text:
Will borosilicate glass absorb 808nm laser light?
Will borosilicate glass absorb 808nm laser light? If it instead refracts, then what is the refractive index? Also, will borosilicate reflect lower frequencies of light like those used in non contact temp. measurements?

The amount of reflection depends on the polarisation and angle of incidence, look up Brewster angles and Fresnel equations. Refractive index depends on wavelength, usually we use Sellmeier equations to preduc=ict this

Borosilicate glasses are a family that has many members with differing properties. You can look up the provider for your glass, and check whether it is transparent in the near infrared. I think it would be in most cases (see here for one example).

SO you have to get that data on a per manufacturer basis.

@kurtnelle yes!

Unless doped, glasses are effectively 100% transparent in visible and near ir but not usually below approx 300nm in uv.  The amount of reflection depends on the polarisation and angle of incidence, look up Brewster angles and Fresnel equations. Refractive index depends on wavelength, (greater at shorter wavelengths in visible region) usually we use the Sellmeier equations to predict this if a value at exactly the wavelength you want is not available.


text:
$K_{eq} $ vs $K_c$
I was wondering if there was a difference between $K_{eq}$ and $K_c$. I think they both refer to the Equilibrium Constant. If I'm wrong, could you please tell me the difference between the two? Thanks!

Welcome to Chemistry SE! Your question appears to be a bit vague. Could you add a bit more on where exactly you saw $K_{eq}$ (most probably in a series of equilibrium reactions)?

My teacher referred to Keq and Kc. When I looked it up (on google), they both seemed to redirect me to the equilibrium constant.

$K_{eq}$ is the generic equilibrium constant.
$K_c$ is the equilibrium constant for a reaction explicitly in solution, for which the mass action expression can be written using concentrations only.
$$\ce{aA + bB <=> cC + dD}$$
$$K_c = Q_c (\text{at equilibrium}) = \frac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b }$$
$K_p$ is the equilibrium constant for a reaction explicitly in the gas state, for which the mass action expression can be written using partial pressures only:
$$K_p = Q_p (\text{at equilibrium}) = \frac{P_{\ce{C}}^c P^d_{\ce{D}}}{P^a_{\ce{A}} P^b_{\ce{B}}}$$
$K$ is the generalizd equilibrium constant, and is otherwise used when reactions are in mixed states. $K$ should perhaps be written with mole fractions $\chi_i$.
$$K = Q (\text{at equilibrium}) = \frac{\chi_{\ce{C}}^c \chi^d_{\ce{D}}}{\chi^a_{\ce{A}} \chi^b_{\ce{B}}}$$
Since concentration ($[i]=\frac{n_i}{V}$) is related to pressure from the ideal gas law:
$$\frac{n}{V}=\frac{P}{RT}$$
$K_p$ amd $K_c$ are related by powers of $RT$, depending on the stoichiometric coefficients.
$$K_c = \frac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}=\frac{(\frac{P_{\ce{C}}}{RT})^c (\frac{P_{\ce{D}}}{RT})^d}{(\frac{P_{\ce{A}}}{RT})^a (\frac{P_{\ce{B}}}{RT})^b}= \frac{P_{\ce{C}}^c P^d_{\ce{D}}}{P^a_{\ce{A}} P^b_{\ce{B}}} (RT)^{a+b-c-d}=K_p (RT)^{a+b-c-d}$$
Similar relations exist between mole fraction ($\chi_i=\frac{n_i}{n_T}$) and concentration ($\chi_i=\frac{c_i}{c_T}$) or pressure ($\chi_i=\frac{P_i}{P_T}$), and $K_c$ or $K_p$ can similarly be converted to $K$ or vice versa.
$$K= \frac{\chi_{\ce{C}}^c \chi^d_{\ce{D}}}{\chi^a_{\ce{A}} \chi^b_{\ce{B}}} = \frac{(\frac{P_{\ce{C}}}{P_T})^c (\frac{P_{\ce{D}}}{P_T})^d}{(\frac{P_{\ce{A}}}{P_T})^a (\frac{P_{\ce{B}}}{P_T})^b} =\frac{P_{\ce{C}}^c P^d_{\ce{D}}}{P^a_{\ce{A}} P^b_{\ce{B}}} (P_T)^{a+b-c-d}=K_p (P_T)^{a+b-c-d}$$

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text:
Is astatine a non-metal or a metalloid?
Is astatine a non-metal or a metalloid?
A metalloid is an element which exhibits some of the properties of a metal as well as those of a non metal.

Is Astatine a non metal or a metalloid?

Unknown
According to the Wikipedia article on astatine, the longest-lived isotope has a half-life of 8.1 hours. The element is so radioactive that large samples would melt and vaporize under the heat generated by its own radioactivity. Such behavior makes it difficult to assess whether astatine has the appropriate traits to be a metalloid.
Metalloids tend to be lustrous, brittle, semiconductors with other properties intermediate between metals and nonmetals. Nonmetals are dull/matte, brittle, and poor conductors of electricity and heat. All of these properties require a visible solid (or liquid) sample to assess, and thus astatine is impossible to assess and classify.
Astatine is usually placed with the metalloids or the nonmetals by different authors based on its position in periodic table, but in truth, astatine is so unstable in its elemental form, we have no idea which one it is.

A metalloid, possibly a metal
Even though nobody has so far seen or been able to test a visible piece of astatine we can extrapolate its properties based on its position in the periodic table. We know that iodine, the element above astatine, has a metallic appearance when viewed under white light, is a semiconductor (in the direction of its planes), and that there is evidence of some delocalised bonding between iodine molecules in crystalline iodine.
Given its position in the periodic table astatine could reasonably be expected to show more metallic character than iodine. Consistent with this expectation, the most common oxidation state of astatine appears to be +1 whereas that of the lighter halogens is –1; and there is some evidence to suggest that astatine forms a cation in strongly acidic solutions. In 2013, based on relativistic modelling calculations, condensed astatine was predicted to be a monatomic metal with a face-centred cubic structure. Metals with this structure are noted for their ductility whereas metalloids are regarded as being brittle.
If astatine does turn out to be a metal then, like other post-transition metals, it could be expected to show appreciable non-metallic character. Trace level experiments on the chemistry of astatine show that it appears to act like a metal in some respects and like a non-metal in others.
The Wikipedia article on metalloids has more information about astatine and its metallic and non-metallic properties. For now, until more evidence comes in, it would be reasonable to classify astatine as a metalloid.

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Overlapping oxygens in crystal structure -- what's going on?
This is a screenshot of  a portion of a metal organic framework (MOF).  These are two oxygens of two NMP solvent molecules.  What I am wondering is why they are so close and overlapping like this.  I checked the CIF file, and occupancy for all atoms in this structure is 1, so it doesn't seem to be the case that there is only one NMP molecule in two equally likely positions.
Could someone explain this?  Thanks

It's a bit hard to answer without having coordinates file (*.res or *.xyz), but the simplest answer that I would expect is that these oxygen atoms belong to two different molecules that are disordered over the same position (i.e. they have only partial occupation on the site that in sum is 1). It's very common for f.e. thf coordinated to alkali-metals.
If you have shelx input for the structure look for the line that defines one of oxygens and see if 6th parameter is 11.0
O1    5   -0.180348    0.245831    0.298215    11.00000    0.02208    0.02526 =
                                               ^^

This means fixed (10) occupancy of 1.
If you have cif-file look for the atom loop
loop_
 _atom_site_label
 _atom_site_type_symbol
 _atom_site_fract_x
 _atom_site_fract_y
 _atom_site_fract_z
 _atom_site_U_iso_or_equiv
 _atom_site_adp_type
 _atom_site_occupancy
 _atom_site_symmetry_multiplicity
 _atom_site_calc_flag
 _atom_site_refinement_flags
 _atom_site_disorder_assembly
 _atom_site_disorder_group
O1 O -0.18035(6) 0.24583(6) 0.29822(5) 0.02794(8) Uani 1 1 d . . .

8th parameter is occupancy (_atom_site_occupancy).
These both examples were for not disordered atoms. If they are disordered then it looks like this:
shelx-ins/res
O1    4    0.297097   -0.198418   -0.064164    10.50000    0.34896    0.34935 =
                                                  ^ occup. 1/2

or
FVAR       0.22777   0.51340   0.51266
...
O1    4    0.551971    0.536866    0.295322    21.00000    0.03100    0.03031 =
                                                ^ occup. bound to 2nd free variable
                                                  here 51%

cif-file:
O1_20 O 0.6673(9) 0.3140(7) 0.1724(3) 0.0434(14) Uani 0.513(3) 1 d PDU C 1
                                                      ^


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What is the mechanism for oxidation of ketones by the Popoff's rule?
Ketones are relatively much harder to oxidise, but they do undergo oxidation reactions at extreme temperatures. Our teacher also told us about Popoff's rule which states that, on oxidation of unsymmetrical ketones, the $\ce{>C=O}$ group remains with the smaller alkyl group. Could you please give me a hint as to how this happens, and the probable reaction mechanism for this?
Also, please give the complete balanced equation when butan-2-one is oxidized to a carboxylic acid.

Your statement of extreme temperature is wrong. With proper oxidants, ketones are oxidized under mild condition.

2-Butanone to propionic acid via the haloform reaction - fast, at room temperature.  You can use chlorine laundry bleach.   doi:10.1021/cr60052a001

@IanFang Is this correct: Ketones undergo oxidation 'to form carboxylic acids' only at extreme temperatures and using a strong oxidising agent? Because ketones do undergo oxidation at mild conditions. For example, alpha-hydroxy ketones oxidises even using mild oxidation agent (Tollens' Reagent). Am I right?

According to popoff's rule, the unsymmetrical ketone on oxidation, C-C bond cleavage and keto group goes with (CH3)2 - CH - group

I am uncertain about the "Popoff's" rule you mention. There are two reactions that can oxidize ketones, and one seems to follow the behavior you are suggesting, but does not form a carboxylic acid. The other reaction does form carboxylic acids, but is more complex.
The Baeyer-Villiger oxidation is an oxidation of ketones to esters using a peracid in the presence of a mild base:
$$\ce{CH3COCH3 + RCO3H->[\ce{Na2HPO4}] CH3CO-OCH3 + RCOOH}$$
The mechanism involves the fragmentation of one of the $\ce{C-CO}$ bonds. Unsymmetrical ketones fragment in a predictable pattern, but not always that the carbonyl remains with the smaller group. The fragment that would be a more stable carbocation (even though carbocations are not formed in this reaction) is the one to move. The migratory aptitude is tertiary alkyl > cyclohexyl > secondary alkyl, aryl > H > primary alkyl > methyl . For exampl, with 2-methyl-3-octanone, the isopropyl group moves because it is attached via a $2^\circ$ carbon:
$$\ce{(CH3)2CHCOCH2CH2CH2CH2CH3 + RCO-O-OH ->[\ce{Na2HPO4}] (CH3)2CH{\bf O}COCH2CH2CH2CH2CH3 + RCOOH}$$
Nitric acid chews ketones apart into two carboxylic acids. Concentrated $\ce{KMnO4}$ in acid also does this. These reactions are a little bit harder to find information on, since they tend to be considered uncontrollable.
The reaction goes through a series of oxidations from ketone to $\alpha$-hydroxyketone to cleaved acids through several enol intermediates.
$$\ce{RCH2COCH2R <=>[\ce{HNO3}] RCH=C(OH)CH2R ->[\ce{HNO3}] RCH(OH)COCH2R}$$
$$\ce{RCH(OH)COCH2R <=>[\ce{HNO3}] RC(OH)=C(OH)CH2R ->[\ce{HNO3}] RCO}$$
If the ketone is unsymmetrical, there is no guarantee that is will cleave predictably. For example, 2-butanone could cleave into propanoic acid and carbon dioxide or two equivalents of acetic acid.
$$\ce{CH3CH2COCH3 ->[\ce{HNO3}] n(CH3CH2COOH + CO2) + (1-n)(2CH3COOH)}$$
A balanced equation for the formation of acetic acid would look like:
$$\ce{CH3CH2COCH3 + 3NO3^- +  -> 2 CH3CO2H + 3NO2^-}$$
A balanced equation for the formation of propanoic acid and carbon dioxide would look like:
$$\ce{CH3CH2COCH3 + 4NO3^- +  -> CH3CH2COOH + CO2 + 4NO2^- + H2O}$$
The permanganate reactions are tougher to balance, since permanganate is a three electron oxidant.

