text:
Which ions do not react with each other in a solution?

Which ions do not react with each other in solution?
A. $Ba^{2+},SO_4^{2-},NO_3^-$
B. $Cu^{2+},SO_4^{2-},OH^-$
C. $Ag^+,Cl^-,H^+$
D. $K^+,Fe^{3+},Br^-$

I have no idea why the answer is D. Please help.

Do you have any other information to draw upon as to why you think it might not be some of the other choices?

The answer by @Daniel is closest to the truth. Each of A, B, and C will produce an insoluble participate, which will fall out of solution, as he notes.  In answer D, bromide ion forms an ionic compound with both potassium ion and ferric ion, which are soluble and so nothing precipitates out of solution.

I believe you mean precipitate?

@jonsca http://chat.stackexchange.com/transcript/message/9912229#9912229

I believe the question is slightly misleading, non of the ions react however some do precipitate.
BaSO4, Cu(OH)2 , and AgCl will precipitate in A, B, and C respectively.

How is Ag$^+$ + Cl$^-$ $\to$ AgCl not a reaction?

The question (and its choices) are bogus. One can make species (and solutions) of all of these combinations of cations with anions.


text:
How to solve this contradiction?
I am getting confused about this the more problems I do on it. My understanding was that only strong acids and strong bases will react to produce water and a salt. Something like this:
Molecular Eq: $\ce{HCl(aq) + NaOH(aq) \rightarrow H2O(l) + NaCl}$
Net Ionic Eq: $\ce{H+(aq) +OH- (aq) \rightarrow H_2O(l)}$
Makes perfect sense, both the acid and base break apart and combine with each other. But then I ran across something like this:
Molecular Eq: $\ce{2 CH3CO2H (aq) + Ba(OH)2 (aq) → Ba(CH3CO2)2 (aq) + 2 H2O (l)}$
Net Ionic Eq: $\ce{2CH3CO2H (aq) + 2OH- (aq) → 2CH3CO2- (aq) + 2H2O (l)} $
The net ionic eq tells me that $\ce{CH3CO2H}$ will not break apart, but the molecular equation tells me that they will break apart and form a salt with $\ce{Ba}$.
I'm so confused by this. Why are they giving two opposing pieces of information? Am I missing something here?

The idea behind this is that a weak acid like $\ce{CH_3CO_2H}$ will break apart, but won't break apart completely. If you have some $\ce{CH_3CO_2H}$ in water, most of it will remain as $\ce{CH_3CO_2H}$, so in the net ionic equation it is written whole, but some will separate into $\ce{CH_3CO_2^-}$ and $\ce{H^+}$ ions. This relationship is quantified by the $K_a$ of the acid. Essentially, the dissociation reaction $\ce{CH_3CO_2H -> CH_3CO_2^{-}  + H^+}$goes both ways: $\ce{CH_3CO_2H}$ decomposes into $\ce{CH_3CO_2^-}$ and $\ce{H^+}$, and $\ce{CH_3CO_2^-}$ reacts with $\ce{H^+}$ for form $\ce{CH_3CO_2H}$. Since the forward reaction happens at a rate proportional to the concentration of $\ce{CH_3CO_2H}$, and the backwards reaction happens at a rate proportional to the product of the concentrations of $\ce{CH_3CO_2^-}$ and $\ce{H^+}$, eventually the ratio $\frac{[\ce{CH_3CO_2^-}][\ce{H^+}]}{[\ce{CH_3CO_2H}]}$ becomes constant, and we call this constant the $K_a$ of the $\ce{CH_3CO_2H}$. For $\ce{CH_3CO_2H}, K_a=1.7\times 10^{-5}$.
Now its time to connect all this back to your question. When you add $\ce{Ba(OH)_2}$ to a solution of $\ce{CH_3CO_2H}$, the $\ce{OH^-}$ ions present from the dissociation of $\ce{Ba(OH)_2}$ react with the $\ce{H^+}$ ions from the (partial) dissociation of $\ce{CH_3CO_2H}$. This lowers the concentration of $\ce{H^+}$, and so more $\ce{CH_3CO_2H}$ dissociates to keep the ratio of concentrations constant at the $K_a$. Therefore, in the presence of enough $\ce{Ba(OH)_2}$, $\ce{CH_3CO_2H}$ dissociates almost completely, even though, on its own, $\ce{CH_3CO_2H}$ dissociates only to a very small extent.

I wouldn't say that the acetic acid dissociates almost completely, but that enough acetic acid dissociates to equal the positive charges of the barium (there may be an excess of acetic acid, which the excess will mainly be undissociated).

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How are arsenic compounds removed from drinking water?
I used to live near a volcano where the groundwater contains naturally occurring arsenic compound. Since $\ce{As2O3}$, $\ce{As2S3}$, and $\ce{As4S4}$ are predominantly occurring in drinking water, what are the processes for removing them?

You may find what you need (happy accident!) in the body of this question http://chemistry.stackexchange.com/questions/472/what-are-the-mechanisms-for-capture-of-ceasiii-and-ceasv-by-magnet.  I don't know that it's a duplicate, I'll let the community decide that.

(but I wholeheartedly agree with Eric's sentiment below, if you are in fact using this as a method to purify drinking water and not just as a lab exercise, consult an expert in this area for your own safety)

The US Environmental Protection Agency (EPA) has a page dedicated to arsenic in drinking water.  It contains a section that suggests ways to remove it from water.
Disclaimer: I have no idea if the arsenic removal methods on the linked page actually work.  You should consult someone trained in the science of water purification for definitive information.

Could you give me the chemical equations for precipitating those Arsenic compounds for my experiments?

I'm sorry, I am not qualified to offer specific suggestions on how to remove toxins thus making water potable.

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What is the average chain length of a glutaraldehyde crosslink between primary amines?
Imagine I'm performing a primary amine to primary amine (i.e. PA to PA) crosslinking reaction at $37~^{\circ}\mathrm{C}$ in phosphate-buffered saline (pH $\approx 7.4$) with glutaraldehyde drawn from a $4 \%$ stock solution stored at $\approx 4~{^\circ}\mathrm{C}$.  I've heard that glutaraldehyde, like other aldehydes, can self-polymerize via aldol condensation under certain conditions.  What might be a good approximation for the number of PA-PA crosslinks that consist of a length $N=1$ chain of glutaraldehyde?


text:
Interpretation of Figure - biofunctionalization
In this paper on page 1985 they say:

Fluorescence analysis confirmed that enhanced protein binding at higher electron dose depends only to a small extent on the slight increase in the lateral dimensions of the nanodeposits but to a much larger degree on their height (Figure 4E)

The amount of adsorbed IgG-Cy3 is largely independent of the actual area of the nominal 100 nm islands but scales with the height of the nanodeposits.

I don't understand this yet: why do I see that the fluorescence is proportional to area in this figure?

The amount of adsorbed protein is proportional to the area of the 100 nm islands. Figure 4E proves it. The authors misinterpreted their own figure.


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Why do you need a strong base for a titration
For a titration, if you want to know the concentration of an acid you use a strong base such as $\ce{NaOH}$. My question is; why you can't use a weak base for this purpose? Is it because the calculation would be tougher? Is it because you'd need more of the base?

There is no reason why you cannot use a weak base to titrate a strong acid.  The calculation is probably considered a bit more complex but I think the real reason is a practical one.  For the sake of simplicity, I swapped your question around just a bit, and the figure below shows the titration of a strong base (blue) weak base (reddish) and very weak base (dashed line) with a strong acid (say, for example 1 M $\ce{HCl}$).

In each case the concentration of the acid is the same (1.0 M) and the concentration of the base is 0.100 M.  The volume of the base titrated is 100 mL.  For the standard strong acid/strong base scenario, one finds a well defined equivalence point.  When you transition to a weak base, however, there is additional information presented in the titration curve, namely the pKb of the base, which is a measure of its strength.  The pKb of the base can be obtained by finding the pH at the half-way mark between no acid added and the equivalence point (and if you know a bit of calculus, this is the point of minimum slope prior to the equivalence point). I've marked the pKb of the reddish titration curve with a dot, and the pKb is 9.  As the base gets weaker (less willing to accept a proton) the region before the equivalence point becomes less well defined.  In an extreme example, I have shown a simulated titration curve of acetic acid (a very poor base indeed!) with a strong acid.  Notice how the equivalence point in this curve would be very hard to pick out if we didn't know where we were looking.
So the bottom line is this: the calculations aren't harder (one uses the Henderson–Hasselbalch equation and this is a standard exercise in a first year analytical chemistry class) but the data one obtains from the experiment are harder to analyze.  And in practice, we do perform weak/strong acid/base titrations when we are interested in determining the strength of the acid or base under consideration.

Because a strong base should be able to titrate nearly any acid, strong or weak. Recall that strong/weak does not mean concentrated/dilute!  It means that, for an acid, it is easy/hard for the acid to donate a proton to the base's hydroxide ion.  If you use a weak base to titrate a weak acid, you may not be able to accomplish anything if the base is too weak, because the weak acid doesn't want to donate the proton, and the weak base doesn't want to accept it.  Even if you use a strong base, it may be the case that your acid is so weak that even your strong base is not strong enough, but this is rare.  This is where you use the so called $\ce{pK_a}$ values of the acid and base to determine if you can get a reaction, which is the titration.

If you use strong base for titration the titration curve will be very steep. It makes it easier to observe the equilibrium point.

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How do you calculate the absorption spectrum of water?
If you look online you can find various graphs of the absorption spectrum of water. The graph below for examples comes from this blog post.

As far as I can tell these graphs come from empirical measurements. Is there a theoretical model that can derive the same or similar results?  Of course if it was easy to implement in software that would be ideal.

You can for examble retrieve line date for a number of different molecules to build up absorption spectra from the HITRAN data base. In the corresponding papers, they also give references and explanations how these parameters are calculated.

The easy answer is no. It is possible to calculate absorption spectra from molecular structures using quantum mechanical calculation programs (for instance DFT using Gaussian) starting from the Schrödinger equation. These programs only work for a relatively small
number of electrons though and therefore usually assume gas phase including a small correction for the surroundings. Liquid phase water, where the molecules are very strongly coupled will be extremely difficult to calculate since it is almost impossible to create a "box" of relevant molecules. I don't have SciFinder or something equivalent, but a search on Google only gave articles calculating X-ray spectra.


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How can I measure the pH of high-viscosity fluids like dough?
I would like to measure the pH value of dough with a pH-Meter, to ensure a pH value of 4.1 (max). Because of the high viscosity I would thin down the dough with demineralized water for two reasons:

I expect more accurate measurements
This reduces pollution of the measurement device.

As far as I understand this will affect the measured value, so I would thin down the dough in a defined ratio of 1:10.
Questions:

Is it correct that the thinning with demineralized water changes the pH value?
The ratio 1:10 should cause an offset of 1 in ph value, right? In which direction? Do I have to add 1.0  to or subtract 1 from the measurement to estimate the real value?

Edit #1
As it seems to be important: We are talking about weak acids produced by fermentation of rye flour (or wheat flour) using sourdough (lactose bacteria + yeast). Some samplings are taken during the fermentation process to check that the pH value is low enough (value of 4.1 or lower).
I'm less interested in the -details- of the theoretical background, more in a practical solution for the issue. The actual real value isn't that important (I need to check, if the pH limit is reached), but it would be interesting how to measure it as there might be future experiments where the value matters...
The use of additional chemicals except (demineralized) water should be avoided for simplicity. Of course the use of tab water would be preferable, but I guess this becomes more complicated because of unknown water hardness and varying ion concentrations.

Measuring pH with 0.1 accuracy in a largely inhomogenious system, with water all over the places interacting with different polymeric compounds like proteins and polycarbohydrates makes no sense. Many of this also can act as a puffer, with changing composition during the fermentation, so it is not the "lets dilute weak acids" homework example. Yes, technically you can read a number a pH meter screen, but it makes not much chemical sense. Also, the pH value of water is not 7, more 5ish, though when mix is with such concentrated mater, it is a good first approximation.

In addition to buckminst’s answer, for weak acids, which are not fully ionized, the pH of the solution depends on the ratio of the concentrations of the conjugate base $\ce{A-}$ to the neutral acid $\ce{HA}$, especially in the pH regime close to the $\mathrm{p}K_\text{a}$ of the acid (the buffer region). Diluting the solution will change the concentrations, but in the same way. The ratio will not change. The Henderson-Hasselbalch equation can be derived from the mass action expression for $K_\text{a}$ of a weak acid:
$$\ce{HA <=>H+ + A-}$$
$$K_\text{a}=\dfrac{\ce{[H+][A^{-}]}}{\ce{[HA]}}$$
$$\text{p}K_\text{a}=-\log{K_\text{a}}=-\log \left(\dfrac{\ce{[H+][A^{-}]}}{\ce{[HA]}}\right)=-\log[\ce{H+}]-\log\left(\dfrac{\ce{[A^{-}]}}{\ce{[HA]}}\right)$$
$$\text{p}K_\text{a}=\text{pH}-\log\left(\dfrac{\ce{[A^{-}]}}{\ce{[HA]}}\right)$$
$$\text{pH}=\text{p}K_\text{a}+\log\left(\dfrac{\ce{[A^{-}]}}{\ce{[HA]}}\right)$$
If the volume increases by a factor of 10, then $[\ce{A-}]$ and $[\ce{HA}]$ both decrease by a factor of 10, and $\dfrac{\ce{[A^{-}]}}{\ce{[HA]}}$ remains constant.
Since the acids in dough are likely acetic or lactic from fermentation; carbonic or tartaric from leavening; or citric or ascorbic from preservatives – all of which are weak acids – diluting the dough might not change the pH.

Yes, we are talking about weak acids from fermentation (sourdough; or: lactose bacteria + yeast). I'll add this info to my question. - Practically spoken, should I: 1. just measure the thinned dough and take the value as it is measured? 2. Take a 2 point or 3 point measurement (2 or 3 different thinning factors) and use them to mathematically extrapolate the real value?

I must admit, that I found that formula on wikipedia but I'm still unable to fully understand the meaning of HA and A-. I only have guesses. I assume A- means "any" acidic ion, whereas "H(+)" means hydrogen. I further assume that the "[ ]"-operator means something like "amount of" (e.g. number of atoms/ions/mol). When I add water (x mol H2O), I'll add 2*x "H+" and x of "O" ... whereas the "O" isn't important to the formula, right?

@StefanK. The [A] means "concentration of A", usually in moles per unit volume, while A refers to a generic acid anion, so HA could be acetic acid $\ce{HC2H3O2}$, for example. The 2 or 3 point measurement (with extrapolation back) is better than a single point measurement any day. Once you have done the procedure a few times, you will have a sense on how the pH of your dough behaves upon dilution. You may decide afterwards that a single point measurement will be sufficient.

Yes, dilution changes pH. pH is defined as the negative log of $\ce{H+}$ concentration, so changing the concentration will change pH.
Not necessarily. For dilute acids and bases, the auto-ionization of water becomes a factor in calculating pH, and will change by less than one pH unit. For more concentrated acids (pH<5 or pH>9 for initial solution), one pH unit is correct. The pH will change toward neutral (pH=7) upon dilution, so it depends on initial pH whether you must subtract or add to make a correction.

If all you want is a practical answer, just try taking a constant mass of dough, dissolve in a constant amount of deionized water, and measure the pH with a standard pH meter. The pH of the measured solution should correlate with the pH of the dough, even if they are not an exact match. The key here is to perform the assay the same way every time, using slightly different amounts of dough or water can change the pH you measure.
You might also consider using a buffered solution of water instead of deionized water. I've seen "pure" water in our lab with pH anywhere between 6 and 7, so a dilute buffer at pH 7 should help get more consistent results. Just keep it dilute so that it doesn't buffer out the pH of the dough.


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Why does changing the temperature shift the equilibrium?
I know that raising the temperature in a reversible chemical reaction causes the equilibrium to shift to the endothermic side.
I know that $\Delta G = \Delta H − T\Delta S$, but I don't know how to put them together to show why the equilibrium would shift.
To summarise, I already know what happens when, and I'm interested in why.

I suggest you to read this answer :
https://chemistry.stackexchange.com/questions/71599/kc-only-effected-by-temperature/71647#71647

I think the best way to think of equilibria intuitively is in terms of rates of reaction. At equilibrium, the forward and the reverse reactions are happening at the same rate.
If you increase the temperature, what happens to the rates of the forward and reverse reactions?
Using the Arrhenius equation,
$$k = A\exp\left(\frac{-E_\mathrm{a}}{RT}\right),$$
you can see that as temperature increases, the rate will increase. The amount it increases depends on the activation energy $E_\mathrm{a}$.
The activation energy on the endothermic side of the reaction will always be larger than the exothermic side. This is because the transition state is always higher in energy than either of the reactants/products, and the reactants for the endothermic reaction are by definition lower in energy than the products! This means that the endothermic reaction will be sped up more than the exothermic reaction. This means the reaction is now out of equilibrium and the endothermic reaction will happen until the reaction is back in equilibrium again.

But when the temperature increases the equilbrium constant approaches 1 indepedent if the reaction is exergonic or endergonic. This seems counterintuitive. We would expect in higher temperatures the formation of the most stable compound. But why when the temperature goes high also the unstable compound is formed? Thanks in advance.

@adosar At high enough temperature the activation energy becomes irrelevant. This means that the most stable compound doesn't matter. At high enough temperature everything is equally stable! You get the highest entropy combinations of products.

Suppose you have some reaction
$$\ce{A <=> B}.$$
The equilibrium constant for the reaction is $K$ and the Gibbs free energy change is $\Delta G$. The equilibrium constant is
$$K = \frac{[\ce{B}]}{[\ce{A}]}.$$
Increasing value of $K$ shifts the equilibrium towards the right, i.e. more $\ce{B}$, and reducing the value shifts it to the left, i.e. more $\ce{A}$. The equilibrium constant and the Gibbs free energy are related by
$$ K = \exp\left(-\frac{\Delta G}{RT}\right)$$
Putting in the expression you give for $\Delta G$ gives us
\begin{align}
 K &= \exp\left(-\frac{\Delta H - T\Delta S}{RT}\right) \\
   &= \exp\left(-\frac{\Delta H}{RT}\right)\,
      \exp\left(\frac{\Delta S}{R}\right).
\end{align}
If we assume the variation of $\Delta H$ and $\Delta S$ is small we can ignore the entropy term because it's a constant, and we get
$$K \propto \exp\left(-\frac{\Delta H}{RT}\right).$$
If the reaction is endothermic $\Delta H$ is positive, so we have the exponential of a negative number and this is less than one. If we increase the temperature we decrease $\Delta H/(RT)$ and the exponential increases. So increasing the temperature makes the equilibrium coefficient bigger, i.e. it drives the endothermic reaction.
To understand why this happens consider what happens when we make the temperature very high.  If you do this the value of $\Delta H/(RT)$ approaches zero and the exponential approaches unity, i.e. the concentrations of $\ce{A}$ and $\ce{B}$ are the same. What we've done is make the thermal energy much greater than the activation energy in either direction, so the rates of the forward and backward reactions are the same.

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text:
Chemical Equilibrium - Why does changing the temperature shift the equilibrium?
I know that raising the temperature in a reversible chemical reaction causes the equilibrium to shift to the endothermic side.
I know that $\Delta G = \Delta H - T \Delta S$ but I don't know how to put them together to show why the equilibrium would shift.


text:
Are fractional oxidation states possible?
To get the oxidation state of carbon (denoted $x$) in octane, $\ce{C8H18}$, I used the fact that hydrogen has an oxidation state of $+1$:
$$8x + 18\cdot 1 = 0 \qquad \Longrightarrow \qquad x = -\frac{18}{8} = -2.25$$
But in that way, I get an oxidation number that is fractional, or not an integer. Is that normal?

We hardly ever get fractional oxidation states, you know.  We see them only in things like Fe3O4 (iron), alkali metal superoxides (oxygen), YBa2Cu3O7 high temperature superconductor (copper), aromatic ions (carbon), trihydrogen cation in interstellar space (hydrogen), ... .

Also you shouldn't just start from the formula, you don't know anything about connectivities for example. In this case it's pretty obvious but there are also compounds with more than one carbon-carbon bond and those are usually not counted when you try to find oxidation states. But answering your question. There are examples where this is postulated but then it is often dependend on the system. Like with non-innocent ligands you can distribute electron density over your system then sometimes weird oxidation states are reported although this is not true.

I also read about an example with a solid state cyanide of cobalt where they had a fractional oxidation state. I never overlooked that system that well but there are things like Mg(I) for example where you have a Mg-Mg bond. If you force it enough you may start doing redox-chemistry somewhere completely else. And then it would suddenly end up without a fractional oxidation state again.

The "oxidation number" is a theoretical value used to do electron bookkeeping and is one way of comparing the number of electrons "owned" by an atom in a molecule or ion versus how many valence electrons present in the atom as depicted on the periodic table.
Your calculation is assuming that all of the carbon atoms in octane have the same oxidation state, which is incorrect. The two terminal carbons have an oxidation state of -3 while the inner 6 have oxidation states of -2. Taking the average of these values gives you an average oxidation state, which in your case is a fractional number.
It is doubtful that the oxidation state calculated in this manner is of any use, since it is based on the assumption that all the carbons in the molecule are the same.

In addition to bob's answer, it's worth noting that oxidation number assumes that bonding is happening mostly ionically, which isn't really the case with alkanes where the electronegativities of Carbon and Hydrogen are very similar.

IUPAC recognizes that there are fractional oxidation states, but asks that you avoid writing them.
IR 4.6.1 says not to write an oxidation state "where it is not feasible or reasonable to define" because:

This avoids the use of fractional oxidation states.
Examples:

$\ce{O2-}$

$\ce{Fe4S4^3+}$

See also IR-5.4.2.2 which says "oxidation numbers are no longer recommended when naming homopolyatomic ions" because "ions such as pentabismuth(4+) (see Section IR-5.3.2.3) and dioxide(1—) (see Section IR-5.3.3.3), with fractional formal oxidation numbers, could not be named at all"
However, in the particular example in the question, not all the carbons are equivalent.  Some will have 3 hydrogens, some 2 hydrogens, and depending upon the isomer, perhaps some with 1 or 0 hydrogens.  So there is no fractional oxidation state in this example.

Oxidation state is a bit of a formalism but fractional states are rare but possible. For instance, Ag2F has silver in +1/2.  There is a fair amount of metal-metal bonding so it's not like you can think of just having a mix of +1 and 0 atoms.  (As you would have +2/+3 in a ferrite.)


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Effect of dillution on titration
When we are titrating acid/base using ph meter we add distilled water to immerse the ph electrode.
Won't this affect the concentration of the acid/base: I mean isn't  this dillution.
Won't  this affect our calcultion of the concentration at the end.

What's important to note in this situation is that the number of moles of the titrand isn't changing. This matters because all acid/base reactions go to completion, so every drop of titrant that falls into the solution will react with the titrand until there are no moles of titrand left regardless of concentration, at which point you've reached the equivalence point.
Adding a little water here or there doesn't really change the pH at the equivalence point, because ideally there are no moles of acid or base present in solution.
The reason this doesn't mess up your measurement for the concentration is because you measured the volume of titrand before the titration, and then recorded the volume of the known titrant needed to reach the equivalence point. From this information, you can calculate the moles of titrant needed to reach equivalence, which in most cases equals the number of moles of titrand present, and with the original volume of titrand noted previously, you can calculate the original concentration of the solution the sample was taken from. voila!

I should note:
titrant--dropping solution of known concentration;
titrand--solution of (originally) known volume, whose concentration you are seeking to find out.


text:
Why is tantalum so unreactive?
What is the fundamental argument behind explaining why Ta metal is so inert and does not usually participate in corrosion? I know that it forms a protective oxide, but nothing explains why in an acidic solution it is resistant to nearly everything (aside from HF). I want to know if it has to do with a bond strength of the metal or the molecular orbitals.
An in-depth citation is required.

I think the answer to your question can be described based primarily on thermodynamic considerations.  Ta has very high fusion (25 kJ/mol), vaporization (760 kJ/mol) and atomization (780 kJ/mol) enthalpies which are a result of strong metal-metal bonding in the solid (see Greenwood and Earnshaw, Chemistry of the Elements).
To further understand the protective nature of the native oxide that forms on Ta, it is fruitful to look at an Ellingham diagram, which describes the stability of a metal oxide as a function of temperature. For more authoritative sources than the Wikipedia page, see a comprehensive Inorganic Chemistry textbook such as those by Shriver and Atkins or Housecroft and Sharpe.  I found this site particularly interesting for purposes of visualizing Ellingham diagrams and used it to make the following plot:

(Apologies for the color scheme, which was not adjustable)
Here I've plotted the free energy for the oxide formation of tantalum oxide along with oxides of iron and aluminum.  The x-axis is temperature.  The large negative values for each of these oxides indicates that they are all thermodynamically favorable at the temperatures plotted.  Room temperature would be very slightly to the left of the intersection of the x and y-axes.  A complete analysis of the stability of the metal oxides would also require consideration of kinetics, but we can use thermodynamics to postulate that the very large negative values for the free energy of formation of the tantalum oxide are one factor in the stability of the oxide film.  I'll leave it to the OP to play around with thermochemical cycles, which should demonstrate why the oxide films are resistant to many types of acids.

I think, there are no secret properties of tantalum. The primer corrosion resistant is thin layer of Tantalum pentoxide $(\ce{Ta2O5})$. It's more case of a reaction of an oxide with fluorides. Here some related quotes from the book Chemistry of Tantalum and Niobium Fluoride Compounds by Anatoly Agulyansky:

As opposed to the oxides, fluoride compounds are characterized by the
  formation of mostly ionic bonds. This peculiarity is related to fluorine's high
  electronegativity.

...

Oxygen and fluorine ions have very similar ionic radii. This steric similarity
  ensures relatively easy substitution of the oxygen in the compounds, by
  fluorine ions. Such substitution opens up huge possibilities for the synthesis of
  oxyfluoride compounds with desired crystal structure and properties.
Fluorine-oxygen substitution, yielding oxyfluoride compounds, can be
  performed without changing the initial oxide crystal structure, provided the
  changes in the anionic sublattice are compensated by appropriate substitutions
  in the cationic sublattice. In these particular cases, the properties of the
  compounds will change due to variations in the character of the chemical
  bonds. On the other hand, fluorine-oxygen substitution can also cause cardinal
  changes in the initial crystal structure. Hence, compounds with desired
  structures and specific properties can be obtained only through controlled
  fluorine-oxygen substitution.

"I think, there is no secret properties of Tantal." I disagree. Many metals, such as aluminum, readily form oxides too. Yet, these are surprisingly susceptible to corrosion in acidic solutions.

That doesn't address the way that magnesium, iron and nickel are all protected against hydrofluoric attack. They all have oxide layers.

Other reactive metals like Tantalum with highly protective oxide layers are Zirconium, Hafnium (though most Zirconium contains Hafnium), Titanium (though it needs oxidising ions or platinum alloying to handle acid solutions), Chromium.
Aluminium and even iron will handle concentrated nitric and sulfuric acids due to protective oxide layers formed. Tantalum is generally the best and most all-around active metal to handle acid and other solutions.


text:
What does $\mathrm{A}_r(\mathrm{H})$ represent when calculating the average natural weight of an elemental atom?
I have a question that explains the weights and abundances of various hydrogen isotopes, and the question concludes by asking to:

Calculate the atomic mass of hydrogen $\mathrm{A}_r(\mathrm{H})$ (with the $r$ in subscript).

The answer simply summed the mass of each isotope multiplied by its natural abundance. However, the text never mentioned the meaning of $\mathrm{A}_r(\mathrm{H})$ and I couldn't find anything online.

To use subscripts etc, use $\LaTeX$ and enclose between $, e.g. for $A_r$, use $A_r$.

The relative atomic mass $A_r$ is 'the ratio of the average mass of atoms of an element ... to 1/12 of the mass of an atom of carbon-12' (from http://en.wikipedia.org/wiki/Relative_atomic_mass).
The atomic mass $m_a$ is the mass of a particular isotope of an atom, and if it is in 'u' (unified atomic mass units) it is relative to 1/12 of the mass of an atom of carbon-12.
So, to get $A_r$ you take a weighted average of $m_a$ in 'u' for each isotope.
Note also that the standard atomic weight is the relative atomic mass of an element using natural abundance on Earth (http://en.wikipedia.org/wiki/Relative_atomic_mass).


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Molecular Imaging - Any surprises to be had?
Molecular imaging, using STM and AFM technologies, appear so far to visually prove most of what we already know about chemical structures, such as VSEPR theory. For instance, here are the ideal structures and AFM of a couple of cyclization reactions:

We've also taken single-atom imaging to pretty scary levels. Behold a field emission electron microscope image of the s and p orbitals of carbon, long relegated to conceptual mathematical images in textbooks:

The question is, have there been any surprises to be had from these images, where something we thought was structured one way was in fact radically different? You can see in the AFM image that some of the bond angles deviate from the ideal regular shapes a little, and it may also be that the electrons in the pi bonds, instead of being regularly distributed, seem to prefer the "far" end of the phenyl rings, but the predicted structures are uncannily accurate. Again, with the orbital imaging, the s orbital isn't perfectly spherical and we don't see the full p-orbital shell, just two "lobes", but again, they're powerful evidence that our theories are right so far.
Has this ever not been the case? Have any of these recent images turned chemistry on its head by showing something totally unexpected?

X-ray cristallography (started in the 1910-1920) has provided ample data long time before AFM and STM provided the images above. Therefore, most of the key elements from the VSEPR theory were already established and/or confirmed through these approaches long before AFM and STM were able to provide direct image of single molecules. A key difference though would be on molecular objects which behave differently when packed in a crystal or in an isolated molecule. This is of high interest for proteins for instance.

The last time I'm aware of when there was a 50/50 chance for such a thing to happen was the determination of the abolute configuration of chiral substances (until then one only knew them relative to each other) by x-ray crystallography analysis of glyceraldehyde in 1951 (see https://en.wikipedia.org/wiki/Absolute_configuration). But this yielded the result that the original guess was correct.

I don't think this is going to put basic quantum chemistry upside down. Because chemistry is complicated, and these tools are "inaccurate". Let me illustrate : it was thanks to spectroscopy of atoms that people discovered the spin of the electron. It then explained a lot of fact that puzzled chemists. A molecule is something awfully complicated, that we can only solve with approximations and powerful computers. So to my mind, molecular observables are unlikely to contradict quantum theories. Now you need the LHC to split particles in order to test the theories.
This guy will tell you that there are surprises to be had about large molecules, the ones from natural sources (or plain coal) that are a pain in the ass to study by NMR,  or enzymes (like this one, studied with single molecule diffraction).
There will be always surprises. But if we could predict them, would they still be surprises ? (sorry that was too long for a 'comment')


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Having problems running DFTB using dftb.org parameter files on Gauss 09 (G09)?
I ran into an issue trying to run dftb.org parameter files on G09 and found questions asked on this but not answers.  I thought I would post the solution here to help anyone else who may run into this problem.

(Remember to note these changes in your method description.)

First, at least for C.01, dftb=readparameters, in the route section, does not work dftb=read does but is not necessary.
Skip a line after molecular descritption and include your parameter files there. Make sure you include all pair interaction files, not just bonded atoms.  It should look something like this:
@C:/G09W/C-C_mio.skf
@C:/G09W/C-H_mio.skf
@C:/G09W/H-C_mio.skf
@C:/G09W/H-H_mio.skf
(I included the mio because we are running more than one parameter set, not a requirement just a name)
The real glitch comes from my final hint.  The DFTB parameters generally have more than 500 points, but the first line in the code is 500 (not 519, or 520, etc).  G09 Rev. D.01 doesn't care about this, B.01 and C.01 do.  You simply need to change 500 ro reflect the correct amount of points and add in the correct order for your atomic numbers (this part is explained on Gauss' website) and you are good to go.
Hopefully this helps anyone trying to run DFTB on G09 for the first time, and helps save some time.


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What is the effect of adding salt to ice?
What will happen to the melting point temperature of ice if some common salt is added to it? How to justfy the answer?

What is your initial impression?  Please show some effort on your part to answer the question.  Hint, see your textbook under "molality".

From http://antoine.frostburg.edu/chem/senese/101/solutions/faq/why-salt-melts-ice.shtml:
Consider that ice, a crystalline solid, forms via weak partial-polar forces between water molecules, known as hydrogen bonding. As liquid water cools, it loses the energy that keeps these forces from attracting and capturing neighboring water molecules, and so these bonds form in a crystalline structure. Exactly what that structure is depends on the temperature and pressure at which the ice forms; your everyday freezer ice as well as most of the stuff outside in January is Ice Ih, which is a very spacious hexagonal arrangement that makes ice less dense than liquid water.
Also consider that even within what looks like solid ice, individual molecules "melt" and "freeze" all the time as they transiently gain enough energy to break free of the lattice, and then lose it and are recaptured. When the rate at which molecules melt and freeze are equal, the ice is stable. This happens at 0.01*C; below this temperature, more water will freeze, while above this temperature, more water will melt. Right at 0.01*C, water remains water and ice remains ice (the other half of this coexistence, besides temperature, is that water like most substances requires more energy gain or loss to change its phase, known as the Latent Heat of Vaporization, and for water that's 334J/mL, 80 times what's needed to cool liquid water from 2*C to 1*C).
Introduce salt, and these dissolved ions interfere with this hydrogen bonding within the ice, preventing it from happening at the normal freezing point of water. Now the melting molecules don't refreeze, because their partial polar charges are more strongly attracted to the full charges of the sodium and chloride ions than to other water molecules. That upsets the balance; more water is melting than freezing at the normal freezing temperature, and so the ice melts. It takes more energy loss within this system before the water is too unenergetic to stay liquid and settles into crystal, releasing the salt, which then finds other liquid water at the boundary layer between liquid and ice.
This creates a vicious cycle; as the salt concentrates within the remaining liquid water, its freezing point is lowered even further as the sodium and chlorine, which won't resolidify as salt crystals until there's just too little water to remain dissolved, will become even more attracted to the remaining water, even further lowering the remaining liquid's freezing point as the water must free itself from even more ions before it can be captured by the ice crystal.
So, an aqueous salt solution, in nearly any concentration, will lower the boiling point of the solution below 0*C, but the minimum freezing point is at -21.1*C, which is the freezing point of a fully-saturated salt solution. Below this temperature, salt that loses its water to freezing can't stay dissolved, and precipitates out as a separate crystal solid suspended within the ice.


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How can melting point equal freezing point?
I don't understand how the freezing point of a substance is the same temperature as the melting point of the same substance.
For example, if liquid water freezes at 0 °C how can ice also melts at 0 °C?

I'd like to see someone answering this considering phenomena in which the melting point is "not the same as" the freezing point, such as supercooling, hysterisis and superheated crystals.

@Alex I agree, but that seems to be a different question that involves concepts that might muddy the waters for the OP.

Because melting point and freezing point describe the same transition of matter, in this case from liquid to solid (freezing) or equivalently, from solid to liquid (melting).
What you may not realize is that while water is freezing or melting, its temperature is not changing!  It is stuck on $0\ \mathrm{^\circ C}$ during the entire melting or freezing process.  It is easier to see this for boiling points.  if you put a thermometer in water and heat it, the temperature will rise until it reaches $100\ \mathrm{^\circ C}$, and then it starts boiling.  And while it boils, it will stay at $100\ \mathrm{^\circ C}$!  All the way until the water has all boiled away.  Now if you could somehow trap the steam (gaseous water) and keep heating it, the steam could have a temperature higher than $100\ \mathrm{^\circ C}$.
So to sum this all up, when matter is transitioning from solid to liquid (melting) or liquid to solid (freezing), its temperature is fixed at the melting/freezing point, which is the same temperature.

The temperature doesn't change in an ideal situation but I believe that the ice would be under the effects of a temperature gradient (the entire 'ice object' wouldn't be exactly 100C).

@LordStryker: Yes, of course you are correct.

I think this is an interesting question where the confusion is mostly due to semantics.
Let's consider an ice cube in your freezer (a typical kitchen freezer has a temperature of -10 C).  When that ice cube is taken out of the freezer and placed in a warm kitchen, heat from the surroundings (air, counter top, etc.) is transferred to the ice cube.  We observe that the temperature of the ice cube increases.  The ice cube stays a cube because the energy of the intermolecular forces that keep the water molecules together is greater than the heat energy added to the ice cube so far.
At the melting point, however, there is enough thermal energy to start breaking those intermolecular forces.  What we observe is that the temperature does not rise, but bonds are breaking and the solid starts to melt.  Once all the solid melts, the temperature of the (now liquid) water can increase when thermal energy is added.
A similar explanation can be used for the reverse process (freezing water) only in this case the thermal energy is being transferred from the water to the surroundings.
So we come to your question, how can melting point = freezing point.  This "point" is the temperature at which the solid and liquid forms of the molecule are in equilibrium.  When we use a term like melts oftentimes we mean melts completely.  In this case, the temperature of the liquid would be just beyond above the melting point.

Thanks, that makes sense now. Could you also help me understand why a stirred mixture of solid and liquid phases of a substance will always adjust to the temperature which is the melting point of the substance? For example, why does a mixture of water and ice always adjust to zero degrees..

@jaykirby The answer involves looking at the energy required to affect the phase change and the energy required to change the temperature of the phases.  Careful with your "...always adjust to zero degrees" which is not entirely true.  An ice cube thrown into Lake Michigan will not change the temperature of the water to 0 degrees.  Relative amounts of ice and water need to be considered as well.

http://en.wikipedia.org/wiki/Triple_point
If a material's equilibrium melting and freezing points were non-identical, you would have a perpetual motion machine of the first kind.  Go head, trace the energy flows for temperature versus specific heats and latent enthalpies of transition.  Cycled divergent vapor pressures at the differing transition temperatures would power a perpetual wind generator.  Ya gotta think about these things. e.g., water below,


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How do you explain the thread sticking to an ice-cube when salt is sprinkled over it?
So this experiment goes like this :

You take an ice-cube
You put a thread on top face of it
You sprinkle some common salt on the ice-cube wherever its touching the thread
You wait for sometime (approx 1 min)
Voila! now you can lift the thread and ice-cube will stick to the end of the thread so you can lift the whole thing altogether

How do you explain this in chemistry?

I'm fairly certain that's actually physics acting there. You may wish to ask it in the physics StackExchange.

I see no relation to physics here :|

@PeeyushKushwaha I'd at least hope that while studying chemistry, you try to see the bigger picture as it relates to physics...

I think "reduction to physics" is unnecessary for this problem.

I conjecture that there are several things happening.

Step 3:  Addition of salt will melt the ice at its immediate point of contact. The string will naturally be surrounded by the highly concentrated brine (salt water).
Step 4:  This is enough time for the brine to get diluted, noting that there is an infinite reservoir of potential water by way of the ice cube.
Step 5:  Once the salt gets sufficiently diluted, then the comparatively large ice cube provides the necessary cold for the (now) very dilute salt water to reform as ice.

Recalling that in Step 3 the salt water surrounded the string, then ice naturally forms around the string in Step 5, hence you can pick up the ice cube with the string.

@Eric Brown has the result right qualitatively.  We can add some more quantitative and chemical information.
Step 3: When salt is added, the freezing point of the water is decreased at that point, so the ice melts to water.  This is a colligative property; i.e. has only to do with amount of substance, and not the particular substance.
Step 4: As the melting water dilutes, its phase shifts back towards pure water, which has a higher freezing point.
Step 5: At some point, the phase shifts back to freezing at 0 °C, and the water re-freezes given the ambient ice at 0 °C.
This Wikipedia article shows you the phase diagram corresponding to my description:
NaCl phase diagram
As I said, my answer merely augments your already selected correct answer, but I thought this might be useful additional information.

Ice at 0 °C would not freeze pure water at 0 °C, much less a salted water. It is important to note that the temperature of the system actually drops way below 0 °C as you add salt.


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What impact does heating or cooling have on the oxidation reduction potential of water?
There are alkalinizing water ionizers on the market that lower the ORP of water. Does heating or cooling that water have an impact on its ORP?

Another question is, how significantly?  What is the context of your question?

I see two questions here.  The one that is stated (temperature effects on ORP) and a hidden one about water ionizers.
First, the temperature effect on ORP.  The thermodynamics of oxidation and reduction reactions is governed by the Nernst equation:
$$E(ORP) = E_o - \frac{2.3RT}{nF}\times log\frac{[reductant] }{[oxidant]}$$
In this equation, R and F are the gas law and Faraday's constants, respectively, $E_o$ is the standard reduction potential of the system of interest, T is temperature, n is the number of electrons transferred in the reaction and the concentrations of the reductant and oxidant are found in the log term.  The coefficient $\frac{2.3RT}{nF}$ in front of the log term can be simplified to $\frac{0.00198\times T}{n}$ and when we perform dimensional analysis on the term we find that it will provide units in Volts.  If we are looking for the largest effect that temperature would have on ORP then we want to use the smallest value of n, which would be 1 electron, further simplifying the coefficient to approximately $0.002T$.  T has to be in Kelvin so if we take our two extreme values as the freezing (273 K) and boiling (373 K) points of water, we find that the difference in ORP between these two extremes (all else being the same) would be about 200 millivolts.
Now to the hidden question about water ionizers.  There's a lot of hype about these systems with little scientific evidence to back up the claims.  There is at least one website useful for exploring the science behind water ionizers that is (I think) accessible to the general public.  A number of claims made by the manufacturers of these products, such as ion streams and water clusters simply don't stand up to scientific experimentation, and consumers need to be wary of the health benefit claims.

Yes. I don't have much time to go into details (sorry) but temperature changes the pKa of water which in turn affects the concentrations of [$\ce{H3O+}$] and [$\ce{OH-}$] and that affects both reduction and oxidation potential of water. The effect it negligible for most intentions and purposes over the 100 C of liquid water.
If you look at Wikipedia page about Pourbaix diagrams on some of them you will see the dashed lines. Those indicate the reduction and oxidation potentials of water as they change  with the pH.


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Why does a silane group react with hydroxyl groups?
I read somewhere that a silane group will react with hydroxyl groups.
Does anyone know why this would happen?

It is well known that silicon forms very strong bonds with oxygen (think sand, $\ce{SiO2}$). Hence, any time one can convert a $\ce{Si-H}$ bond into $\ce{Si-OH}$, it will happen.
I believe that silanes are extremely air sensitive due to the moisture content.

I don't know about H-Si groups, but alkoxysilanes are surprisingly unreactive in the short term. I suspect there may be some sort of kinetic barrier in place. If you mix a methoxy/ethoxy silane 1:1 with 10M NaOH, it takes a good minute for the reaction to take place. Once it does, though, it's incredibly incredibly exothermic. You still wouldn't want to leave it out uncapped for long though. Source: cleaned the results off of my fume hood ceiling about 10 hours ago.

@chipbuster Interesting, seems to be some sort of significant kinetic barrier. I wonder why that is.

@NicolauSakerNeto A kinetic barrier must be at least as high as the energy of the endothermic (up-hill) product.

Hydroxyl radicals are strong proton acceptors, due to an oxygen with additional electrons, and will therefore abstract hydrogens from anything that has hydrogens, e.g., silane or methane. This attack can be enhanced, depending on the electronic properties of the atom sitting at the center (silicon or carbon). The more stable the atom, the more stable the 3-hydrogen radical will be, the more thermodynamically favorable the hydrogen abstraction reaction will be.
In the case of silane combustion, this leads to the formation of $\ce{^.SiH3}$ radicals, which react with molecular oxygen ($\ce{O2}$) in a chain-branching step. Depending on whether the temperature is low or high, this step can occur along various different pathways. It generally occurs following the reaction:
$$\ce{^.SiH3 + O2 -> Si(O)OH + H + H}$$
but the intermediates change depending on temperature and the energetics of the reaction.
You can find more information in a couple of papers:

Babushok, Tsang, Burgess Jr., Zachariah, "Numerical study of low- and high-temperature silane combustion." Twenty-Seventh International Symposium on Combustion (1998) pp. 2431-2439.
Kondo et al., "Ab Initio Energetic Calculations of Elementary Reactions Relevant to Low-Temperature Silane Oxidation by Gaussian-2 Theory." Journal of Physical Chemistry A (1997) pp. 6015–6022

Just to be clear for future readers, hydroxyl radicals and hydroxyl groups are quite different things. Your answer is certainly correct (and we are very appreciative of sourced answers!), but it may not directly answer the OP. I suspect they were talking about hydroxyl groups in molecules, in which case Eric's answer (though rather short) is correct thermodynamically, and kinetically it probably has to do with silicon's larger size which would stabilize a five-coordinated transition state where the hydroxyl group attacks the silicon atom in a silane directly.

Excellent point. I pounced on "hydroxyl" because silane so readily combusts, and hydroxyl radicals play an important role in any combustion process, but I neglected "groups." Oops!


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Why is ethanol written as C2H5OH not C2H6O?
This is balanced equation of fermentation of glucose:
$$\ce{C6H12O6(aq) -> 2C2H5OH(aq) + 2CO2(g)}$$
I know that the result of glucose fermentation is ethanol ($\ce{C2H6O}$). What represented in the aforementioned equation is slightly different. I would like to know what $\ce{C2H5OH}$ is?
What is the OH functional group doing here? Does it make ethanol acidic? Does it mean that ethanol is diluted with water?

It is better to write $\ce{C_2H_6O}$ as $\ce{C_2H_5OH}$, because there is more than one compound with the same overall formula, and emphasising the hydroxyl group $\ce{-OH}$, which is what characterizes an alcohol makes it ethanol and nothing else.  So when you ask, "What is OH functional group doing here", it is exactly that the product is an alcohol, characterized by a hydroxyl group. And yes, alcohols are acidic in the right situations, given that their hydroxyl groups do produce acidic hydrogen ions, and yes, that is why it's so soluble in water.

I confused compound formula with the actual formula of ethanol. Compound formula is $\ce{C2H6O}$. But the actual molecular structure is $\ce{CH3CH2OH}$. The oxygen is attached to hydrogen which it in turn is attached to Carbon.


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How do indicators work when they mixed with each other?
Methyl orange, bromothymol blue and phenolphthalein indicators were mixed together to form a solution.

Over what pH range would the solution be yellow?
I cannot understand the question. If two or more different indicators are mixed, do they still work as indicators? Because each their colours will blend with each other and result to formation of a new colour, how can we know what the color will be?

The indicator molecules form their color by absorbing light, so they combine by the subtractive color mixing model.

In general, the chemistry of one indicator does not usually affect another as far as I know, but the colors will mix as you would expect. It gets tricky here because the question shows the colored ranges but leaves a blank in the transition pH ranges, but I would guess the answer they're looking for is from 3-7.5

They don't show the colors because in the transition region, because it's a mix of colors. The color of a solution of multiple chromophores is the (subtractive) combinations of their colors. For example, protonated bromothymol blue is yellow, and deprotonated is blue. At the pKa (~7.1), you have equal amounts of protonated and deprotonated forms, so the colors would mix just like you had equal amounts of unrelated blue and yellow chromophores/pigments. (i.e. green)

The mixture of indicators should still work because most indicators are weak acids, and the presence of other weak acids does not interfere with any given weak acid.  If you had an indicator that happened to be a weak base, you would likely have a problem, but that is not the case for your solution.
Here's a nice video from the CHEM Study series on exactly this topic, including a thorough discussion on mixing the indicators.
Acid Base Indicators CHEM Study
In terms of the specific answer to your assigned question, I think @Gavin Kramar is spot on.  I'm just adding this answer for some additional information.

Looks to me like it would be:
orange from 0 to 3,
yellow from 3 to 7.5,
green from 7.5 to 10,
purple-ish? from 10 on up (what's green plus fuschia?).
One could actually go into the lab and give it a try. . .just sayin.
I have actually done bromothymol blue and phenolpthalein mixed, and it performed as expected.  Several years ago I used methyl orange and phenolpthalein, and again it did as expected.  Have not tried all 3.


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Energy required to break single bond vs double bond vs triple bond of a carbon atom in an organic molecule
Assuming every other conditions are the same, what would be the comparison of energy required to break single bond, double bond and triple bond of a carbon atom in an organic molecule?

365 kJ/mol for the single bond, 598 kJ/mol for the double bond, and 813 kJ/mol for the triple bond. These are usually listed as bond enthalpy or heat of formation in scientific table books. If you don't find the one you need you can calculate it in steps according to Hess' Law.
If you want the energy required to break one bond as opposed to a mol of bonds, you divide the number I gave by Avogadro's number, which is the number of atoms or molecules in a mol of a compund or element.

noting that these are "suggested values" for bonds that "some people" "typically" ascribe to being "single", "double", "triple."   (Basically, there is some subjectivity in divisions due to bond energy)

Let it be organic or inorganic compound
More the no. of bonds between 2 atoms more the energy required to break it.
This is because intermolecular force of attraction increases as the no. of bonds increase. So to break the bonds one needs more energy or so called more bond enthalpy.


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Is melting/boiling point of ionically bonded substance higher than of covalently bound?
Is the melting and boiling point of ionic bond usually higher than covalent bond?
I know that compounds with ionic bonds are usually solid at room temperature, so I want other answers than this. (and this question can also be about why compounds with ionic bonds are usually solid in room temperature.)

Neither ionic nor covalent bonds have a "melting point". The melting point is a macroscopic property of a compund or element, while bonds are phenomena on the mircoscopic (atomic scale) level. Did you mean melting points of compounds incorporating ionic bonds as opposed to compounds incorporating solely covalent bonds?

@TanithRosenbaum - your comment is a good starting point for an answer that distinguishes between interatomic (intramolecular) forces and intermolecular forces, which govern the macroscopic properties.

In case of ionic compounds, there is a complete transfer of electrons from one atom of the element to another atom of the another element. So, there exists a strong electrostatic force. In case of covalent compounds, electrons are not completely dragged, the atoms of different elements are attracted by a weaker electrostatic force. Thus, ionic compounds have atoms bound by stronger electrostatic force than in case of covalent compounds. So, in general it is easy to break covalent bond than ionic. Thus, melting and boiling points of ionic compounds is greater than covalent compounds.

The melting point of diamond (under pressure) is not lower than typical ionic compound. I guess one need to phrase the word "usually" in certain sense....

He should have said discrete covalently bonded whereby the actual bonding that defines the phase is physical or Van Der Waals. All network covalent molecules (especially 3D) have high melting points.

The ionic bond is usually stronger than the Van der Waals forces between the molecules which are covalent solids. Covalent Bonds are not broken during melting. Ionic Bonds are broken.

The answer relates to the strength of the interactions between the component units that make up a crystal or a solid.
The reason why anything is a solid at a given temperature is, crudely, that the interactions between the units that make up the solid (atoms, ions or molecules) are stronger than the amount of thermal energy available at that temperature.
Diamond, for example, is a solid because each unit carbon is bonded to its neighbours by a strong carbon-carbon bond that takes a great deal of energy to break. Non-polar molecular compounds, like candle wax (in reality a mixture but imagine a pure long chain hydrocarbon to keep it simple) are held together by the intermolecular forces between the molecules (often called dispersion or Van der Waals forces). These are much weaker than chemical bonds and depend on the surface area of the molecules. So small non-polar or not-very polar molecules tend to have low melting points. Hexane is a volatile liquid. The shape and symmetry of the molecule matters a bit to these interactions so toluene is a liquid down to -95°C but benzene, which is very similar to toluene but far more symmetric and slightly flatter only melts at about 6°C as the molecules pack together better encouraging stronger interactions. Polar molecules have stronger interactions because the molecular dipoles interact and this produces stronger forces than the non-polar dispersion forces. So phenol (very similar in size to toluene but far more polar) melts at about 41°C.
Simple ionic solid are made from charged ions. Table salt crystals consist of Na+ and Cl- ions. The key interaction between these is electrostatic and this is a strong force that takes a lot of energy to break. The melting point is about 800°C. Permanent electrostatic interactions are strong especially if the components can get close (remember, it's an inverse square law). This explains why not every ionic compound is a solid. There are ionic compounds that are liquid at room temperature. These tend to consist of molecular ions that are large and sometime irregular so forcing structures where the ionic force has to act over much larger distances and, therefore, be much weaker. An example is [BMIM]PF6 which has a melting point well below room temperature:

These ionic liquids are an interesting area of research in chemistry as they sometimes enable much less wasteful chemical reactions in industry (see wikipedia).

One set of comparisons is:
Why it take so much energy to turn sodium chloride molten, whereas candle wax melts virtually upon touch?
"Ionic compounds" tend to form crystals.  You may have seen pictures of a crystal lattice of sodium chloride:  cubic packing with a sodium being surrounded by many chlorides.  It takes a lot of energy to break up all of those bonds.
In the case of wax (also a solid at room temperature), there are not these strong, "ionic" interactions between molecules. Things are more "layered" and disordered rather than crystalline. (Though there is definitely inter-molecular bonding Otherwise it would never be solid!!!)

It looks to me like sodium chloride is sort of a big frat party. All the boys are attracted to all the girls (you choose which element is which), so they're all holding hands with each other. To break NaCl into smaller bits, you have to break those handholds (melting, requiring lots of energy) or give them something else to hold (such as one end of a polarized water molecule).  By contrast, sugar molecules are a bunch of close knit family groups. The members of each family are holding on to each other, but not attracted to any other family. To melt sugar, by getting the families to move around relative to each other, does not require much energy. When sugar is dissolved in water, the water molecules easily fit in among the sugar molecules, but again, elements of a given family are not attracted to outsiders.

Although the analogy is nice, it neglects much of the actual chemistry behind this effect. Perhaps you could go into more detail about why ionic and covalent compounds have these properties.

In the case of ionic compounds, there is a complete transfer of electrons from one atom of the element to another atom of the another element. So, there exists a strong electrostatic force.
In the case of covalent compounds, electrons are not completely dragged, the atoms of different elements are attracted by a weaker electrostatic force. Thus, ionic compounds have atoms bound by stronger electrostatic force than in case of covalent compounds.
So, in general it is easier to break covalent bonds than ionic bonds. Thus, the melting and boiling points of ionic compounds are greater than covalent compounds.

Um, mind me, this is not why the OP's claim is [usually] correct. It totally depends on what bonds we compare. And in a molecule with intramolecular covalent bonds the bonds that break for the boiling aren't the covalent bonds, but intermolecular forces, namely, Van der Waals forces. Your answer is, all in all, incorrect. How would it explain the melting/boiling points of diamond?


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Analysis of a mixture of carbonate salts
We've got a mixture of carbonate salts. It's a buffer, used in fish tanks. How might we go about identifying the components and their concentrations?
I suspect it'd be hard to separate some of this. Is this correct? If so, why?

How about using the old fashion precipitation and dissolve. Like using AgCl and NH3 solution to tell some of them apart?

I was thinking particularly about sodium and potassium carbonates. I don't know how I could differentiate them via precipitation; do you?

As this currently stands, it's better as a comment. Just one question isn't a good answer according to SE standards.

@Jeffrey I have converted your answer to a comment, if it can be extended to fir our format, I'll be happy to undelete it.


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How many elements in food?
A slightly deranged curiosity: if we had a 3D printer able to “print” anything starting from “cartridges” of pure chemical elements, how many of the 118+ possible cartridges (or of the 98ish for non-artificial elements) would we actually need to print most - or all - foodstuff?
(OK, one could have phrased the same question in a more mundane way, I agree.)

maybe you can help answer the question:  can you find a resource that contains an elemental analysis of "foodstuffs"?

I am confused: I was under the impression that this was exactly the gist of my question and that, not being a chemist nor an experimental scientist myself, somebody could be able to help here. (Moreover, English is not my first language: isn't “foodstuff” a correct word?)

It's a correct usage of the word.  I doubt you could get a definitive answer, beyond "you need all of them because all elements are in virtually all naturally-occuring composite substances."  You may want to consult a nutritionist if you want to know what is necessary, though they may point out that elements make molecules, and there are myriad molecules in food.

Thanks for the pointers. It was a curiosity arisen while speaking with friends, none of which a chemist or thereabouts. We could just figure out that (is that right) we could probably leave out noble gases and the most radioactive elements. As for the molecules, our approach bypassed the problem by just assuming the fictional 3D printer would cater to it, if we were providing it with the necessary elements. Right, we shall go for a complete, 118-element kit...

Slightly more seriously, I am quite sure about C, H, O, N, and S, and several more, but I don't even know enough to understand whether in, say, a steak “almost all” elements are present, or just a few. But your remark that “you need all of them because all elements are in virtually all naturally-occuring composite substances” points to the former. Is it a quote?

It's an interesting question--if you've ever seen Star Trek, they have those "food makers" and they have to get their ingredients from somewhere.  Good luck, maybe someone would have a more definitive answer.

For the record, I am not a biochemist, a nutritional expert, or any sort of person who is required to know this for my profession.
That said, as far as I know, to get the vast majority of nutritional food substances, you'll need to be able to produce at least:

sugars and starches
lipids
amino acids
cellulose
various other non-polymeric organic molecules of various levels of complexity
salts
proteins

These together will give you a reasonable approximation of food which will be most of what you need to stay alive, without necessarily going into the complexity of actually producing a better analogue of dead tissue "under the hood".
Sugars and starches are entirely composed of carbon, hydrogen, and oxygen. The same goes for lipids, cellulose, and the majority of the organic components you'll need. For amino acids, and the other organic bits, you'll need to add nitrogen and sulfur.
You also need phosphorus for DNA, RNA, ADP, and ATP, as well as a whole bunch of other signalling and respiration-based molecules, and bone material.
You need sodium and potassium ions which are needed in living tissue for nerve signalling, and you also need calcium ions to make bones.
Many plant materials seem to contain small amounts of boron, though the biological role of it isn't concrete as far as I know.
As far as I know, the proteins are where the metallic awkwardness comes in. A lot of proteins have a reaction site containing a metal, metal-sulfur, metal-porphyrin, or another metal-organic molecule complex, which can come in various metallic flavours. For example, haemoglobin requires an iron-porphyrin complex, while chlorophyll requires magnesium-porphyrin, and a lot of respiration-based proteins seem to involve iron-sulfur complexes.
I'm fairly confident you could make reasonable analogues of most food with C, H, O, N, S, P, Na, K, Cl, and Ca, with relatively small additional quantities of B, Fe, and Mg. You might suffer after a while from micronutrient deficiencies, though.
(I'll probably find that I've forgotten something important somewhere.)

Well, the human body has a bunch more, like Molybdenum. I assume this has to come from food some time. But you're probably right, the majority of food products can be made without the trace elements.

I think the answer is 26 - as suggested from this periodic table of life

One can estimate that the raw materials in the human body amount to approximately $3.50 (so you probably can't sell off your little brother for more than a buck and a quarter).


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Can natural gas appliances run on Hydrogen?
Can natural gas appliances run on Hydrogen? If so, does the appliance have to be modified and how significant of a modification would it need?
Thanks

When I was young, late 60s or early 70s, something similar to the reverse happened in the UK.  We used to use town gas / coal gas.  This was a mixture of gases but contained a lot of hydrogen.  In the run up to switch to natural gas, an engineer visited.  I don't know what he did but it didn't take long.

Pure hydrogen, as opposed to hydrocarbons, degrades metals.  $\ce {H_2}$ molecules are tiny, so they wedge themselves into the relatively large metallic crystal structures and form hydrides.  This is the entire reason nickel-metal hydride (NiMH) batteries work - the hydrogen is stored in an iron-hydride matrix. An appliance that might last 20 or 30 years burning natural gas might not make it past ten years burning $\ce{H_2}$, and a failure may take the form of a dangerous leak.

While hydrogen degradation in pressure systems is a real problem it is very unlikely that this is a significant problem in appliances as pressure is too low for it to be a big issue. Possibly somewhere in the distribution infrastructure it might be worse, but that isn't the question.

I'm not entirely sure of this, feel free to correct me.
This probably depends on the type of appliance. I'll try two, the internal combustion engine and the stove.
Internal combustion engine
The nice thing about an internal combustion engine is that it takes in gas and oxygen, isolates it, and then ignites it. Besides, it uses a spark to ignite it. Hydrogen is more explosive than natural gas, so the isolation is quite useful here. And the reaction with air is more easily triggered by a spark rather than by heat.
On modification needed to make it work would be to recalibrate the apertures of the oxygen and hydrogen feeds such that the gas mixture is sucked in in a 1:2 ratio.
Secondly, there needs to be a cleaner way of disposing the water formed. Internal combustion engines assume that the exhaust is gas and particulate matter -- which can be compressed and expelled entirely, and isn't adhesive. Water can't be compressed. This makes it easier to expel, but some of it may adhere to the walls. As water, oxygen, and electricity are being supplied, this may quickly rust the walls of the cylinder. So there needs to be a way to protect the cylinder  from rusting. Also, the engine will just stop working if the spark plugs get shorted by water. There needs to be a way to prevent that from happening, too.
Actually, a hydrogen internal combustion engine exists, but it looks like they got there by starting from scratch, not modifying a regular internal combustion engine.
The stove
The only issue I see here is that the aperture on a normal stove is too large. Oxygen may enter and end up exploding the whole setup.
In such a case, having one pipe (with a very small bore) for each individual flame may work better. This would probably have to be lit automatically -- I don't see how one can safely light a set of such pipes without there being an individual knob for each pipe.

Nowadays, the more efficient way of using hydrogen for fuel is in a fuel cell. This way, expensive hydrogen doesn't get wasted, and is instead reused (the energy input comes from other sources)

Again, what is this obsession with the problem of water? Hydrocarbon combustion already produces lots of water and appliances and engines cope with that without any problem. Also, the water produced from hot combustion is vapour (it is, after all, very hot!) which is as compressible as any other gas.

A few addition comments.
The production of water would create a problem, as if being used as a  heat source, how is the water going to escape? Steam and condensation can cause many problems, not to mention; there is the issue of steam and the safety risk to end user. All these factors also cause metal pipes, machinery and appliances to (potentially) rust. On top of the points mentioned above.
The Hindenberg is an interesting example of the flammability of Hydrogen and the effects of it's products when combusted.
As an point of interest, I have provided a link for a table of flammable gases:
Hydrogen's "Lower Explosive or Flammable Limit" (LEL/LFL) (%) is 4, in the order of methane and propane 5 and 2.1 respectively, bit it's "Upper Explosive or Flammable Limit" (UEL/UFL) (%) is much higher 75 as compared to 15 and 10.1 for methane and propane, respectively. Making it more highly combustible at concentrated levels, leading to the difficulties in containing such explosions, as mentioned in the other answer.
http://www.engineeringtoolbox.com/explosive-concentration-limits-d_423.html
I still remember clearly a video our Chemistry Lecturer showed us of the Hindenberg and how the sheet of water (produced from the combustion of Hydrogen) saved lives. That was over 30 years ago, so it made an impression and it's a subject I find interesting.

If you're already burning natural gas, that also produces water. I also don't think air infiltration into natural gas systems is common; they're generally kept at a modest pressure (~0.25 psi) so the UFL doesn't seem as relevant as the LFL (leak from the supply into a room). Embrittlement (mentioned by Meredith) and maybe an unanticipated air/fuel mix (mentioned by Manish) are far more relevant.

Note that there is used for welding a mixture acetylene+oxygen, not hydrogen+oxygen, even if the latter would be more economic. But hydrogen diffuses into iron metallic lattice and reacts with present carbon to methane. Methane is then trapped within the iron lattice, what makes the welded connection brittle.

The production of water is not a problem. All burning hydrocarbons create a lot of water and it doesn't seem to be a current problem. And the flammability of hydrogen doesn't create fundamentally different problems to other gaseous fuels: the whole point of a fuel is that it is flammable.

The primary problem when switching to different fuels in appliances or engines is getting the fuel/air mix right
Previous answers contain a disturbing number of misleading hypothetical reasons why hydrogen can't work. Many are just plain wrong. For example, the idea that dealing with the water produced is ludicrous given that hydrocarbon fuels produce plenty of water without any major issues.
Hydrogen is a viable fuel if the devices are adjusted to use it (whether the transmission infrastructure needs to be modified is a different question).
The key issue is that most engines and appliances (eg gas cooker or boilers) are set up to use a specific gas/air mixture that optimises the combustion for the specific fuels that is being used. If you change the fuel, that mix changes (the mix being the specific amount of air that is mixed with the fuel to achieve optimal combustion).
So engines need to be adjusted to use hydrogen not hydrocarbon (which might involve hardware and software changes since hydrogen is a gas and petrol/gasoline needs to be vaporized but the air/fuel mix will be different).
Cookers and boilers have simple hardware to control the mixing of air and fuel. So changing to a different fuel needs that hardware to be changed. The UK has experience of doing this for gas-driven domestic appliances when the fuel changed from "town" gas (consisting of a mix of mainly CO and Hydrogen produced from coal) to natural gas (consisting mainly of methane) in the 1960s. The changeover required the hardware in cookers (which means the parts on burners that mix air and fuel) to be swapped out for parts optimised for methane rather than town gas.
So the answer to the question is: yes, but not without hardware changes.

At ambient temperatures , hydrogen gas does not enter metals in significant amounts . The confusion is that NACENT hydrogen  from corrosion or electrolytic action enters steels readily and can cause failure in minutes. I more or less made a  career from dealing with this situation.

It should be noted that current natural gas burning appliances also produce water. In fact high efficiency furnaces produce a lot of water that is actually acidic, so it is very corrosive. Hydrogen would not produce acidic water. Water production from combustion is actually not a problem.
The hardware changes needed to use hydrogen instead of natural gas would be significant. The "holes" (apertures) needed for hydrogen would be very small compared to natural gas. These apertures would probably have to be precision machined to work correctly, which means costly to produce. But probably not as costly as producing fuel cells to use instead.


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Why is methoxide a strong base?
Why is methoxide strong base? In case of halogens, I get why one is weak base, and one is strong base, but in the case of ones with oxygen, this seems to be hard to figure out.

Methoxide ($\ce{CH3O-}$) is the conjugate base of methanol. Methanol is very weak acid (e.g. its dissociation constant is very small), so its conjugated base is very strong.

Further reasoning can be that there is no delocalization for the free electron pair of oxygen because of the way the orbitals are oriented, so that makes it less stable.


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Entropy and racemization?
The process of racemization of an optically compound is entropy driven and hence nature prefers a racemic mixture to an optically active one, which is consistent with the idea of the intuitive understanding of entropy as the degree of randomness or disorder. Then, how can the entropy change of a process of conversion of an optically pure compound to its racemic mixture be computed theoretically or atleast estimated accurately enough so that it can be compared for two different reaction outcomes or the spontaneity of a reaction be estimated?
Moreover, which has higher entropy, a racemic mixture or a mixture of all possible stereoisomers of a compound including the meso form? In the later case, there is more randomness and hence should possess greater entropy. Can we treat a mixture of stereoisomers as a mixture of two different compounds and then use their mole fractions (or activities) to calculate the entropy of mixing which in case of gases is $-nR(x_1\ln {x_1} +x_2\ln{x_2})$.

There is really no way to address this question based upon the completely untrue idea that entropy is the same thing as randomness.  Entropy is the distribution of states available.  You  need to quantify whether or not a change in entropy increases or decreases the number of states available.


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Polyethylene terephthalate reaction with tap water
I have searched the internet and I found that reusing PET no.1 is dangerous. Why it is dangerous and what is the equation of the reaction between it and water tap ? And why it is not danger to when it is used for the first time in water bottles but it is danger when reusing it ?
Thanks!
Links I have found useful:
http://greenliving.nationalgeographic.com/plastic-drink-bottles-safe-reuse-drinking-water-3110.html
http://www.plasticsinfo.org/Main-Menu/MicrowaveFood/Need-to-Know/Plastic-Bev-Bottles/The-Safety-of-Polyethylene-Terephthalate-PET.html

It is probably not a chemical reaction that's the problem, it's that the container should be cleaned before it's refilled, and an unclean container might or will have bacteria.  While you're thinking 'refilled' might be in a matter if minutes, someone labeling the bottle is thinking about it sitting on a shelf with the cap off for months before reuse.  At that point there could be any number of unhealthy contaminants.

But as I understood from your words that it is safe to reuse these bottles if only they are cleaned. Is that means it can be reused ?

You might try http://biology.stackexchange.com .
In short - do not cultivate bacteria more than necessary. If you can wash and dry, do it, if not, don't. If it smells, definitely do wash and dry. If it does not help and you can afford it, throw away.

PET bottles are perfectly safe for repeated use; they will not degrade or undergo any chemical change.
However, it IS very important to thoroughly wash (with soap and water), rinse and DRY the bottle so there is no moisture left in the bottle where bacteria might grow.  This is true of any type of bottle, but especially one with a narrow neck where it can be difficult to thoroughly dry the inside of the bottle.


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How to disarm and correctly dispose of tear gas spray?
I have a can of spray for self-protection which has expired way back in 2008. It hasn't been used at all, and it doesn't seem a good idea to just throw it away without disabling it somehow first. I'm looking for ways to disarm it with minimum risk to myself and others. Obviously, I can just spray it away, but I'd rather not if I can avoid it.
The active ingredient is 2-chlorobenzalmalononitrile.
It's not water-soluble, but I was thinking of some combination of cooking oil and alkaline agent (baking soda, for example) to neutralize it. I'm still doing research on it, and I won't be doing anything rash just yet :-)
Any particular advice how to neutralize this spray?
If jurisdiction is somehow important, assume European Union.

Do not open, do not try to neutralize it yourself! Bring it to Recycling center (DE: Wertstoffhof). If your city is 100k+ there should be one. That's the place where you bringing your old furniture, the rubbish after renovations, and so on. They have a part for disposal of hazardous waste (DE: Sondermüll). What usually comes in there are paints, old car batteries, old pesticides, etc.... Another possibility - if you have some friend at chemistry school of some local university, give it him. Each Chemistry faculty has some "disposal" facility.

Good idea about Chemistry faculty, we have one in the university. I have to check with them if they can take care of this.


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Can a spark ignite hydrogen?
Can a spark ignite hydrogen? Also, what about the tip of a cigarette? I've heard that hydrogen is very explosive. Thanks

H2 is not explosive. Appropriate mixes of H2 with O2 or possibly Cl2 can be explosive.

Can a spark ignite hydrogen? Also, what about the tip of a cigarette?

Yes, if the $\ce{H2(g)}$ concentration in oxygen or air is greater than about 4% not only will it burn, but most likely explode. You should avoid all heat sources when dealing with it, including direct sunlight. Just so you know, hydrogen safety datasheets include words such as "extremely flammable" and "DANGER" in red capital letters.
Besides its high flammability, another thing that makes it so dangerous is that you can barely see its flames when there's a high oxygen concentration, as seen below. It also reacts violently (A.K.A. explodes) with most oxidizing compounds, so you should be careful where you are storing it.
A rocket engine at full thrust. There's a barely visible bluish flame below it, that's the hydrogen-oxygen mix used as fuel.

Actually hydrogen is not at all explosive. The mixture of hydrogen with oxygen (also air) is. As you know hydrogen is stored as liquid (high pressure cylinders) and it's perfectly stable. You can mix it with any amount of argon heat it as high as you want and nothing will happened. OK, if you reach temperature as high as on the Sun then it will explode, but that's an nuclear fusion (i.e. physics not chemistry). On the other side the mixture of hydrogen with oxygen can very easily by ignited by an electric spark or a cigarette.

Even at the Sun's core temperature Hydrogen is incredibly stable. After all, the Sun hasn't burned even half its Hydrogen after all these billions of years, and that's despite the presence of C,N and O catalysts.

This is slightly misleading. If you ha a room completely filled with hydrogen then, yes, a spark would not ignite the hydrogen. But few such rooms exist. What if you have a balloon filled with pure hydrogen? Then that is highly flammable and can be ignited with a spark despite the hydrogen being pure. It won't explode but it will burn. The same balloon filled with an oxygen/hydrogen mixture will explode violently.

Hydrogen is flammable only when mixed with oxygen, hence this can be mixed with compressed oxygen in petrol engines (hydrogen gas & oxygen gas) to be used as fuel.

I have accidentally blown up a small container with a mixture of hydrogen, oxygen and the water I was using to extract the gasses. Well I was using DC electricity to get the gasses and there was a spark that happened in the small container and it was a violent explosion and so loud people were driving around looking to see where it came from. I was lucky not to be hurt. There were pieces of the plastic container embedded in the wood framing of my garage. It wasn't much gas either.


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What is the nature of water?
Is water neutral in nature or amphoteric in nature? I understand it as an neutral compound in complex compounds, but I've been told that it is amphoteric, too.

Both. Having a pH of 7 it's by definition 'neutral'. But as a polar molecule and its ability to spontaneously dissociate into H and OH ions, water is often called the 'Universal Solvent' and can therefore be considered amphoteric.

$\ce{H2O}$ is a both a neutral molecule and an amphoteric molecule — the two are not mutually exclusive terms. An amphoteric molecule is simply one that can act as either an acid or a base, while a neutral molecule is one in which the total number of protons is equal to the number of electrons, such that there is zero net charge. Water can act as a Brønsted acid (i.e., an $\ce{H+}$ donor), as per the following reaction:
$\ce{H2O + NH2- -> OH- + NH3}$
However it can also act as a Brønsted base (i.e., an $\ce{H+}$ acceptor):
$\ce{H2O + HCl -> H3O+ + Cl-}$
When proton transfer is involved, as in the Brønsted definition of acids and bases, the more precise term is amphiprotic. However, water can also act as both a Lewis acid (an electron pair acceptor) and Lewis base (an electron pair donor). Notably, as a Lewis base, it is not limited merely to proton abstraction, but can act as a nucleophile with any number of electrophilic molecules.


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Extracting Zinc (oxide) from baby powder
I have a bottle of baby powder that is 15% Zinc oxide and the rest is talc. Is there an over-the-counter way of purifying zinc oxide from it? I'm trying to make some sunscreen from it.

This might be not worth it, because you will need special equipment.
Description totally theoretical.
Mixture consist of $Mg_3Si_4O_{10}(OH)_2$ - tale powder, $ZnO$ - zink oxide powder, and starch. 81% to 15% to 4% it's most common variation with 15% zink oxide, there also 10% mixtures, but process will be the same.

You need to get rid of starch. Best choice to convert it to dextrine, because starch in not fully soluble in water. You need to heat it in oven on a foil for about 2 hours, with temperature $200^{\circ}C$. It will have no effect for other components, zync oxide might colorize to orange, but when heat is gone it will be white again. Dextrine will be yellow color and it's fully soluble in water.
Mix new mixture with water, and pour it out. Now you have only $Mg_3Si_4O_{10}(OH)_2$ and $ZnO$ in you mixture. Both of them not soluble in water.
Then you need to obtain or build equipment for sublimation - sublimation apparatus. Because compounds have different melting point Talc - $800^{\circ}C$, zync oxide - $1975^{\circ}C$, you can get them one by one. You will need to freeze and dry your mixture first.
Heat over $800^{\circ}C$ to obtain talc, left over structure will be zinc oxide.

I'm not sure about sublimation from start, with starch inside, but this method in theory should work.

It sounds great till the 800 degrees part. Could I potentially throw it all in HCL and then into sodium carbonate solution? Or will I get some magnesium oxide from the talc as a result?


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What is the Molecular Formula for CH3-C≡C-CH2-CH3
I am taking an Organic Chemistry class in my college and I was wondering what the Molecular Formula for CH3-C≡C-CH2-CH3 was.
It would also be nice to have the class of compound.

Seems like 2-Pentene http://en.wikipedia.org/wiki/Pentene

@RyanArmstrong, 2-pentene is not the correct name for this compound. The triple-bond makes it an alkyne, not an alkene. Jerry has correctly identified the compound in his answer below as 2-pentyne/pent-2-yne.

Are you sure you didn't mean the name of this molecule? Because strictly speaking, the molecular formula of the molecule you posted would be $\ce{C5H8}$. Pent-2-ene has molecular formula $\ce{C5H10}$.
The molecule you asked about has a triple bond, meaning that it's an alkyne and in this case, pent-2-yne. If you want the wiki page, it's here.
Count the number of carbon atoms and the number of hydrogen atoms to be sure.


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What is the energy released in chemical reactions?
I have heard that chemical reactions create energy.  I wanted to know how this energy was created (specifically light energy) and how this energy came about.  I wanted to know if the energy produced was made of particles (for example, light energy is made of photons).
My answer I already knew about heat, kinetic, potential, and sound energy.  What I was really looking for was light energy, which is caused by a phenomenon known as chemiluminescence.

Your question is a bit vague.  There are many forms of energy and nearly all of them can be the result of a chemical reaction or interaction with chemicals.  I'm also not sure I understand what you mean about your either/or: subatomic particles (e.g. electrons) play a pivotal role in chemical reactions, but "devoid of physical form" - I don't follow.

Energy is usually expressed as a "wave" or other vague term.  I'm asking for a precise explanation of what it is, and I want to know if the energy from a chemical reaction is a "wave" or if it's more tangible, like radiation.

@Stopforgettingmyaccounts... you seem to be very unclear on energy. Energy can be potential energy, which is basically due to interactions between particles. For example, if you keep two electrons close the energy is potential energy. