE. Just don't write "all natural, organic, and chemical free" on it and our heads won't explode.

I'm voting to close this question as off-topic because it's not actually about solid chemistry. Instead it depends on local/national laws.

There are no "laws" that govern cleaning products...unfortunately. Companies can poison us with anything and do not even have to tell us what they are using. Im just trying to get solid information to market honestly.   Thank you.

@jerryhere I don't find the claim 'no artificial ingredients' to be informative at all. The distinction of 'artificial' vs. 'natural' has no bearing on safety. The most potent toxins are from nature, yet unsafe. A washing machine is artificial, yet your product depends on its use. I suggest you find a better way to market your product than a 'no artificial anything' claim.

Yes, it is artificial. Sodium carbonate is a substance that occurs in nature, even though any of it is most likely manufactured.
Sodium percarbonate, on the other hand, does not occur in nature and it has to be manufactured. It is an oxidiser, and oxidisers don't occur in nature normally. That's a problem for you, because oxidisers (aka bleach) are desired in laundry applications.

Thank you Michael.  I felt that would be the answer but wanted to confirm with others that know more than me in chemistry.

Would Oxalic acid be considered the same....artificial?

@jerryhere what if it is a substance that occurs naturally, but the one you would probably use is bought somewhere and manufactured? There are so many substances that occur in nature, but it's easier and cheaper to manufacture them artificially. Does that make them any worse? Is the fact that a substance is "not artificial", whatever that means, make it any safer or better? Absolutely not. You need to have strict definition of what is "artificial", and we're not the place to look for that answer.


text:
Making a non-CN dissolved gold solution ideal for electroplating
I've had success electroplating $\ce{Ni}$ and $\ce{Cu}$, but interestingly never $\ce{Al}$. My focus here is on plating with $\ce{Au}$, specifically onto an $\ce{Ni}$ surface.
In general from my own work and others' suggestions, I have found applying a voltage (direct current) of 5 V through a relatively dilute solution of relevant metal, e.g. 2.53 g $\ce{Ni(NH4)2(SO4)2*6H2O}$ dissolved in 81.9 mL 0.42 % acetic acid at ~60 °C makes a very nice, lustrous plating of $\ce{Ni}$ on a clean $\ce{Cu}$ surface after ~90 s of submersion.
I have been able to dissolve $\ce{Au}$ in warm, concentrated $\ce{HCl}$ and minimal $\ce{HNO3}$, and have heard that you can dissolve it in $\ce{HCl}$ and $\ce{H2O2}$, which makes sense because $\ce{Cu}$ does the same.
I'm looking for an ideal solution here, starting with pure metallic gold. I want to avoid any mistakes because I only have 1 g to work with. By default I'm going to try a comparable concentration of dissolved $\ce{Au}$ to what I mentioned with $\ce{Ni}$, but any advice is appreciated – especially a specific method for making an "ideal" solution.
I've found that several commercial solutions contain $\ce{(CN)4AuK}$, but I really want to avoid using cyanide.
This paper by Paul A. Kohl is immensely helpful and interesting, but I'm wondering if anyone has had any success with something simple.
It makes mention of using sulphide, thiosulfite, thallium, etc.

The problem is your amount of gold that you have. Normal chemical baths contain several grams per liter. So you will need to operate in small volumes. You can use this article - Chemical Papers
June 2012, Volume 66, Issue 6, pp 617-620
Date: 23 Dec 2011
Etching and recovery of gold from aluminum substrate in thiourea solution, that is using thiurea. Be aware of the low stability of this bath. The bath will degrade quite fast, but that is common problem of all non-cyanide gold baths.
As the article deals with a gold dissolution, I would suggest to reduce the gold back to a solid state. Use the article or this Hydrometallurgy
Volume 55, Issue 2, March 2000, Pages 137–152, Gold electrowinning from aqueous–alcoholic thiourea solutions and then do the electrodeposition.


text:
Routes of formation of esters with highest yield
Esters can be formed in two ways according to the syllabus I'm following:
(1) Mixing an alcohol with a carboxylic acid in the presence of sulfuric acid; and
(2) Mixing an alcohol with an acid chloride.
My question is:
Which method produces the highest yield of esters?
I think method (2) does because the $\ce{Cl-}$ will attach with the $\ce{-OH}$ in alcohol. However, I'm not exactly sure of the reason. Can someone explain it clearly with full details please?

Please visit [this page](http://chemistry.stackexchange.com/help/notation " latex="" at="" chemistry.se""="), this page and this one on how to format your future posts better. || Excuse my ignorance, but what's the difference between 02 and 2?

Hold up, how do you get an ester from just an alcohol and an acid chloride? Did you mean mixing alcohols with some other compound in the presence of acid chloride?

There is a method to form esters by adding an acid chloride to an alcohol. I actually meant that.

Esterification with acid chlorides are usually done in presence of weak bases like pyridine.

Option 2 generally gives higher yields.
This is because option 1 (Fischer esterification) usually results in an equilibrium mixture which contains significant amounts of products and reactants, limiting the yield. This is mitigated by using a stoichiometric amount of a concentrated acid catalyst, such as sulfuric acid, which is also a strong dehydrating agent, and therefore removes the water from the equilibrium as it forms, favouring the products. Additionally, the reaction mixture can be heated to evaporate the water, or you can use a large excess of one of the reactants, typically the alcohol because it also acts as the solvent.
Conversely, option 2, more easily gives high yields because the reaction is more or less complete, with very little reactants left at the end. This is because the reaction produces hydrogen chloride gas instead of water, which is lost from the reaction mixture, driving the reaction forward. The downside of this option is that acid chlorides are generally more dangerous than carboxylic acids and are harder to prepare, or more expensive to buy. Aditionally, hydrogen chloride gas is produced which must be disposed of safely.
Another, often used method, is esterification using acid anhydrides. These are intermediate in reactivity between carboxylic acids and acid chlorides and are easier to use than acid chlorides, but still provide excellent yields. A classic example of this is the acetylation of salycylic acid with acetic anhydride to produce Aspirin.

Also, for methyl esters one of the cleanest methods is to use diazomethane.
Of course, diazomethane is pretty nasty, but there are nice kits that can safely generate it in small quantities.


text:
How do we identify left handed and right handed crystals?
I was going through this article about how Louis Pasteur studied and explained the absence of optical activity in Racemic acid. Here's an extract:

When Pasteur next examined crystals of the same
  salt of racemic acid, he found that they, too, were chiral
  but that some of the crystals were left-handed and others
  were right-handed. The crystals were related to one
  another as an object and its non-superimposable mirror
  image, and they were present in equal amounts. With a
  magnifying lens and a pair of tweezers, Pasteur carefully
  separated these crystals into two piles: the left-handed
  ones and the right-handed ones.

My guess is that the left/right handedness refers to the molecular spatial alignment. So,

How exactly can you assign the property of individual molecules to an entire crystal?
How can you determine if a crystal is left/right handed just by looking at it?

related http://chemistry.stackexchange.com/questions/7489/how-does-walter-white-make-pure-crystal-meth-using-a-non-stereospecific-reaction

as I was writing my answer, I also found this related question: http://chemistry.stackexchange.com/questions/6439/why-only-chiral-molecules-rotate-plane-polarized-light-and-how-does-it-rotate-li

Yes, you are right that the handedness refers to the molecular 3D arrangement. Such handed molecules are stereoisomers known as enantiomers.
I think that the first part of your question will make more sense following an answer to the second part. So with regard to the part about discerning handedness by looking at it:
(I couldn't get to the article right now due to network issues, so I hope I'm not repeating too much.)
A molecule is chiral when one of the atoms in it has four unique groups attached to it.
Chiral molecules observably rotate plane-polarized light, either clockwise or anti-clockwise (see here for a picture of what plane-polarized light is).
Pasteur observed this along with the cancelling effect of a racemic mixture. As in any scientific endeavor, Pasteur was building upon the efforts of previous work--in this case, that of Jean-Baptiste Biot in studying tartaric acid. Pasteur's careful work enabled him to deduce that the naturally occurring or biologically produced form of tartaric acid was enantiomerically pure because it rotated light and the mirror image crystals also rotated light opposite directions. This was in contrast to the racemic acid which did not rotate light since it was a 50:50 mixture that cancelled the rotation in either direction.
Bryn Mawr College has a helpful explanation of how enantiomers interact with light:

You have probably [not] learned in lecture that chiral molecules have optical rotations and that enantiomers rotate plane polarized light in equal and opposite directions... This behavior upon exposure to plane polarized light is not some magical property that defies the general principles of enantiomers. The reason enantiomers interact differently with plane polarized light is that plane polarized light is chiral. Though polarized light is a planar phenomenon, it is actually comprised of two superimposed helical forms. You can think of the electrical component as oscillating in such a way that there are two electrical vectors rotating in a cork screw fashion in opposite directions away from the source of the propagation at the same rate. Since a helix is a chiral shape , the two versions of the light constitute an enantiomeric pair. If you resolve the vectors of the two superimposed helices, you end up with oscillation in a single plane...Enantiomers interact differently with the two forms of light. The basic idea is that when plane polarized light encounters enantiomerically pure, chiral molecules, one form of light will slow down more than the other as a consequence of the interaction. This results in one helix coming out of sync with the other. When the vectors are resolved, the resultant plane is rotated with respect to the original position. This change in position is what we mean when we say a compound has an optical rotation.

So looking at a crystal being exposed to polarized light should allow you to observe whether it rotates light to the right or left. I'm not sure how easily detectable to the eye the rotation is, but polarimeters are instruments which are used to measure the angle of rotation.


text:
Oxidation of acetic acid in wine?
When wine is exposed to oxygen, acetic acid forms because of an oxidation reaction.
The wine then loses acetic acid. Why? Is it because it is oxidised itself to form carbon dioxide?

doesn't the alcohol (ethanol) in the wine oxidise in the presence of oxygen? Which produces ethanoic acid, or, Acetic acid . And hence the vinegar(y) taste of gone off wine.

Yeah it does. But we did an experiment and measured the ethanol concentration in wine as it is exposed to oxygen every day for a week and our concentration decreases but then it increases. My teacher said this is right but I can't think about why it happens. I think it has something to do with the co oxidation of acetic acid or decarboxylation.

I am not sure why the ethanol concentration would increase again, maybe that should be a separate question. However, I am aware of a reaction that may occur after the acetic acid concentration reaches a certain level. this is ethanol + acetic acid = ethyl acetate. this would give one explanation of the fall in [acetic acid] after the rise.

The oxidation/reduction reactions of
ethanol <-> ethanal <-> acetic acid (ethanoic acid)

are at equilibrium. Oxidation (presence of $\ce{O2}$ or other oxidizing agent) and reduction (addition of $\ce{H}$ to the $\ce{C-O}$ carbon) are in constant flux. The system can be pushed to the right if wine is exposed to air for long periods of time, as you found, but at a certain point it will reach equilibrium, in which case the excess of acetic acid will reduce (marginally) back into ethanal and/or ethanol.
A better explanation may be that since the B.P. of ethanol (78 dec. C) is significantly lower than acetic acid (118 deg. C), the vapor pressure of ethanol is greater and it has a higher affinity to convert to a gaseous state. This also would shift equilibrium to the left to produce a lower post-equilibrium concentration of acetic acid.
Acetic acid does have the possibility of decarboxylating, but this is usually in excess heat. In this case, a $\ce{CO2}$ molecule breaks off of the acetic acid leaving $\ce{CH4}$, which would also be generated as a gas.
Finally, it is possible that you're mistaking "loss of acetic acid" for an actual increase in pH, irrelevant to the presence of whole acetic acid molecules. It could be that other materials in the wine "soak" protons to increase pH and leave ethanoate counter ions ($\ce{CH3COO-}$). You should clarify how you're actually quantifying the concentration.
I wouldn't expect the "decrease" of acetic acid to be significant, compared to the amount that was produced in the first place. I would also expect your system to re-equilibrate after a while longer. It would be interesting to see what happens next.


text:
Are bulkier molecules more likely to react?
I have heard a theory from someone and just wanted to confirm if its true or not. Is it true that more bulkier an organic compound is, more likely it is to react? So does this mean that higher the molecular weight of an organic compound, more will be the probability of it to react?

Generally bulky substituents are working the other way round.

This is actually an interesting question but needs some work. Reaction rates are functions of collision probability. Collision probability depends on many factors; the commonly known factor is concentration but it would also depend on molecular size (collision radius) and diffusivity (inversely proportional to molecular weight). Also, these relationships are not always linear.Things would differ for non-well mixed systems. So, you should state all the assumptions clearly.

the problem is, i am a computer science student, my research field is bioinfo, i do not understand so many factors, any case that is valid for most compounds reacting would do me a lot of help. I don't know that, which is why i asked the question

The relative rates are going to depend not on molecular weight so much but the structure of the molecule.
The specific reaction is also very important.
Nucleophilic substitution reactions (some of the first reactions most students learn about in an organic chemistry course).
Let us compare 1-chlorobutane and 2-chloro-2-methylpropane, both $\ce{C4H9Cl}$. However, the 1-chlorobutane is a primary alkyl halide (not very bulky) and 2-chloro-2-methylpropane is a tertiary alkyl halide (bulky).
The bulkier compound reacts faster in a solvolysis reaction:
$$\ce{(CH3)3CCl ->[\ce{H2O}] (CH3)3COH}\ \ \mathrm{fast}$$
$$\ce{CH3CH2CH2CH2Cl ->[\ce{H2O}] CH3CH2CH2CH2OH}\ \ \mathrm{slow\approx no\ reaction}$$
The reason that the tertiary halide reacts faster in this reaction is that its mechanism involves the formation of a carbocation intermediate and the extra carbon groups at the tertiary center provide more inductive stabilization to that carbocaion than the single alkyl group attached to a primary position. In general, in reactions that generate carbocation intermediates, tertiary substrates will react faster.
However, in many reactions bulkier molecules have steric interactions that negatively effect the rate. In a nucleophilic displacement with $\ce{NaI}$ in acetone, 1-chlorobutane will react faster.
$$\ce{(CH3)2CCl ->[\ce{NaI}][\mathrm{acetone}](CH3)3CI}\ \ \mathrm{slow\approx no\ reaction}$$
$$\ce{CH3CH2CH2CH2Cl ->[\ce{NaI}][\mathrm{acetone}]CH3CH2CH2CH2I}\ \ \mathrm{fast}$$

bulkier as in with higher weight?

Usually in organic chemistry I have seen the interpretation bulky = takes up more space and not bulky = heavier.

is there any example of two compound say A and B such that A is more bulkier than B but B has a higher molecular weight than A

Sure, I consider anything connected to a tert-butyl group to be bulky in the steric sense, for example $\ce{(CH3)3COH}$, where as a functional group at the end of a linear alkane is not (but could have a higher molar mass): $\ce{CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2OH}$.

is there any way in which this bulky property can be quantified? Like any value which would tell how bulky the molecule is

@Rishika - this sounds like a great question to ask!

done! lets see what response i get

@BenNorris If you think of reaction rate as a function of collision probability, then "bulkier" molecules have higher chance of colliding  because of their size (collision radius). However, collision probability also depends on the diffusivity which would be higher for lighter (low molecular weight) molecules. I am not an expert in this area but I am sure there are elaborate theories on this topic.


text:
Miscibility of Molten Metals
While reading about the size of atoms and ions of the Group 1 elements in the textbook "Concise Inorganic Chemistry" by JD Lee, I came across this line:

The Li+ is much smaller than the other ions. For this reason, Li only mixes with Na and above 380 Celsius, and it is immiscible with the metals K, Rb and Cs, even when molten; nor will Li form sunstitutional alloys with them. In contrast the other metals Na, K, Rb, Cs are miscible with each other in all proportions.

I am unable to understand why the elements cannot form alloys with Li and what makes the other elements miscible with each other in all proportions.
I observed that the metallic radius of Li is 1.52 $\unicode{xC5}$ and that the difference between its radius and that of Na is about 0.34 $\unicode{xC5}$. But the difference between the radii of the other successive elements in this group is also about the same, so it doesn't help explain miscibility.
I would really appreciate some insight and clarification.

By the way Lithium has a small degree of solubility with other metals such as Aluminium and Magnesium.

Questions like this are not prone to easy answers. But, lets look at what is known about the liquids, then at the various binary diagrams.
Lithium - Melting point 453K, density at melt is 0.512 g/cm$^{3}$, or 0.0737 mol/cm$^{3}$
Sodium - Melting point 371K, density at melt is 0.926 g/cm$^{3}$, or 0.0403 mol/cm$^{3}$
Potassium - Melting point 336K, density at melt is 0.828 g/cm$^{3}$, or 0.0211 mol/cm$^{3}$
Calcium - Melting point 1112K, density at melt is 1.376 g/cm$^{3}$, or 0.0343 mol/cm$^{3}$
Clearly Li has by far the smallest atomic volume in the melt.
The Li-Na binary, as you point out, shows a miscibility gap, tilted towards the Li side, but peaking around 300C. The K-Li system is less well studied, but it also shows a miscibility gap which peaks at perhaps a similar value of 300C.
In sharp contrast, the Ca-Li system exhibits an intermediate compound (Li$_{2}$Ca) and can be described with a 'virtually ideal' liquid phase (C.W. Bale and A.D. Pelton, Bulletin of Alloys Phase Diagrams 8(2) 125-127 (1987)).
Given that the molar density of liquid Ca lies between those of Na and K it is clear that you cannot ascribe the observed miscibility gaps solely to atomic radii.
To further complicate matters, lets keep looking at more binary diagrams. The K-Na diagram shows an intermediate compound, KNa$_{2}$ and no miscibility gap.  KNa$_{2}$ has the same crystal structure as Li$_{2}$Ca (both are MgZn$_{2}$ prototype). The Ca-Na binary shows -- surprise! -- a miscibility gap towards the Na side, peaking at 1200K. So, Ca and Na, even with fairly similar atomic volumes, do not want to mix in the liquid, while Ca and Li are perfectly happy to.
Bottom line - atomic radii do not really mean that much when predicting binary phase diagrams.  Even with comparable radii, you would be surprised at the weirdness than can be found on phase diagrams. As one example, Ag-Cu (both fcc metals) do not form a continuous solid solution, and have a slightly positive enthalpy of mixing in the liquid (not enough to form a miscibility gap).  Yet, Au-Cu (Ag and Au have similar atomic radii) form a fcc continuous solid solution (with intermediate ordered fcc-based phases at lower temperatures), and has a strongly negative enthalpy of mixing in the liquid.
Have fund exploring binary phase diagrams (and ternary, ...), but don't rely on simple heuristics.

Metal atoms that have comparable radii (in the condensed metallic state) often form alloys, because you can just exchange them, without loosing too much bonding energy.
The Li atom would leave too much space around itself in a K melt, and a K atom in a Li melt would break the bonds between the Li atoms around it.
Metals with different radii instead often form stochiometric intermetallic phases, where the different atoms have their own designated place in the crystallographic lattice.


text:
Acid Accident Treatment
What are the general ways to treat an accident involving highly concentrated or similarly dangerous acids?
I have a lab instructor who recently debated with our group about treating an acid spill on the skin. According to him, one should immediately rinse off the acid with running water for about 15-20 minutes. He says that one should never apply bases to the site because an Acid-Base is highly exothermic and would only increase the damage.
However, I clearly remember learning once that one should apply baking soda (sodium bicarbonate) to a bee sting or ant bite in order to neutralize the acid and relieve the burning.
I also know that the dilution of an acid in water is done by adding the acid slowly to the water containing vessel because of the heat released in the process. And especially when preparing a dilute solution of sulfuric acid, one wears safety goggles and gently stirs the solution to prevent splashing.
So which is it? And does it depend on the severity of the burn or type acid?

And ant bee stings are not "highly concentrated or similarly dangerous acids" so there's no reason to suppose that folk remedies for stings have any application to chemical spills. (Well, OK, stings are "similarly dangerous" if you're allergic to them but that's not what you're asking about.)

What's the source of your "very inexperienced" prejudice?

@Nit He sounded very unsure of himself and is new around here. So the description is just based on judgement.

@tennispro1213 A lot of the most competent scientists I know tend to "sound unsure of [themselves]." Clearly, he knew what he was talking about here. Maybe you should have a little faith that he got to where he is with a lot of experiences along the way.

My Lab First Aid book$^{*1}$ tells the following:

The local therapy mainly consists of the instant, intense rinsing (at least 10 minutes) with tap water (ideally 15 to 20 °C). Don't try to neutralize the chemical burn, because the heat generation could lead to further tissue damages.

This means, that your lab instructor is absolutely right.
The rinsing water should wash the cold water away faster than the heat from the acid base reaction could harm you. It's not for nothing that there are emergency showers and eye douches in every lab.
I have a friend who is a chemical lab assistant and once had to carry concentrated sulfuric acid in a big glass beaker. Suddenly the bottom of the beaker broke off and all the acid poured over her etc. She not even thought of anything else than getting out of her clothes and under the next emergency shower.

$^{*1}$ The book is: R. Rossi, Erste hilfe bei akuten Notfällen - Begleitheft zu Jander - Blasius, S. Hirzel Verlag, 2005, p. 11f.. The text is in german and I translated it.

Rinsing is the most proper way and should be totally automatic (experience acquired by suffering with normal heat burns). Later "neutralizing" with diluted bicarbonate solution should not harm, but first rinse. Do not think what chemical I have to dilute in which concentration, etc... every second matters!

I once asked a fire fighter if they use any kind of chemicals for decon-p, and he basically said "Nah, we just put as much water on the person as we can. Dilution neutralizes everything."

Speaking from experience.
Minor spills (drops) on skin of concentrated nitric or sulfuric acid are nothing to speak of. Wash them off with plenty of water for a minute or two and you are fine. I'm pretty sure that this is the case for phosphoric acid as well. Hydrochloric acid is a bit more troublesome, mosly because it fumes intensively, but still nothing spectacular.
A common aftereffect of small nitric acid spills is local skin yellow coloring and a quickened loss of upper layer with fresh skin below. Again, nothing to panic about.
These acids cause proteins to denaturate, forming strong barriers for further diffusion into skin. As such, short contact with skin is safe as long as the acid is immediately washed off and the skin is undamaged. Diluted (0.1-0.5 M) hydrochloric acid is so tame, it is pretty safe to wash briefly glassware in it without gloves.
There are, however, acids, that are trouble. One is glacial acetic acid. It diffuses rather readily through lipid membranes and may cause deep, troublesome chemical burns requiring special treatment.
Also, note that while our skin is surprisingly resilient, our mucous membranes, and eyes in particular are not. Even one unlucky drop may require special help.
So, summing up.
Wash off sulfuric acid, nitric acid and hydrochloric acid, but as long as the contact is brief and small, it is nothing to care about. Longer and/or wider contacts should get medical attention.
Wash longer acetic acid, formic acid.
Generally avoid (and I mean it) hydrofluoric acid and derivitives (toxin, rapidly diffusing through sking) and $\ce{HClO4}$ (may form explosives on storage)
BY ALL COSTS PROTECT YOUR EYES. THEY CANNOT BE REPAIRED.

To add to @PH13's answer, according to the Texas A&M University's page Safety Guidelines, that rinsing is crucial, from the site:

Chemicals may cause severe burns. Immediately flush skin with cold water. Remove contaminated clothing (including socks and shoes) and rinse at least 10x as long as the chemical was in contact with the skin.

However, as all chemicals, including acids, have differences in how they react, it is crucial to be aware of the first-aid guidelines laid out in the Materials Safety Data Sheet (MSDS):

Read the MSDS sheets and product labels to learn the first aid procedures, health hazards and physical hazards for the chemicals used in each exercise.

This should be done as part of the lab preparation and hazard assessment, also to have on hand in case of emergencies such as a spill on skin or in the eyes.

Skin is fairly resistant to acids. Rinse it off with water for a minute and you're good. Everything that has already penetrated the skin (nitrous gases, HCl, HF(!)) will not be removed by washing afterwards. Further cooling might keep the swelling and inflammation down, of course.
There is ONE exception: Wipe off conc. $\ce{H2SO4}$ with a dry piece of cloth (or with anything dry) before applying water. Otherwise you'll get a severe burn.
Bases take longer to wash off, they enter the skin more easily (by dissolving it).
Never try to neutralize anything, it takes much longer than jumping under the emergency shower, is totally useless and likely dangerous, too.
But how does an "inexperienced" lab instructor discuss this with you? This is a subject for a very serious, compulsory lecture by the responsible head of laboratory, before anyone as much as opens a lab door!

He's still at the student level and since the actual instructor wasn't available yet, my group decided to talk to him...also, I thought of the example of sulfuric acid the whole time when he was explaining. I didn't realize it was an exception. Thanks a lot!

Conc. sulfuric acid is not so much an acid as an extremely vigorous desiccator. That makes it very dangerous for your eyes. On dry skin, it sucks the last bit of water out of the topmost dead layer, and then basically nothing. Proteins take long to dissolve in acids. If you want to see how evil the stuff really is, put a drop on a piece of cotton or a paper napkin. And do that in the fume hood and remove anything flamable from the vincinity.

There’s another key point why rinsing immediately and for a significant time is better than attempting to neutralise and it does fully support the heat argument.
When you attempt to neutralise acid spills, e.g. with hydrogencarbonate, the solid will stay on your skin and react only there. That means, the heat is generated right where you already spilt the acid and probably already have a mild burn. And there is nowhere the heat could go because none of the reactants will move.
If you neutralise with lots of water, yes, a lot of heat will be generated. But this heat will heat up the water which flows away. So even though you are generating a lot of heat, you are also creating an equally efficient heat-removal mechanism; and because you should rinse at least five minutes this will mean much less heat damage can be done to the skin in question.

Bee stings are very different in this regard. The amount of acid is very small and it has already somewhat penetrated the skin through the action of the sting (as opposed to just being on top as with an acid spill in the lab). The neutralisation will not generate as much heat because there is not that much acid, but rinsing with water is much less effective because the water is much less effective at actually getting to the acid itself.
Therefore, for two different accidents two vastly different treatment methods can both be correct (as long as you use the right one at the right time).


text:
SN1 (lim) and SN2 (lim)
I noticed (lim) is sometimes added to the SN1 or SN2 notation. I suppose it means limiting, but still do not understand it completely. Can someone please explain how is SN1 and SN2 different to SN1 (lim) and SN2 (lim)?

Text from the first row:
Existence of an intermediate with a lower coordination number
No intermediate
Existence of an intermediate with a higher coordination number
Text from the first collumn:
Stoichiometric mechanism
intimate-mechanism
Ingold-Definition
Dependance of the speed on the nature of the entering group
Text from the last row:
The speed does not depend on the entering group.
The speed does depend on the entering group.

Do you have an example which you could add in text (preferably) or as picture?

For those of us who cannot speak German, it will be useful if you can put the text from the image (above) into your question, which will enable translation to users' native languages upon viewing it.

I think, but this is just an educated guess, that the lim notation refers to limit, meaning an ideal nucleophilic substitution, while the other notations refer to mechanisms that behave more like the parent mechanism. There is a point where SN1 and SN2 cannot be determined anymore, which is somewhere in the middle.

This probably has to do with the rate-limiting step of the reaction mechanism, i.e. the slower step.

I think long story short: the difference between (lim) and not is that the dominant step drives the limiting case more strongly than the non-limiting case.
There are two aspects to a substitution reaction:

Stoichiometric mechanism: the sequence of elementary steps that take us from reactants to product

Associative (A): the bond formation between the metal and the entering ligand i
Dissociative (D): the bond breaking between the metal and the exiting ligand is the important step, i.e. the intermediate has a decreased coordination number
Interchange (I): there is no evidence of an intermediate

Intimate mechanism: what affects the rate constants of particular steps?

The reaction rate is affected by the entering group (a)
The reaction rate is affected by the leaving group (d)

You can see that SN1 and SN1 (lim) both have dissociative intimate mechanism, which means that the breaking of the bond between the metal and the leaving group is the rate determining step. In SN1 (lim), the leaving ligand bond breaks first forming an intermediate with a lower coordination number; the entering ligand then forms a bond with the metal. However, SN1 is an interchange stiochiometric mechanism which means that there isn't a true intermediate where the leaving group has totally left. In this case you could imagine that the leaving metal-ligand bond merely weakens or 'stretches' before the incoming ligand bonds to the metal.
Similar principle applies to SN2 (lim) and SN2. In SN2 (lim), an intermediate with higher coordination number is formed when the entering ligand bonds with the metal. For SN2, no such intermediate step is observed. You could say that the incoming ligand simply begins to form a bond to the metal before the leaving group bond starts breaking.
(http://nptel.ac.in/courses/104106063/Module%205/Lectures%208-10/Lectures%208-10.pdf for image above)


text:
Palladium catalyst
Recently I found the following question by sam:

I'm supposed to produce dry and pure oxygen using the decomposition of hydrogen peroxide with a manganese dioxide catalyst.
  $$\ce{2H2O2(aq) -> 2H2O(l) + O2(g)}$$
The gas will be collected in a bottle by downward displacement of water. How can I safely move the oxygen gas through the drying agent without mixing in atmospheric gas?

Based on that I am asking me, if and how palladium could catalyze the reaction?

Your title states sth about Pd, your first question says sth about Pd but below you only write sth about MnO2 and how to move the gas. Where is the connection?

This question is suspiciously similar to this one

I don't think that it is a duplicate. It's about what a change of the catalyst would bring to the system.


text:
Bromate forming in open water
According to this article, the city of Los Angeles (in the USA) is facing a problem where bromate, a suspected carcinogen, has been forming in their open water supplies.  They believe that the formation of bromate has to do with exposure to sunlight, and are now distributing millions of "shade balls" into water supplies in order to block the sunlight.
Given that an overwhelming majority of water on this planet is exposed to sunlight, why are there not high levels of bromate present in all water, especially shallow bodies of water where the surface area is maximized?  Or is it?

My knowledge of this is limited, so I'm not going to provide a full answer. However, I did read something on this same news article which said something along the lines of... There are several ways bromate can be generated, the most common being the combining of bromide, which occurs naturally in ground water, and ozone. It can also occur when water is treated with chlorine and subsequently exposed to sunlight. Hope this helps towards a proper answer.

The water referred to in the article was being stored in a reservoir after being treated with chlorine.  It is the combination of chlorine and sunlight that facilitates oxidation from bromide to bromate.
So, no, this is not applicable to natural bodies of water, only bromide containing water which has been artificially treated with chlorine.
See Bromate ion formation in dark chlorination and ultraviolet/chlorination processes for bromide-containing water for more information.