Update:
When I wrote this answer, I had never heard of Popoff's rule. No textbook I own mentions this rule, and a Google search about it brings up this question as the top hit (and similar questions at other sites like ask.yahoo.com as the other hits). I now know that Aleksandr Popov published a paper in Liebigs Annalen in 1872 describing a variation of the reaction that would become the Baeyer-Villiger oxidation. This article is behind a paywall for me, and the first page preview confirms I would not be able to make much of it. I only know a small amount of German and the PDF sadly looks like a low quality copy of a copy of a copy.
However, from the title "Die Oxydation der Ketone als Mittel zur Bestimmung der Constitution der fetten Säuren und der Alkohole" I can parse what the paper was about. Roughly this paper was about "The oxidization of ketones - a means for determining the constitution of the fatty acids and alcohols". Thus, his method involved the oxidation to the ester and hydrolysis of said ester in one step. I don't know what reagents Popov used, but I'm pretty sure it was not a peroxyacid (since this reagent will not hydrolyze the ester). I have previously answered a question about migratory aptitude in Baeyer-Villiger reactions. Since I cannot read Popov's paper, nor can I find any authoritative resource on his rule (on- or offline), I have to assume that the known migratory aptitude for the Baeyer-Villiger reaction is the same as Popov's rule.

Isn't the major product(2 acids) predicted by popoff's rule?

Even if you spoke german, this is old language and hard to understand (and the quality does not help either). However, here is what he used to oxidise: $30~\mathrm{g}~\ce{K2Cr2O7}, 20~\mathrm{g}~\ce{H2SO4}, 100~\mathrm{g}~\ce{H2O}$ for $6.1~\mathrm{g}$ butyl phenyl ketone (I think), followed by azeotropic distillation. Apparently they form (after some more steps) benzoe-, isobutyric- and acedic acid. (However this might happen...) I did not read much further as it pained my eyes and brain. I hope this unfogs the mystery of the paper a little. (BTW: there was a previous version in russian.)

Popoff's rule: just pop off the smaller substituent when oxidizing a ketone!

Confirming what @Mart said, although I have read enough of these papers to not find them hard (and I am fascinated by the old spelling variants). Popov states a rule at the beginning that if a phenyl or methyl (or sometimes ethyl) ketone is oxidised, then the oxidation breaks the $\ce{C-C}$ bond on the other side of the ketone; so propyl methyl ketone would give acetic acid and propionic acid. Not sure if that is the rule or not.

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Why is it hard to reduce titanium dioxide?
Titanium is expensive, although it is not rare and titanium dioxide is a ubiquitous component, it's part of such mundane articles as toothpaste, paper or plastics.
The reason for that is that the processes that extract titanium from titanium dioxide (which occurs naturally and just needs some inexpensive purification) are quite costly. The established industrial method for this is the Kroll process. It works by first chlorinating the titanium to $TiCl_4$ and then moving the chlorine over to some $Mg$, yielding $MgCl_2$ and pure $Ti$. Then follows a costly remelting process under vacuum to further purify the metal.
Recycling titanium scrap is even harder, and removing the other components from the alloys is not entirely possible, so recycling only yields titanium based alloys, not the pure stuff (which is not necessarily a problem since it's very rarely used in its pure form).
What specific chemical properties of titanium make it so hard to purify? For example, the chlorine process at the beginning of the Kroll process takes place with coke in a fluidized bed reactor, very similar to the reduction of $Fe$ in a blast furnace, but if the carbon takes care of the reduction, why is that not sufficient? Why is the chlorine route needed?
Research into titanium production dates back to the late 1880s, and the Kroll process, patented in 1940, made it feasible for mass production, but remains costly and has not been improved upon since, despite much effort. How would a chemist look for suitable reactants during such research?

Titanium dioxide reacts directly with carbon when heated to form Titanium Carbide (see Wikipedia page on Titanium - History). Maybe a fluidized bed reactor could prevent that reaction in some way.

@kaliaden, it appears the chlorine (rather than the reactor) does, grabbing the titanium and leaving the carbon to pair with the oxygen, lest it will spend the rest of the reaction alone. I'm looking for the actual scientific explanation for that.

Reasons why Titanium is hard to extract and purify:

The +4 oxidation state of $Ti$ is the most stable (due to its noble gas configuration) and so reducing $TiO_2$ to $Ti$ is quite difficult.
$Ti$ has chemical properties which are similar to its neighbouring elements (Vanadium, Zirconium, Chromium) and the presence of compounds containing these elements in its ores makes purification even more difficult. For example: Vanadium and Zirconium also form volatile halides ($ZrCl_4$)and oxyhalides ($VOCl_3$) which have to be separated from $TiCl_4$ by fractional distillation in Kroll's Process.

I have tried to explain why the chlorine route is taken using only thermodynamics (I'm not absolutely sure about this but these reasons are what I could come up with).
Consider these reactions which could take place in the fluidized bed reactor:

$\ce{TiO_2(s) + 2Cl_2(g) + 2C(s) -> TiCl_4(g) + 2CO(g)}$
$\ce{Ti(s) + C(s) -> TiC(s)}$

Comparing the Gibbs' Energies of these reactions can tell us which one is spontaneous since if the change in Gibbs' Energy is negative, the reaction will be spontaneous.
And the equation for change in Gibbs' Energy is:
$\Delta{G} = \Delta{H} - T\Delta{S}$
At normal temperatures, reaction 2 would occur more readily than reaction 1 since $TiC$ is extremely stable (reflected by its very high melting and boiling points) while $TiCl_4$ is unstable (it hydrolyses immediately on contact with air or oxygen to form $TiO_2$ and $HCl$). And so it is almost impossible to reduce Ti using C at low temperatures.
At higher temperatures (Kroll's Process is carried out at around $1000^oC$), $T\Delta{S}>>\Delta{H}$ so we can neglect $\Delta{H}$.
For the first reaction, since all the products are gaseous (while some of the reactants are solid), $\Delta{S}$ is positive. Therefore $\Delta{G}$ will be negative and the reaction is spontaneous and will occur easily.
For the second reaction, since the number of molecules is reduced, $\Delta{S}$ is negative. Therefore $\Delta{G}$ will be positive and the reaction becomes unfavourable.

Notes:

For the states of the reactants and products, I compared the melting and boiling points of the reactants and products with $1000^oC$ which is the temperature at which Kroll's process is carried out.
Actually, from what I read, there are many possible products for the first reaction. depending on the reaction conditions, a mixture containing $CO$, $CO_2$ and $COCl_3$ can be formed. But all these products are still gaseous so it doesn't affect the explanation.


text:
Carbon atoms in triple bonds
I was trying to know what is the probability for the carbon atoms to rotate .I know that in single bond it is 0 % and I think it is 100 % in double bond but   how about the triple bond ?

You might need to clarify what you mean by carbon atoms "rotating".

Triple bonds allow free rotation. Was that what you meant?

Actually, it's the opposite. Most Carbon-Carbon single bonds can rotate without much difficulty. Ethane, for example, has a (potential energy of conformer vs dihedral angle) graph like this:

The energy barrier isn't that much here, and ethane probably rotates without much effort.
Propane and all will have higher energy barriers, and will rotate less often. Even cyclic compounds like cyclohexane undergo ring flipping
However, not all single bonds can easily rotate a full turn. Systems like these:

have a hard time rotating about the central bond due to steric repulsion of the groups on the biphenyl system (which makes the two compounds shown above noninterchangeable). These systems can only rotate a small angle, and rarely go 360° (it may be possible to force this to happen by providing the system with energy).
And, of course, there are systems where it's obvious that there cannot be much rotation. For example, in bicyclo[2.2.2]octane, most bonds will rotate very little, and it is impossible for any bond to rotate a full 360° without breaking any bond.

Now, single bonds can rotate because they are sigma bonds. A sigma bond is formed by overlap of one lobe of an orbital from each of the two atoms:

(this is ethane)
There's nothing preventing the carbons from rotating here.
On the other hand, double/triple bonds have pi bonds (in which two lobes each overlap) as well as sigma bonds:

(this is ethene)
Here, it is impossible to rotate without breaking the pi bond. Note that this has some importance in how our eyes see -- one of the mechanisms in eyesight is a double bond in rhodopsin being broken (by incident light) into a single bond, rotating, and reforming (which leads to some nerve impulse).

A way to  visualise this is by holding a pencil between your fists. You can easily rotate your hands. Now, hold three pencils (a bit spaced out) between your fists -- you will no longer be able to rotate your hands.

Finally, to answer your question: The probability/average time depends on the compound and the surroundings. For double/triple bonds, it's more or less 0%.

Wrong. Rotation around triple bonds is possible. See $\ce{Ru2Cl8^2-}$. If an electron is excited from $\delta$ to $\delta^*$, the quadruple bond turns into a triple bond and the $\ce{Ru#Ru}$ bond rotates by $45^\circ$.

This comment is totally off the mark.  Transition metal-metal bonds are not anywhere close to a good model for carbon-carbon bonds.

Manish's Answer is appropriate for single and double bonds, however:
Rotation about C-C triple bonds is a nonsensical concept.  There is no observable change in geometry.  It is impossible to determine whether or not any rotation happens.

See my comment under ManishEarth’s answer


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How to get the valencies of elements?
How to find the valencies of elements by using its distribution of electrons?
Please explain the method in simple words. Do you have to study the valencies or is there a simple way of remembering?
PS: There is supposed to be some method of getting the valency after knowing the distribution of electrons. For example the element sodium, $\ce{Na}$ has a distribution of electrons: 2, 8, 1. What is it's valency?

related https://www.quora.com/How-do-I-find-the-valency-of-elements.

Valency is a concept that describes the power of an element to bind to other elements. There is an official definition in the IUPAC goldbook:

valence
  The maximum number of univalent atoms (originally hydrogen or chlorine atoms) that may combine with an atom of the element under consideration, or with a fragment, or for which an atom of this element can be substituted.

So in principle it is not possible to know the valency just from the electron configuration, but you have to also know the chemistry it performs. It is also noteworthy, that according to this definition all elements have only one value for it's valency.
In the first period, according to the definition, hydrogen is monovalent, i.e. $\ce{H2}$ or $\ce{HCl}$. Helium on the other hand is technically divalent, since $\ce{HHeF}$ was theoretically predicted, see wikipedia. But it is somewhat safe to assume helium is inert and therefore zerovalent.
In the second period, where the octet rule holds, it is the number of unpaired electrons in the valence shell, with the exception of carbon. Unfortunately this is also the period which behaves most different from the rest of the periodic table. So the best way here is to just learn them as exceptions: $\ce{NaCl}$ (mono-), $\ce{BeCl2}$ (di-), $\ce{BCl3}$ (tri-), $\ce{CCl4}$ (tetra-), $\ce{NCl3}$ (tri-),(1) $\ce{Cl2O}$ or $\ce{H2O}$ (di-), $\ce{FCl}$ or $\ce{F2}$, $\ce{Ne}$ is zerovalent.
Below the second row it becomes a little bit more predictable. For most main group elements it is the number of electrons in the valence shell. This obviously coincides with the highest stable oxidation state.
Some examples may illustrate that. For the first group all binary compositions of the alkaline element $\ce{A}$ of the composition $\ce{ACl}$ are known. This is also the maximum of chlorine atoms any of these elements can bear, hence they are all monovalent.
 The same applies to the earth-alkalines, $\ce{Ea}$. Here all binary chlorides of the form $\ce{EaCl2}$ are known, hence they are divalent.
For the triels $\ce{M^{(III)}}$, all of the chlorides of the form $\ce{M^{(III)}Cl3}$ are known. However not all of them are very stable, i.e. $\ce{TlCl3}$ (wikipedia). So they are trivalent.
For the tetrels, $\ce{M^{(IV)}}$, the trend continues. The binary chlorides $\ce{M^{(IV)}Cl4}$ are known, even $\ce{PbCl4}$ is quite stable below $0~^\circ\mathrm{C}$ (wikipedia).  For Flerovium only predictions have been made, but it's believed to be very unstable in its tetravalent state due to the inert pair effect.
For the pnictogens, $\ce{M^{(V)}}$, all binary chlorides up to and including antimony are known. However, also $\ce{BiF5}$ is known, hence all elements can be considered pentavalent.
For the chalcogens, $\ce{M^{(VI)}}$, all binary fluorides with coordination number six are known. Most prominent is sulfur hexafluoride, for which a chloride counterpart exists. The elements can therefore be considered hexavalent. The tetra chlorides are also stable for all elements of this group, in that case the elements can be considered tetravalent. However, more common are compounds where the chalcogens are divalent, i.e. in the respective hydrogen compounds, $\ce{H2M^{(VI)}}$.
The halogens $\ce{Hal}$ are just amazing, and once you have looked into interhalogen compounds you will most likely remember a lot of them. For chlorine, the interhalogens $\ce{ClF_{\{1,3,5\}}}$ have been reported, making it pentavalent. For bromine, the same holds true: $\ce{BrF_{\{1,3,5\}}}$. Only $\ce{BrCl}$ has been observed. However this makes bromine also pentavalent. For iodine the colourless gas $\ce{IF7}$ has been observed, making it the only heptavalent element of the series. For astatine only a few compounds are known, and the current believe is, that it is monovalent.
For the noble gases, $\ce{Ng}$, there are a few compounds known, and their valency increases throughout the group. While argon can be considered at most divalent due to the discovery of the highly unstable $\ce{HArF}$, the divalent nature of Krypton is undoubtedly proven by $\ce{KrF2}$. Xenon is the most reactive and potent noble gas and it can be considered hexavalent according to the above definition, as of $\ce{XeF6}$ does exist. However, also $\ce{XeO4}$ is known, which would make it octavalent according to a different definition.(2)
Transition metals are a whole different story and it is difficult to find a generalised rule or observation for it, since the IUPAC definition is very strict on the matter. All elements have to be looked at in detail. It will certainly also exceed the purpose of this answer.
Just to illustrate the problem have a look at manganese. There are various compounds known, such as $\ce{MnCl2}$ (divalent), $\ce{MnF3}$ (trivalent) and even $\ce{MnF4}$ (tetravalent). Binary hydrides have not yet been reported. However, there are oxides of various composition, ranging from divalent to heptavalent.

Notes

The molecule $\ce{NCl5}$, pentavalent, has been theoretically suggested, see wikipedia and references within.
A more modern interpretation of the valency concept allows elements to have more than only one value. It is therefore always depending on the compound in which it is looked at. A very brief summary can be found on wikipedia.

The most characteristic chemical property of an element is its valency. There are numerous measures of valency each with its own area of usefulness and applicability. Simple definitions refer to the number of hydrogen atoms that can combine with an element in a binary hydride or to twice the number of oxygen atoms combining with an element in its oxide(s).

(N. N. Greenwood and A Earnshaw: Chemistry of the Elements. Second edition. Butterworth-Heinemann, 1997. Chapter 2, p. 27.)

Look at it this way: The atom needs to have a complete octet in its outermost shell, and will give away/accept electrons to achieve this goal. Sodium, for example, can give away one electron (+1 valency) or accept 7 (-7). Generally accepting more than 4 electrons is hard, so you don't see the latter happening. Similarly, oxygen has $2,8,6$ -- it can accept 2 electrons (-2) or give up 6 (+6, never observed)
Note that the octet rule is not universally applicable. Elements can have multiple valencies, not all satisfying the octet rule. To understand this you need to understand orbital theory first (which will be hard to explain in the scope of one post).

In addition, I'd advise [OP] to take a look at the Electron Configuration chapter of his book, or at this Wikipedia page about it.

There's actually no need to remember the electronic configuration for this, you just need the group number for simple application. Like for metals, the valency is the number of electrons in the outermost or valence shell, and for non-metals and the metalloids the valency is 8 minus the number of electrons in the valence shell. In the strictest sense, valency is like combining power or capacity for bond formation.
Using group number, sodium for example has a valency of 1 since its in Group 1. Chlorine has a valency of 1 because it's in group 7 (8 - 7 valence electrons = 1). As kappi has mentioned, this doesn't apply all the time since only the second period of the periodic table follows the octet strictly.
Using the electronic configuration for this pupose is a bit confusing though.


text:
I'm looking for a substance that can be used as an adhesive, but that will burn without producing any ash
I need to 'adhese' paper to plastic, but I need the adhesive to be flammable and quick burning, but without producing any ash or too many toxic fumes.
I know this is an absurd question, but does anything like this exist?
Thanks so much

You may also want to ask it on DIY.SE

Nitrocellulose(flash paper). Using acetone or something you get a gel. Caution recommended.

Thanks for your answer, nitrocellulose is my 'paper' layer - will the acetone gel thing still work or will it mess up the nitrocellulose paper? Thanks

Try dampening the plastic with acetone or the gel. You need to test to get the best way to do it.

It would be nice if you could list the properties / explain why they occur. Not necessary, though :)

I don't know much. What I told you is from experiments long ago before everything was a terrorist threat.


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Why does the melting point get lower going down the Alkali Metal Group with increase in atomic number?
Why does the melting point get lower  going down the Alkali Metal Group with increase in atomic number?

You have to understand that melting point is related to the bonds in a metal between the metal cation and the 'sea of electrons'. With an increase in atomic number, you have an increase in electron shells.
As the radius of atoms get larger down the group, you could say that the force holding them together is spread over the greater area and hence, the metal cations are more weakly bonded. And coming back to my first point, when the bonding is weaker, the metal's melting point will decrease.

As you go down any group, there are extra electron shells (for example Lithium has 2 shells, Francium has 7) which cause electron shielding. Because the force of attraction must extend further and through these shells to the outermost shell, the force becomes less so there is less strong bonding occuring.
Thus, it is easier to break these less strong bonds, and because the melting point is a measure of the point at which all the bonds holding a metallic structure together are broken, it will decrease because this point is reached a lot sooner as the bonds are not as strong.

melting points of all group (i) elements is dependent on the strength of the metallic bond. In metallic bonding, the group (i) cations in the metallic lattice are attracted to the delocalised electrons. Down the group, the number of delocalised electrons and the charge on each cation remains the same at +1 but the cationic radius increases so the attraction between the cations and the electrons in the lattice get weaker down the group and so does the strength of the metallic bond

Why does the melting point get lower going down the Alkali Metal Group with increase in atomic number?

Ok, you're looking for a generalization of what happens to the melting point as you go down a column in the periodic table.
Think of the metal atoms as cations $\ce{M^+}$ and anions $\ce{M^-}$ packed into a crystal like NaCl. However for the metals think of the metal atoms switching + and - charges at a high frequency.
So as the atoms get bigger, the energy that is holding the crystal together gets smaller because the "ions" are further apart.
Note that this simple explanation doesn't explain "real" chemical bonding. So there are other layers of truth under this simple model.

Shell structure of an atom
Different crystalline lattices and hence energy "within" the crystal
Ionic , covalent, and metallic bonding

So the "real" truth is gleamed from the interactions of simple models into a more complex system.

You have to know that elements with high lattice energy has high melting and boilng point.And since lattice energy is inversely related with radius of atoms it increases down a group and hence the melting point increases.
I hope this will help you!

This doesn't follow. Atomic radius increases down the group so according to you lattice energy should decrease and so melting point should decrease.

The alkali metals are not ionic solids.

@bon although this answer doesn't make much sense, the melting point does indeed decrease down the group..


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What is the difference between graphite oxide and graphene oxide?
I am considering making a sheet of graphene using the LightScribe technique and I have read that I need graphite oxide. However when I look up graphite oxide for sale, graphene oxide pops up.
Are these two oxides similar? What is the difference?

And while I am at it, what is the difference between CO2 and graphite/graphene oxide?

I love your answer. I got the terms mixed up and have been searching for hours myself. I believe the term graphene/graphene oxide/graphite oxide/graphene nanoplatelets should all be defined clearer. Not a big fan of companies selling graphene nanoplatelets at USD 160 per gram as their products often do not have the properties of graphene. Anything over should not be sold as a product with a graphene it its name.

In short, graphite is several graphene sheets piled one above another.
Graphene is made up of one single sheet of carbon atoms arranged in hexagonal pattern (like a honeycomb), and graphite are several such sheets, each sheet linked to another by weak intermolecular forces, which gives the graphite its lubricative properties.
EDIT:
Similarly, graphene oxide is a layer of graphene with oxygen atoms, $OH$ groups and $COOH$ groups attached to it, while graphite oxide are layers of graphene oxide.
Because of the presence of $COOH$ groups, graphite oxide can disperse in basic solutions (or even water!) to give graphene oxide.
$CO_2$ is simply carbon dioxide, one carbon atom doubly bonded two two oxygen atoms.
By contrast, graphite oxide is like a huge field of carbon with the groups I mentioned earlier while carbon dioxide is like a small ball.

I need Graphite Oxide too, should I buy Graphene Oxide? I understand the difference but I'm a bit confused!

I'm hardly an authority on graphene structures (only been reading the literature for about a year), but let me try to contribute.
Unfortunately it seems that research into graphene and its derivatives is progressing at such a pace that it can be hard to standardize notation. I have actually previously had a small discussion on the internet regarding the difference between graphene oxide and graphite oxide.
First, let's talk about graphene itself (for the sake of completeness). Originally the term meant a honeycomb sheet with single atom thickness, however it seems to have slightly evolved in usage to mean "graphite thin enough to show peculiar properties". According to Geim and Novoselov (The rise of graphene, 2007), there is a distinction to be made between single-layer graphene, bilayer graphene and few-layer graphene (3-10 layers) based on their electronic structure. Anything above 10 sheets thick already is electronically sufficiently similar to bulk graphite that it can be called as such. However, there also seems to be another term on the rise for stacks over 10 layers thick, with a still small but not well-defined upper limit, called graphite nanoplatelets (or confusingly, graphene nanoplatelets, depending on who you ask).
Regarding graphene/graphite oxide, I follow the interpretation suggested by Ruoff's group (2011). As far as I understand, when dry or dissolved in a highly concentrated, non-basic solution with little stirring, the individual oxidized sheets tend to stay stacked, and one may speak of graph*ite* oxide. If the oxidized sheets are in a basic, dilute, and/or vigorously stirred or sonicated solution, or if the sheets are supported on a substrate and proven to be composed of few layers, then they are unstacked (exfoliated) and one can speak of graph*ene* oxide. Theoretically, there is no composition difference between graphite and graphene oxide, since it's merely a matter of aggregation (The chemistry of graphene oxide, 2010). The difference would be like that between bulk gold and colloidal gold; the composition is the same, but they are chemically different. I have been informed that it is also common to refer to the stacked oxidized sheets as graphene oxide, and that separate sheets are called exfoliated graphene oxide. I don't agree very much with that standard, though, since it doesn't make too much sense to me.
To make things increasingly complicated, it seems that what most people considered to be graphene/graphite oxide is actually neither, but a mixture of oxidized graphitic sheets with highly oxidized polycyclic molecules, collectively called tannic acids or oxidized debris, given they appear as side-products during the process of vigorous oxidation of graphite (Rourke et al., 2011). These debris can make up as much as 30% of the dry weight of graphite oxide prepared via Hummers and Offeman graphite oxidation, and they have also been reported as side products of the oxidation of CNTs with HNO$_{3}$. It could well be that many properties previously attributed to graphene/graphite oxide are in fact strongly altered by the presence of these debris, which stay adsorbed to the graphene sheets via $\pi - \pi$ interactions and hydrogen bonding. For example, graphite oxide is only appreciably soluble in polar solvents such as water, DMSO and NMP because the debris act as a surfactant of sorts, stabilizing the suspension. When removed, the purified graphene oxide sheets are actually quite insoluble in most solvents, much like the non-oxidized sheets. Also, perhaps the existence of previously undetected debris is part of the reason why there are still debates on the proper model for the structure of graphene oxide.
In short, graphene and graphite oxide, while not exactly the same, are terms used interchangeably, and it's possible to go from one to the other reversibly by adding/removing solvent (as long as no concentrated base is added, or dilute base with heating). "Graphene" is a trendier term though, so that's usually what oxidized graphite will be advertised as. Though in actuality, due to the presence of debris, without purification it is really neither!
Edit: Ever since I wrote this answer, it seems that efforts to begin standardizing the nomenclature of graphene-related carbon nanostructures are underway (free access). Quoting from the article:

Graphene oxide (GO) – chemically modified graphene prepared by oxidation and exfoliation that is accompanied by extensive oxidative modification of the basal plane. Graphene oxide is a monolayer material with a high oxygen content, typically characterized by C/O atomic ratios less than 3.0 and typically closer to 2.0.
Graphite oxide – a bulk solid made by oxidation of graphite through processes that functionalize the basal planes and increase the interlayer spacing. Graphite oxide can be exfoliated in solution to form (monolayer) graphene oxide or partially exfoliated to form few-layer graphene oxide.

Note that the definitions proposed in the article will not necessarily take hold, and even if they do the transition will take some time. Be wary of nomenclature mismatches during this period.
Also, in my post I mention the presence of oxidized debris in graphite oxide. Research is still ongoing to determine their nature and properties.


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Difference between the terms "autocatalysis" and "(branching) chain reaction"
Is there a technical difference between the terms "autocatalysis" and "branching chain reaction"? They both seem to refer to something very similar, namely sequences of reactions in which one of the intermediates is also one of the products, leading to the possibility of exponential growth. Is one of the terms more specific than the other, or are they generally understood to mean the same thing?
There do seem to be some differences in how the terms are used. I've never seen the formose cycle, for example, referred to as a chain reaction, which does seem to imply that the term "chain reaction" means something more specific. If this is the case, what is the defining difference between the two terms?
Here is a diagram of the formose cycle (from the German Wikipedia entry, because it has a much better diagram than the English one) - is there a specific reason why it would not be referred to as a chain reaction?

An autocatalytic reaction is a reaction where the product catalyzes the reaction. An autocatalytic reaction takes the following form (or more complex variations of it).
$\ce{A + B -> 2B}$
An autocatalytic reaction must produce the exact species $\ce{B}$ that cataylzes the reaction. More about autocatalytic reactions, including examples, can be found at this question and this answer.
A chain reaction is one where every step of the mechanism produces another activated intermediate of some kind that can continue to react under the same conditions. Chain reactions continue until they terminate by means specific to the type of chain reaction. Chain reactions may appear to be autocatalytic, but there is a net consumption of both reagents, and the end product does not tend to cataylze the reaction:
$$\ce{A + B}* \ce{ -> C}$$
$$\ce{C + B -> D + B}*$$
Net:
$$\ce{ A + B -> D}$$
In the generic example above, the second step produces more $\ce{B}*$, which can reenter the first step. Overall, however, the reaction is not autocatalytic, since no more $\ce{B}$ is being produced and the concentration of $\ce{B}*$ is likely unchanging (steady state). All chain reactions have this feature. While the reactive species may appear on both the reactant side and the product side, there is no net change in the number of reactive species present in the reaction.
A classic introductory chain reaction is the radical halogenation of an alkane like methane. This reaction's mechanism is divided into three regimes: 1) Initiation (generation of radicals where there were none previously), 2) propagation (the chain part), and 3) termination (in which there is a net reduction in radicals, usually by recombination).
Initiation:
$$\ce{Cl2 ->[hv] 2Cl*}$$
Propagation:
$$1) \ \ \ \ \ \ce{Cl* + CH4 -> HCl + CH3*}$$
$$2) \ \ \ \ \ \ce{CH3* + Cl2 -> CH3Cl + Cl*}$$
Termination:
$$\ce{CH3* + Cl* -> CH3Cl}$$
$$\ce{2CH3*  -> CH3CH3}$$
The above example is a cyclic chain reaction. Chain reactions can be linear, for example chain-growth polymerization. In this kind of chain reaction, two species, where one is the activated end of a polymer chain $\ce{P}*$ and the other is a new monomer $\ce{M}$, react to form a new activated chain end $\ce{P}'*$ that is one monomer unit longer. $\ce{P}'*$ can then reenter the reaction and react with another $\ce{M}$. Again, this reaction is not autocatalytic. The number of activated chain ends is not increasing (or decreasing).
$$\ce{P}* +\ce{M -> P'}*$$
$$\ce{P'}* +\ce{M -> P''}*$$
The activated chain ends, $\ce{P}*$, can be radicals, anions, cations, the products of ring-opening, or organometallic complexes.

Many thanks for the answer - I will read it carefully later. But just a quick comment for now: I guess I should have been more specific in my question and said branching chain reaction.

You mean like nuclear fission of uranium? I suspect in that case then you have a chain reaction that is also autocatalytic.

Nuclear fission, combustion, PCR, ...

sorry for the brief responses yesterday, I was pressed for time. A branching step in a chain reaction is defined as one that increases the number of activated intermediates, as opposed to a propagating step, which keeps the number the same. I guess my question is really about what counts as an "active" intermediate - is it just anything that can continue the reaction (in which case autocatalysis and branching chain reactions are probably the same thing), or does it mean something more specific?

Would it be correct to describe the Formose cycle as a chain reaction? I've edited a diagram into my post. In this case the "initiating" and "terminating" steps are not shown, but would it be correct to describe it as a chain reaction in which all the reactions shown are propagating steps, except for the decomposition of the aldotetrose into two glycoaldehydes, which is a branching step?


text:
Acids / Bases - Need some clarity for final exam
I'm studying for my final exam, and these are the questions I got wrong on the mid term, could anyone help correct me? I know people are against giving answers for homework, so I must add that is now exam period for Universities, so this is only for my further understanding!
Thanks so much!

My attempts will be in this format

1)The pH of a 0.2M unknown base X OH is 8.15
a)Calculate [OH-]

10^(-8.15) = 7E-9M

b)What is the % disassociation

(7E-9/0.2M) x 100 = 3.5E-6%

c)What is the Kb of the base?

This one I don't even know where to begin..

2)Calculate the pH of the following solutions
a) 0.10M KCN, Ka (HCN) = 6.2E-10 (Hint: KCN is the salt of a weak acid and a strong base)

x = sqrt(Ka*c)
x = sqrt(6.2E-10 * 0.1) = 7.8E-6
-log(7.8E-6) = pH = 5.1

Welcome to Chemistry.SE! For this type of question, we would like you to help us help you. What progress have you made on these problems? What don't you understand? It's possible your wrong answer is a simple arithmetic mistake instead of a conceptual problem. As it stands, we don't like questions that seem to suggest we should do your work for you.

In addition to what @BenNorris said, in the future, please ask only one question per post. (You may ask more questions if they are closely related, though). You've already gotten an answer, but if you want this reopened, just tweak it so that you ask one question and show what you have tried, and I'll reopen it :)

1a) First use $pOH = 14 - pH$
Then use: $pOH = \log([OH^-])$
b) Simply use: $\dfrac{\text{dissociated}\ [OH^-]}{\text{original}\ [OH^-]}$
c) Use $K_b = \dfrac{[X^+][OH^-]}{[XOH]}$
2) Dissolving $KCN$ in water gives:
$\ce{KCN \rightarrow K^+ + CN^-}$
The following equilibrium reaction is thus formed
$\ce{CN^- + H2O \rightleftharpoons HCN + OH^-}$
Which is:
$K_b = \dfrac{[HCN][OH^-]}{[CN^-]}$
From the $K_a$, you can get the $K_b$ through $K_w\ =\ K_aK_b$ so that:
$\dfrac{K_w}{K_a} = \dfrac{[HCN][OH^-]}{[CN^-]}$
Let the amount of $CN^-$ that reacts to be $x\ M$, thus $[CN^-]$ becomes $0.1-x$ and you get $x\ M$ of both $[HCN]$ and $[OH^-]$. Solve for $x$ to get $[OH^-]$ and from there, $pH$ should be easy to calculate.
3) Use $Ksp = [Ba^{2+}][F^-]^2 = 1.7\times 10^{-6}$ and solve for $[Ba^{2+}]$.

Thank you so much! To clarify, my error on the first question was I used pH instead of pOH. So I did 10^-8.15, when I should have done 10^(- (14-8.15) )?

@BrettD Yes, since by taking the pH, you are actually finding the number of moles of $[H^+]$. You could still continue if you used the other equilibrium expression: $K_w = [H^+][OH^-] = 1.0\times10^{-14}$.

for 2) should be $6.2\times10^{-10} = \dfrac{[0.1-x]^2}{[x]}$

@DavePhD Oh my, that was long ago hmm... I'll opt for another way of doing that. Updated. Also had a typo in there.


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Using Arrhenius equation to calculate the ratio of two reaction rates

For two reactions,
$$\begin{align}
\ce{A ->[$k_1$] B} \tag{1} \\
\ce{A ->[$k_2$] C} \tag{2}
\end{align}$$
you have determined that the activation energies are $\pu{71.9 kJ/mol}$ for $(1)$ and $\pu{142.8 kJ/mol}$ for $(2)$. If the rate constants are equal at a temperature of $\pu{321 K}$, at what temperature will $k_1/k_2$ be equal to $2$?

I know I have to use the Arrhenius equation
$$k=A\exp\left(-\frac{E_\mathrm A}{RT}\right),$$
and I've tried solving for $T$ after filling in my data into the formula as ratios, ie $k_1/k_2$, $E_\mathrm{A}^{(1)}/E_\mathrm{A}^{(2)}$, etc, but I can't seem to get a coherent answer. Any guidance on figuring this out would be much appreciated.

To a first approximation, the activation energy of a single given reaction is a constant at all temperatures (see reasons here, although more precisely, there is some temperature variation). However, the activation energies for both reactions need not be the same.
If you consider both reactions at $\pu{321 K}$,
$$\begin{align}
\ln k_1 &= \ln k_2 \tag{1}\\
\ln A_1 - \frac{E_{\mathrm A}^{(1)}}{(\pu{321 K})R} &= \ln A_2 - \frac{E_\mathrm{A}^{(2)}}{(\pu{321 K})R} \tag{2}\\[4pt]
\ln A_1 - \ln A_2 &= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{(\pu{321 K})R} \tag{3} \\[4pt]
&= \frac{\pu{71.9 kJ mol-1} - \pu{142.8 kJ mol-1}}{(\pu{321 K})(\pu{8.31446 J K-1 mol-1})} \tag{3} \\[4pt]
&= -26.56 \tag{4}
\end{align}$$
Now, denote the desired temperature at which $k_1  = 2k_2$ as $T^*$. At this new temperature, we have
$$\begin{align}
\ln k_1 &= \ln 2 + \ln k_2 \tag{5}\\
\ln A_1 - \frac{E_{\mathrm A}^{(1)}}{T^*R} &= \ln 2 + \ln A_2 - \frac{E_\mathrm{A}^{(2)}}{T^*R} \tag{6}\\[4pt]
\ln A_1 - \ln A_2 - \ln 2 &= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{T^*R} \tag{7}
\end{align}$$
Here we can reuse the previously calculated value for $\ln A_1 - \ln A_2$, and we can also plug in the same activation energies:
$$\begin{align}
T^* &= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{R(\ln A_1 - \ln A_2 - \ln 2)} \tag{8} \\[4pt]
&= \frac{\pu{71.9 kJ mol-1} - \pu{142.8 kJ mol-1}}{(\pu{8.31446 J K-1 mol-1})(-26.56 - 0.693)} \tag{9} \\[4pt]
&= \pu{313 K}. \tag{10}
\end{align}$$

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


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Smallest nearest neighbor distance among elements
Of all the elements in periodic table, which one has smallest nearest neighbor distance? I tried searching on net, but could not find any reference. My earlier guess was one which goes into FCC with smallest unit cell dimension should be one. But I am not sure.
In an FCC, if unit cell dimension is a, then nearest neighbor distance will be a/sqrt(2) because nearest atom to an atom on the edge of the cube will be face atom. But this is for FCC. For SC of size a, nearest atom will be at a distance of a. So, for different elements, there will be different nearest neighbor distance based on its unit cell geometry and unit cell length. I want to know which combination of these result in shortest nearest neighbor distance.

Both FCC and HCP packings have nearest neighbours as close as possible (in nature at least, the simple cube is even closer but doesn't exist). Their nearest neighbour distances in terms of $a/R$ are $2\sqrt{2}$. Here $a$ is the length of a side of the unit cell and $R$ is the radius of the atom the cell consists of.
When you are looking for the smallest nearest neighbour distance this means that you are looking for the smallest $a$ in an FCC or HPC packing. You can translate this into: you are looking for the atom which forms an FCC or HPC packing and has the smallest value for $2\sqrt{2}R$ so for $R$.
Now, I don't know all the options but for example $\ce{Cu, Au, Ag}$ form FCC structures and $\ce{Zn}$ forms an HCP structure (see e.g. here). You can look up the atomic radii and find that Zinc has the smallest: $122$ pm and will thus have a nearest neighbour distance in the lattice of $345$ pm.

An FCC face center atom touches face center of other faces so how can we prove the distances to be a/√2.


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Balancing a chemical equation, but with an incorrect mole ratio
I'm wondering if it's possible to balance a chemical equation with an incorrect mole ratio.  I have the unbalanced equation $\ce{AgNO3 + Cu -> Ag + Cu(NO3)2}$, I have a $2:3$ ratio of $\ce{Cu}$ and $\ce{Ag}$ that I must plug in.  Is this possible, because it doesn't look like it.

The balanced reaction is $$\ce{2AgNO3 +Cu -> 2Ag +Cu(NO3)2}$$
Firstly, it's not necessary that all the reactants are depleted in the reaction.
There is the concept of a "limiting reagent" -- the reagent that gets depleted first and drives the extent of the reaction.
In this case, the limiting reagent is $\ce{AgNO3}$, and you will have half a unit of $\ce{Cu}$ left over.
In the general case, the limiting reagent need not be the one in the least quantity. For example, for $\ce{2A +B ->C +D}$ , if $\ce{A:B}$ is 3:2, then $\ce{A}$ is still the limiting reagent as it will get depleted first.

But I have to write an equation with the mole ratio of 2:3

@Someone: Not possible unless you modify the reaction.

The balanced reaction is this:
$$\ce{2AgNO3 +Cu -> 2Ag +Cu(NO3)2}$$
Which is normally what you would write and then just mention that your reactants ratio is different so you have a limiting species. However, if you really want the 3:2 ratio inside your the reaction equation you could write this:
$$\ce{3AgNO3 + 2Cu -> 3 Ag + 3/2 Cu(NO3)2 + Cu}$$
where I have just added equal amounts of Cu and Ag on the left and right and explicitly put in the copper that is left over. Of course by standard convention (leaving no fractions in the reaction equation) you would actually have to write:
$$\ce{6AgNO3 + 4Cu -> 6 Ag + 3 Cu(NO3)2 + 2Cu}$$
So now you have a reactants ratio $\ce{Ag:Cu}=3:2$ and you can see that this will result in left over copper, because the silvernitrate is the limiting species.

But I can't add 2 Cu to the other side.

@Someone : why not? if you put in 6 AgNo3 and 4 Cu that is what comes out: 2 Cu left over


text:
How can I know the colour change of this reaction?
After you react rotundone ($\ce{C_{15}H_{22}O}$) with water it loses all of the double bonds (except the carboxylic bond, $\ce{C=O}$) and gets two of the functional group $\ce{-OH}$ (alcohol). Two compounds are formed, one (A) where both the alcohol groups are tertiary and another (B) where one is tertiary and the other is primary.
I'm asked to state and explain the colour change of both A and B when they react separately with acidified potassium dichromate(VI) solution.
Now I can see that compound A would have no colour change because the $\ce{-OH}$ groups are tertiary so the potassium dichromate won't react or oxidise them.
I can also see that there will be a colour change on compound B because one of the $\ce{-OH}$ groups is primary, so the potassium dichromate will oxidise and react with it.
Here's my problem:
I'm expected to know that when compound B reacts, it changes from orange to green. How would I know that rotundone is orange in the first place? And what makes it go green?
I know that I'm expected to use intuition as we definitely aren't required to know the colour of rotundone.

Potassium dichromate (VI) is the species that turns from orange and to green when reduced. Most of the organic substances I came across for testing purposes are colourless, so I would assume that compound B would be as well.
It is just like you are expected to know the colour change of potassium manganate (VII) from purple to colourless when it is reduced.
Those two oxidants were used frequently in my labs, and I'm pretty sure you'll be using them often as well.
I hope this answers your question! :)

Yes thank you. Helps me now know that I need to learn a bit more. Although it's AS Level so we probably won't be using them much - did you assume I was an undergraduate?

@Olly_price I didn't assume anything. I just thought that these things were taught everywhere as I did A level and didn't go further than that ^^; (and that's a trend I've seen with other people I've answered concerning redox reactions). It would really help if you did some labs with those reactants. Seeing and using them leaves an impression which isn't easy to forget :)


text:
Can you carbonate a non aqueous liquid?
We have a home carbonator for makeing seltzer. It blows $\ce{CO2}$ under high pressure into water. Could I use it to carbonate corn oil? Is $\ce{CO2}$ soluble in a no aqueous solution?

There are some things human beings were never meant to carbonate.

@RichardTerrett: Unfortunately we could never get funding for the Large Soda Stream Collider, designed to carbonate fundamental fundamental particles.

Boldly to carbonate everything, which nobody carbonated before.....

$\ce{CO2}$, being non-polar (due to the 180 degree bond angles),  will probably be more soluble in solvents that are non-polar as well, although the quadrupole moment might cause higher solubility in polar solvents as well (thanks to @georg for this addition). Also temperature will play a significant role, the higher the temperature the lower the solubility.
These qualitative remarks however only take you so far. If you really want to know the solubility in a specific liquid you will have to look it up or, if that is not possible, test it yourself. In this paper they test the solubility of $\ce{CO2}$ in different types of crude oil and make predictions based on the structure of the oil.
I cannot find anything specific for corn oil, although the reverse process of dissolving palm oil in supercritical $\ce{CO2}$ might set you on the right track to find what you want to know.
Just a fun addition: don't ever try to carbonate milk as these guys do - it will get messy! (it is in Dutch, but you can still watch what happens). Click here an English video with similar footage

""CO2, being non-polar"" This is wrong! CO2 has a quadrupole moment, which enhances solubility in polar solvents.

@Georg - You are right that CO2 has a quadrupole moment, but that does not classify it as a polar molecule. Indeed, you are right, it would make it more soluble in polar solvents due to this. - However, a molecule is called non-polar if there are no positive or negative poles because $\ce{CO2}$ has its bonds exactly at 180 degrees the oxygen atom on the left pulls just as hard as the oxygen atom on the right thus leaving it without poles hence 'non-polar'. See e.g. http://chemistry.about.com/od/chemistryglossary/a/nonpolardef.htm

Come to think about it a bit more. In the formal physics definition a quadrupole moment would indeed mean that the molecule is polar. However, the standard definition used by chemists for polarity is that the dipole moment is 0 so that would make CO2 a non-polar molecule. See this post by @Manishearth for details http://physics.stackexchange.com/questions/59886/difference-between-non-polar-and-dipole-moment-vec-mu-0

$\ce{CO2}$ is also soluble in various organic liquids, but your mileage may vary
From an excellent comment by Greg E. that he didn't opt to turn into an answer:

CO2 is a non-polar molecule, so its solubility will be partially a function of the polarity of the solvent in question. Furthermore, solubility of gases is very sensitive to temperature. Solvation of a gas is entropically disfavored (lower degrees of freedom for the gas molecules), so at some point a further increase in temperature will take primacy over the heat of solution (enthalpy change due to solvation) and any further increase in temperature will actually result in more CO2 coming out of solution. Additionally, higher temperature will increase the vapor pressure of solution.


text:
Why is lead sulfide found in nature, whereas lead oxide is less common?
This question probably applies to other heavier metals as well.  The only rationalization I can figure out is that lead in the +2 oxidation state (which is most common) is a borderline soft acid and prefers to take electrons from a soft sulfur instead of trying to react with a hard oxygen.
Is this the right direction?  Thank you in advance.

Lead(II) and sulfide ions are large and polarizable, making them soft in the context of hard/soft acid/base theory (HSAB), which states that ions with similar hardness will form stronger bonds.  The oxide ion is hard and therefore has a tendency to form weaker interactions with Pb(II).
HSAB theory has been discussed on Chem.SE in this question and this answer.

@OscarLanzi I agree. It's been a while since I posted this answer, and have no recollection why I focused on Pb(IV), especially since Pb(II) would tend to be softer than Pb(IV).

I guess that a long time ago there was a lot of $\ce{H2S}$ in the atmosphere and then the $\ce{H2S}$ could have reacted with that lead oxide to form  $\ce{PbS}$?
On wikipedia it says (under Lead(II)oxide,reactions) :

The red and yellow forms of this material are related by a small change in enthalpy:
PbO(red) → PbO(yellow) ΔH = 1.6 kJ/mol
PbO is amphoteric, which means that it reacts with both acids and with bases. With acids,it forms salts of $\ce{Pb2+}$ via the intermediacy of oxoclusters such as $\ce{[Pb6O(OH)6]4+}$. With strong base, $\ce{PbO}$ dissolves to form plumbite(II) salts:[6]
$\ce{PbO + H2O + OH- -> [Pb(OH)3]-}$

It could have reacted with $\ce{H2S}$ following this reaction : (in water)
$\ce{Pb2+ + H2S -> PbS + 2 H+}$

Ores were/are formed deep in the earth crust. Oxygen has only a minor influence in case of ores brought in contact to air by mining. Exaples are cinnabar oxidized to elementary mercury in ores close to surface and oxidation of pyrite leading to acid effluents fron abandoned coal mines.


text:
How to calculate the pH of a buffer after HCl was added?
A buffer is made from $50~\mathrm{mL}$ of $1.0~\mathrm{M}$ benzoic acid, $K_a = 6.3\cdot10^{-5}$, and $50~\mathrm{mL}$ of $1.0~\mathrm{M}$ sodium benzoate.
a) Calculate the pH of this buffer
For this do I simply use $-\log(6.3\cdot 10^{-5})$?
b)Calculate the new pH when $0.010~\mathrm{M}$ $\ce{HCl}$ is added to $100~\mathrm{mL}$ of this buffer.
This one I do not understand.

For (a), the Henderson-Hasselbalch equation, $$\ce{pH} = \mathrm{p}K_a + \log([\ce{A^{-}}]/[\ce{HA}]),$$ comes in handy. Because your molarities and volumes of the acid and its conjugate base are equal, this indeed reduces to simply $\ce{pH} = -\log(6.3 \cdot 10^{-5})$.
For (b), the volume of $\ce{HCl}$ added is required, as the concentration of the solution alone is not sufficient information. The standard practice is to assume that $\ce{HCl}$ (being a strong acid) reacts fully with the conjugate base in your buffer solution to produce an equal amount of the conjugate acid (i.e., if $x$ moles of $\ce{A^{-}}$ are consumed by $\ce{HCl}$, $x$ moles of conjugate acid $\ce{HA}$ are produced). Therefore, you can use the Henderson-Hasselbalch equation to recalculate the $\ce{pH}$, subtracting the moles of $\ce{HCl}$ added from your conjugate base, and adding that some number of moles to your conjugate acid.

The information I gave was all the info given on the midterm we should be able to answer it with that info.

@BrettD, you need to know the volume of HCl added, not just its concentration. The 0.010M merely expresses that the solution contains 0.010 moles HCl per 1 liter volume. We need to know the volume added at that concentration to determine how many actual moles of HCl were delivered.

@GregE. The guy was called Hasselbalch ;)

@Martin, indeed he was, thanks for the correction.


text:
Why are the oxide materials insulating?
Why are the oxides usually insulating?  It would be nice to get an explanation with chemistry terms (e.g., electronegativity, electron affinity, etc...) and a different explanation involving the language of band structure (i.e. energy gaps).

Without defining the temperature which you have in mind, the question is meaningless. Oxides are not metals, in the first place, and as non-metals most of them have a rather big gap in electron band. So, most of them are isolators at room temperature and semiconductors at appropriate themperatures or when doped with some foreign elements.

He didn't even say whether he meant thermally or electrically insulating- which one?

I meant electrically insulating and at room temperature.  Perhaps my question would be best worded as, "Why do oxides tend to have large band gaps?"

I don't see how would one able correctly describe band properties without band structure, based on some chemical concept rather limited. Maybe some simple, and somewhat wrong explanation like "they have ionic bonds due to large electronegativity difference between metal and oxygen". Note that there are small bandgap oxides, semi-metals, even metals. Some, like $ReO_3$ are highly conducting that cannot be explained by electronegativity.

To have conductuing valence zone requires to have infinite amount of ovarlapping valence orbitals with lesser amount of electrons. This is not usually possible for oxides, because they contain direct contacts O-O, but not M-M. O-atoms in oxides has completely filled electronic shell. Pure metals has electronic shells with much less electrons then available orbitals so they can allow electrons to move freely. This is the difference.

I am aware our constrains here, but I think it would be better to use at least the concept of unit cell, so the explanation can go beyond "infinite number of orbitals" and such. I hope you noticed that even a single atom has infinite number of orbitals, and that the reason of insulation is NOT the lack of overlap in most insulating materials.


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Isopropyl alcohol forming peroxides
I read in many MSDS that Isopropyl alcohol tends to form peroxides (by reacting to air OR light) which may explode when they reach a given concentration.
For example: http://nj.gov/health/eoh/rtkweb/documents/fs/1076.pdf or http://www.hvchemical.com/msds/isal-70.htm as well as many others.
I own a few liters of isopropyl alcohol (a few bottles) and I'm getting concerned.
It came in standard transparent (polypropylene I suppose) bottles with a plastic bag wrapped around it (I suppose to prevent contact with air). I only keep one bottle ready to use, all the others are as when I bought them (with plastic bag wrapped around it), stored in a relatively cool (<= 25°C) dark area of my house.
This is at home and I couldn't find any fire resistant storage cabinet in my area.
Now, I'm not sure if I understand the datasheet correctly, but as far as I understood the problem lies with the opened bottle (which I open and close regularly) and not with the air-tight never opened other bottles, right?
Therefore a simple solution may be to say: after one year I empty the opened bottle, trash it away and buy another one. Should this be enough?

Is it pure isopropanol? For example, most "rubbing alcohol" that you would by at the supermarket or pharmacy is 30-80% (the remainder water). The presence of water decreases the peroxide danger.

Pure. Isopropanol >= 99.9%.

I think this may be an overreaction.  First, MSDS are designed for workers working with industrial quantities in industrial conditions.  Second, the danger comes from concentration of the peroxide from evaporation or, more typically, distillation.  Small quantities of peroxides in solution are not usually an explosion hazard.  Third, I have never seen isopropanol listed in lists of hazardous peroxide prone chemicals as used by lab safety departments (ethers are the typical risk here).  The closest I have seen is sec-butanol, which is the secondary alcohol from butanol, and I've only seen this listed once, in a list that seemed much more cautious than some other lists.  Isopropanol is the secondary alcohol from propanol.  Given your storage, cool and dark, I think you are fine.  Remember that 91% isopropanol is sold in drugstores, though these may contain stabilizers and you larger bottles may not.
But having said all this, Lighthart's brief answer is correct though probably unlikely.  If you DO see any crystals, don't even try to open the bottle (crystals in the threads are the most common source of explosion).  But again, seems unlikely.
If you are still concerned, there are peroxide test strips you can buy that are very easy to use and can tell you what concentration of peroxides are present.

I bought my isopropanol directly from a chemical supply. It's isopropanol > 99.9%. Would your remark still apply?

Sure, and Ben Norris' answer essentially agrees with mine. Peroxides did form from ancient bottle after concentration by distillation. And he also recommended test strips.  So at the moment, I'd say you're fine.

Please pardon me if I sounded alarmist.  Chemists work all day to make awesome crystals, and most are curious when some high quality crystals actually form.  It is natural to investigate closely

Regular peroxide test strips are reputedly insensitive to the APEX/TATP https://en.wikipedia.org/wiki/Acetone_peroxide that is supposed to form https://www.uksachsen.de/aktuelles/detailansicht/moegliche-bildung-von-explosiven-peroxiden-in-ueberlagertem-isopropanol-propan-2-ol

Isopropanol is not one of the notorious peroxide formers, but it will eventually do so in the presence of oxygen (with or without light - the process is autocatalytic).
When I was a graduate student TA in an organic chemistry class, isopropanol was a distillation "unknown". However, the "unknown" bottle had probably been in existence for 20 years and was periodically refilled. Some of the isopropanol had been around in the presence of air for a long time. While the bottle never developed the crystals typical of ether peroxides, peroxides did form. Our proof was the explosion of three distillations using the stuff. Fortunately no one was injured.
There is a qualitative test for the presence of peroxides. Buy yourself some KI Starch test paper. KI reacts with oxidants to produce $\ce{KI3}$, which binds with starch to form a blue-purple complex. This chemistry is also the basis of the famous iodine clock reaction.
$$\ce{ROOH + 2H+ + 3I- -> ROH + H2O + I3-}$$
$$\ce{I3- + starch -> [I3-starch]^-} \ \ \text{(blue)}$$

I assume therefore that never opened bottles are not a concern at all, right? Once opened a bottle lasts about a year so it should be fine from what you've been saying, right?

The unopened bottle has headspace. Unless otherwise labeled, that headspace is probably air. Unopened bottles can be dangerous, too. When in doubt, dispose of it.

There is a very small headspace. But if they're only 1-5 years old?

Not quite enough.  If you ever see crystals in the bottle, year old or otherwise, evacuate and call the bomb squad.

It sounds alarmist; But I assume that you just left out that it is improbable that peroxide crystals form. (That's said elsewhere.) It is not alarmist to say that in case peroxide crystals did form, it's time to be alarmed.

I give you my sincerest hopes you never see the crystals, yes.


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What is the multiplicity of this signal in the ¹H NMR of glutamine in D₂O
The 2 Hydrogens near the carbonyl (the lower 2 $\ce{H}$s are not bonded to the nitrogen) should be the most deshielded so I think they appear as the rightmost peak in the NMR.  Are these 2 $\ce{H}$s a doublet of doublets?  I think this because they are split by 2 different $\ce{H}$s.

That's glutamine and not tryphophane in your molecular formula

sorry, I meant glutamine

You're thinking about deshielding the wrong way around, the most deshielded signal appears at the highers ppm value, so to the left. In this case it would be the signal at around 3.8 ppm, as the signal at 4.7 ppm is residual water.
The two signals at the right are multiplets, if you take a look they're obviously caused by rather complicated coupling patterns that you can't easily resolve. The reason for that is that the protons you see in this spectrum are not magnetically equivalent, even the CH2 protons are not magnetically equivalent. This leads to the complicated couplings you can observe here.

thank you.  Also, is it true that the H on the dash corresponds to the triplet at about 3.7ppm?

@user176105 That one is the H-alpha and it is indeed at 3.7 ppm. And it is a doublet of doublets with a very similar coupling constant, which looks like a triplet.

thanks again.  just to clarify, if i write doublet of doublets on the beta hydrogens that would be wrong?

The signal at ~3.8ppm is looks like a triplet[t], C-alpha
The signal at ~2.5ppm looks like a triplet of doublets [td], C-gamma.
The signal at ~2.2ppm looks like a doublet of doublet of doublets [ddd], c-beta.
The 2.2 protons are assigned 2.2 because they are the farthest from carbonyls and thus the least deshielded by these functions.
The ddd signal indicates clearly that both protons on C-beta are non-equivalent, because you need 2 3-bond couplings to make a ddd when combined with the 2-bond geminal coupling.
Given the C-beta protons are non-equivalent, the triplet is actually a superimposed doublet of doublets, and the td is actually a similarly superimposed double of doublets [the superimposed-apparent triplet pair and then...] of doublets.

thanks.  so even though the 3.8ppm peak appears as a triplet should I still write doublet of doublets?

In my research days I wrote that as (dd [apparent t]) but you should confer with your faculty member about her/his preference.  In my teaching days I tended to accept either asnwer.


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Mol file of Morphine
I am using the Morphine mol file and I'm attempting to create a matrix which contains a table of the atoms and how they bond similar to a ball and stick format.
In the mol file bond block listed bellow what do the type place holders mean?
  7  1  1  0  0  0  0
  1  2  1  0  0  0  0
  2 12  1  0  0  0  0
  1  3  2  0  0  0  0
  3 14  1  0  0  0  0
  2  4  2  0  0  0  0
  4 18  1  0  0  0  0
  3  5  1  0  0  0  0
  4  6  1  0  0  0  0
  5  6  2  0  0  0  0
  7  8  1  0  0  0  0
  7  9  1  0  0  0  0
  7 22  1  1  0  0  0
  8 10  1  0  0  0  0
  8 11  1  0  0  0  0
  8 20  1  1  0  0  0
  9 12  1  6  0  0  0
  9 13  1  0  0  0  0
 10 15  1  1  0  0  0
 10 14  1  0  0  0  0
 11 16  2  0  0  0  0
 13 17  1  6  0  0  0
 13 16  1  0  0  0  0
 15 19  1  0  0  0  0
 21 15  1  0  0  0  0
 22 21  1  0  0  0  0

This is the format I'm trying to create but not all the atoms have the full amount of bonds. What is wrong?
C   7   2   3
C   1   12  4   12
C   1   14  5   5
C   2   18  6   18
C   3   6   3   6
C   4   5   4   5
C   1   8   9   22
C   7   10  11  20
C   7   12  13  12
C   8   15  14  15
C   8   16      16
O   2   9   2   9
C   9   17  16  16
C   3   10      10
N   10  19  21
C   11  13  11  13
O   13
O   4   4
C   15
H   8
C   15  22  22  22
C   7   21  21  21

The bond block has the number of the atoms followed by the number of bonds mol files support up to three bonds. Not all the atoms are supposed to have all their bonds filled.

Ah, so you had a problem and solved it yourself? That's great! (also, welcome to Chemistry.SE). It would be nice if you elaborated your answer though.. :)

The self-accepted answer given here is wrong.
Each line in a bond block of a MOL file represents one bond between two atoms.
A line in a bond block has the following compact format:
111222tttsssxxxrrrccc
With other words, each entry (=different number or letter) in such a line can take up to three spaces. If it needs less. e.g. when the atom number is smaller than 100, the data is right-aligned.
111 is the number of the first atom
222 is the number of the second atom
These atom numbers refer to the position of the atom in the atom block in the same file, where the cartesian coordinates and the atom type (element) are given.
ttt is the number decoding the bond type: 1=single, 2=double, 3=triple, etc.
sss describes the stereochemistry, 1 is a single bond pointing up, 6 is a single bond pointing down, etc.
The latter are used in the flat 2D MOL file of morphine cited in the question, e.g. for the only hydrogen atom in the file: a bridgehead H-atom.


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Common Elemental Termolecular Reactions
Are there any common elemental (single-step) reactions with three reactants?
Everything I've found either says that they're very rare, or can be broken down into two bimolecular steps with combined reaction rates far greater than the termolecular step.
What reactions require an indivisible termolecular step?

Termolecular reactions almost never happen.  The chance of three molecules "colliding" at the same instant is so small as to be ignorable.
On the other hand there are "almost" termolecular reactions when one molecule hits another and "sticks" to it, and then a third molecule hits the two that are stuck together.  This most often happens when one of the molecules is part of a solid.  But this is really two sequential bimolecular reactions.


text:
What is the freezing point of sulfuric battery acid?
As we all know, the freezing point of absolute sulfuric acid is 10 °C. The average lead-acid battery (sulfuric) uses ~30% sulfuric acid, and has a freezing point much lower than 10 °C.
What constituents contribute to this effect? What is the diluent used in these batteries? What would the freezing point be?

Welcome to Chemistry.SE. Could you link to a reference for the freezing point of sulfuric acid. I must not be in the "we all" you mention.

Maybe this wiki article can give you some insight?

Freezing point of sulfuric acid: https://www.google.ca/search?q=freezing+point+sulfuric+acid

@BenNorris You and me both.

Batteries have water-sulfuric acid solutions.
There freezing point of water-sulfuric acid solutions varies as a function of concentration in an unusually complex way, having plural local minima and maxima.
This is because the solid has eight identified phases:
$\ce{H2O}$
$\ce{H2SO4 * 6 H2O}$
$\ce{H2SO4 * 4 H2O}$
$\ce{H2SO4* 3 H2O}$
$\ce{H2SO4 * 2 H2O}$
$\ce{H2SO4 * H2O}$
$\ce{H2SO4}$
$\ce{H2S2O7}$
with mixtures of the various phases reaching local minimum freezing points at eutectic points.
The freezing point can be as low as -71 °C near 31% (on an $\ce{SO3}$ basis) and as high as 35 °C near 90%.
See [1] or Fig. 2 from [2]:

References

Gable, C. M.; Betz, H. F.; Maron, S. H. Phase Equilibria of the System Sulfur Trioxide-Water. Journal of the American Chemical Society 1950, 72 (4), 1445–1448. https://doi.org/10.1021/ja01160a005.
Ohtake, T. Freezing Points of H2SO4 Aqueous Solutions and Formation of Stratospheric Ice Clouds. Tellus B 1993, 45 (2), 138–144. https://doi.org/10.1034/j.1600-0889.1993.t01-1-00006.x.

Also see: https://modernsurvivalblog.com/alternative-energy/lead-acid-batteries-state-of-charge-versus-freezing-temperature/

The general effect you're talking about is freezing point depression, one of the basic colligative properties. There are several flavors of lead-acid batteries, so there are different additives that can modulate the freezing point.  The actual composition of the electrolyte varies with the charge state of the battery, so that means the freezing point also varies with how charged it is. I think all of the lead-acid batteries use water as a diluent (it will always be present as it's a species in the electrochemical equilibrium.) From the Wikipedia article on "lead-acid battery":

Due to the freezing-point depression of the electrolyte, as the battery discharges and the concentration of sulfuric acid decreases, the electrolyte is more likely to freeze during winter weather.

What constituents contribute to this effect?

This phenomenon in general is called "Freezing-point depression", and is the process in which adding a solute to a solvent decreases the freezing point of the solvent. It can all manner of solutions. The freezing point depression depends solely on the concentration of solute particles, not on their individual identities, and is thus called a colligative property.
Now, why does this happen? The presence of a solute results in the lowering of chemical potential of solvent; since the lowering occurs even in ideal solutions where enthalpy of mixing is zero (and also since it doesn't depend on the properties of the solute) one can probably guess that it is an entropic effect.
A rough qualitative argument can be made as follows:
We know that liquids possess greater entropy than solids, thus when we freeze something we are going from a "high entropy state" to a "low entropy state".  The presence of a solute makes an additional contribution to the entropy of the liquid, and thus this enhanced molecular randomness opposes the tendency to freeze, and consequently we need a lower temperature to freeze.
As an aside, a similar discussion can be applied to the elevation of boiling points of solutions (compared to pure solvents)

What is the diluent used in these batteries?

Typically, it is plain old water

What would the freezing point be?

This is perhaps the most difficult question to answer. If one were to do such a calculation, there are least three complications I can think of.
First, most chemistry textbooks give a simple relation to calculate the depression of freezing point, namely $\Delta T_f = ik_fb$, however, this simple empirical relation just holds for dilute solutions. A more complicated treatment is necessary to deal with concentrated, non-ideal solutions.
The second problem is that the concentration of the electrolyte itself changes during the charge-discharge cycle.
And finally, as @DavePhD has pointed out, the freezing point of water-sulfuric acid mixtures doesn't follow a simple trend.


text:
Ksp of Manganese Phosphate
I can't find the $K_{sp}$ value for Manganese Phosphate ($\ce{Mn_3(PO_4)_2}$) anywhere online. Has it been measured? What is it?

Well, the MSDS entry tells me that the solubility is 50%. Converting, we get 0.5 g per gram of water, or 0.0005 g per liter of water. This is $s=1.4\times 10^{-6}$ moles per liter of water.
$K_{sp}=(3s)^3\times(2s)^2=6.14\times 10^{-27} \rm M^5 $

We should get 0.5 g of manganese phosphate per gram of solution (the remaining 0.5 gram is water). Per liter, approximately 500 gram (.5 * 1000) is manganese phosphate and not 0.0005 g (.5 / 1000) - there are 1000 g of water in 1 L of water.


text:
How different are molecule with same connectivity
Let us represent molecules by graphs (atoms are vertices and bonds are edges).
Let $A$ be the adjacency matrix of the molecular graph. $A_{ij} = b$ where $b$ is either binary: 0 when atom i and j are not bonded else 1 (Repr. 1). Or, further we may take b from the set $\{0,1,2,3\}$ depending on whether the no bond (0) or single(1) double(2) or triple (3) bond is present (Rper. 2). Here were are not distinguishing the atoms - for our purose they represent a node in a graph.
Let us consider only non-isomers and molecules with at least 10 atoms. How different will be two molecules with same graphs (with respect to Repr. 1 or Repr. 2) in terms of their properties or particularly about drug-like properties. My interest is to know how faithfully the topological indices characterizes the molecular activity. Specific or general examples of pairs of  non-isomer molecules with same graph but very different property could be useful. It would be helpful if I can find enough examples to convince myself about the usefulness or otherwise of topological indices which are calculated from the grpahical representation outlined above. Topological indices are popular in drug discovery research, but I am not very much convinced about their utility since most the results published are not statistically significant (use of very small data size - from 15 to some hundreds molecules, among other weakness) nor rigorously validated.
Thank you.

In my experience, calculated parameters like topological indices are not viewed as significant if considered alone. It's a strategy for finding trends or correlations and, subsequently, candidates from  large volumes of chemical structures. Many studies will publish trends they find in a target chemical groups which are not meant to apply to every chemical group.
An example of how all these seemly useless data is useful - see the Merck Kaggle Challenge.
https://www.kaggle.com/c/MerckActivity
http://blog.kaggle.com/2012/11/01/deep-learning-how-i-did-it-merck-1st-place-interview/
Statistical significance is harder to comment on. I work with between 8 and 20 million data points and the idea of how to define statistical significance is evolving in cheminformatics as it evolves in big data analytics. It gets much more complicated because standard methods, such as the t-test, fail.
NOTE
I just realized there might be some fundamental confusion. Mol files include atomic information as do most graph based chemical representations. They're stored as properties of the nodes. An algorithm can choose to use or ignore this information. Not all topological parameters ignore atomic information.

Carvone, $\ce{C_10H_14_O}$:  R-(–)-Carvone is spearmint, S-(+)-carvone is caraway.  Are the stereoisomers' graphs different?  Quantitative indices generate a paradox that well-managed research fairs poorly in basic discovery (absent insubordination).
In counterbalance, a single failed reaction is a setback but a million failed reactions are a combinatorial library.


text:
Iodine clock with moving bands
Has anyone seen a version of the iodine clock reaction where instead of the system oscillating between clear and blue it shows blue bands travelling down the vessel?
I ask because someone not unknown to this site has made a version of the experiment with such moving bands and I was wondering if it was worth sending a write up of it to the Journal of Chemical Education or some similar journal.
Clarification:
Apologies if I was unclear, I am asking the phenomenon has been published before or if it is unpublished but already well known.
I know the BZ reaction can produce all sorts of patterns, but I have never seen travelling bands in the iodine clock. A quick Google has found no mention of the phenomenon. I am not asking "is it worth publishing", which I believe is against site rules anyway. I'm just asking about prior art.

Uhhh... I'm not sure what you're saying or what your question is, exactly. Did you get that reaction to run and would like to know if it's worth writing a paper on it? Did you see it and would like to do a writeup for the researcher who got the reaction to work like that? Did you hear someone did it and would like to write a paper without even having seen it yet? Or something else entirely? Also, who's "someone not unknown to this site"? :)

@TanithRosenbaum: He means me ;-) I basically had modified the Briggs–Rauscher oscillatory reaction for fun, and it resulted in cascading bands in a beaker. John suggested I publish it in JCE; I was skeptical that it was big enough to be published.

@ManishEarth: actually, now I think about it, if all you did was add layers of the reagents prepared at successively later times this isn't very interesting. I was assuming the horizontal banding arose spontaneously or at least without excessive manual manipulation.

@JohnRennie: Well, it wasn't excessive manual manipulation. I let the normal reaction take place fully. Whatever was left, I poured some selected reactants on in a certain order (quickly, but carefully). That was all the manipulating involved, the bands happened automatically. There is no manipulation during the entire duration of the video :)

@ManishEarth: well if you can reproduce it I reckon chemistry teachers across the world would like the recipe :-)

I gotta say I would love to have a worked out procedure for this at my disposal. Know a few chem teachers personally, and I'm rather certain most would positively love it. So I'd say go for it. Besides, worst that can happen is that the journal or the peers say "no". No biggie, then you send it to another one. ;-) It's definitely worth publishing IMHO.

Whilst not specifically an iodine clock reaction, Belousov-Zhabotinski reactions are oscillatory and can produce moving bands, spirals and concentric circles of colour. This works best in thin films (limiting convection) as the top commenter on the linked video points out.
There's a great thesis by Peter Hantz which looks at these kinds of reactions in detail (the first bit is in french, the rest is in english) and specifically covers Liesegang banding, where an oscillating chemical reaction in a mobile phase causes bands of precipitation in a gel column. As Hantz points out on page 22, the gel is required to prevent convection from destroying the bands, however the viscosity of the gel also freezes the precipitate in place as it forms. This would seem to be close to what you are interested in, however.
Here's a picture of Liesegang bands in test tubes from wikipedia:

Ah, but the Belousov-Zhabotinski are supposed to be hard to get right (not something you'd do in a first year undergrad lab). Any idea if a modified version of the Briggs-Rauscher that creates cascading bands has been published before? If not, is it worth publishing in an educational journal like the JCE? (it's something quite easy to do in the lab, and quite fun once you get it. Plus you can try to make it faster.)

@ManishEarth - I'm not sure. If it's a novel reaction I would say it's worth publishing or at the very least blogging. I was thinking that a Briggs-Rauscher reaction in viscous media (for instance, K-Y Jelly) or a beaker full of polyacrylamide water beads with reagents injected at different points might produce some neat effects.

I didn't use any special media, just the normal reagents (not a novel reaction per se, it's just the process of mixing the reagents). But yeah, sounds like something nice to try :) Not sure if it would make any difference, though -- there's nothing moving in the oscillatory reaction, you just have the impression of motion (like Christmas lights). Then again, I can see how it may make a difference in the initial stages.

@ManishEarth - I understand that, but playing with viscosity and percolation may provide new opportunities for messing with the phase of the oscillation as a function of position as well as creating concentration gradients of reactants, which should affect the rate of oscillation as a function of position.

Ah, right, didn't think of that. Good insight! :)

@RichardTerrett: thanks :-) I knew about the BZ reaction but I'd never heard of the phenomenon in the iodine clock.

@JohnRennie - I think that Liesegang banding is broadly analogous to the phenomenon Manish has described in the iodine clock, however if moving fronts are occurring in fairly inviscid media rather than a gel, that's pretty neat and worth sharing in some way. Also, JohnRennie and ManishEarth you should check out this - it's an art project by Dominik Strzelec modelling a BZ type reaction in 3D. Not sure on the specifics.


text:
Why are elements on the right side of the periodic table nonmetallic and gas at room temperature?
Elements on the left side of the periodic table tend to be solid and metallic, elements on the right side of the periodic table are nonmetal and tend to be gases at room temperature, and the semi-metals are in between.  Why is that?  Why are the elements on the right side nonmetallic and in a gas phase?

1st, bromine, iodine and astat arent gases, are they? 2nd the hard core "Reason" is that Meyer and Mendelejew decided to start with the alkalies on the left. (of course if we were writing from left to right like arabs...) The periodic scheme kind of cyclic.

The elements on the left-side of the periodic table are relatively electron deficient (i.e., they have few valence electrons), and due to their comparatively low effective nuclear charges (the net positive charge of the protons minus the shielding core electrons below the valence level), their electrostatic hold on these electrons are weak. Consequently, the electrons of metal atoms tend to be highly delocalized, not forming localized covalent bonds or possessing tightly held lone pairs. In metals, the electrons are effectively shared between many atoms, loosely bound in what is often described as an electron "sea," and the atoms of metals arrange to achieve maximally efficient packing in a crystal lattice, through which the electrons (particularly the most extensively delocalized electrons in the s and p orbitals at the highest main energy level) move freely. The tight packing of the crystal lattice accounts for the density and hardness of metals, while the extensive delocalization of electrons partially explains their malleability and conductivity.
Elements further right on the period table, on the other hand, have higher effective nuclear charges and stabilize electrons more effectively, leading to localized covalent bonding and the formation of molecules (vs. the ionic bonding and crystal lattices of metals). Electrostatic intermolecular forces of attraction are generally weaker than ionic bonding or delocalized metallic bonding, so non-metals and molecular compounds are more likely to exist in the liquid or gaseous phase at comparatively low temperatures than metals or ionic compounds.

Thanks!  I'm a physicist, so I usually think of metallic and nonmetallic behavior in terms of band structure and band gaps, but this helps too!


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Where do the rate laws for the Chlorine Dioxide-Iodine-Malonic Acid reaction come from?
After reading a section on this particular oscillatory reaction, something started bugging me about it.
However, let me lay out the basis of my question.  From what I've researched, Lengyel and Epstein proposed a model that captures the overall behavior of the system. There are a few different versions of the model in regard to the dimensions, but this seems to be the most popular:
$\ce{MA+I_2 \rightarrow IMA + I^- + H^+;\hspace{1mm} \frac{d[I_2]}{dt} = -\frac{k_{1a}[MA][I_2]}{k_{1b}+[I_2]}}$
$\ce{ClO_2 + I^2 \rightarrow ClO_2^+ + \frac{1}{2}I_2; \hspace{1mm}\frac{d[CLO_2]}{dt} = -k_2[CLO_2][I^-]}$
$\ce{ClO_2^- + 4I^- + 4H^+ \rightarrow Cl^- + 2I_2 + 2H_2O}$
$\ce{\frac{d[CLO_2^-]}{dt} = -k_{3a}[ClO_2^-][I^-][H^+]-k_{3b}[CLO_2^-][I_2]\frac{[I^-]}{u+[I^-]^2}}$
Personally, I am much more familiar with the mathematics than the chemistry.  First, are these rate laws empirical or were they derived from physical laws? Pardon me if this is a naive question, but I am not as familiar with chemistry than the other hard sciences. In basic cases, rate laws are a straight multiplication of the concentrations of the present species. However, that neglects situations involving, for example, autocatalycity.  My suspicion is that it is a mixture of processes that I'm not familiar with and simple observation.
Obviously, these reactions don't violate the laws of thermodynamics. My assumption is that the overall free energy of the system continues to decline - meaning that the system eventually reaches equilibrium.  Unfortunately, I've been at a loss to find good sources on oscillatory chemical reactions.

From what I know the rate law of a reaction can only be determined experimentally. I'm not sure if there are any exceptions.

@kaliaden You raise a good point. Rate laws can only be determined experimentally.  What I should have said is that the general framework of rate laws may be determined before experiments. For instance, A+B->C (assuming an elementary step) rate law has the general form of r=k[A]^a[B]^b. "k" can only be determined by experiment - thus forming the full rate law. That said, my question is really if a general framework was developed empirically, or there is a an principle forming a basis for the development of the rate laws in my post.

I do not understand what Your question is. Do You know it Yourself?

@Georg "First, are these rate laws empirical or were they derived from physical laws?" And second: how is the reaction described thermodynamically? (the last paragraph).

authors: Алексей Владышевский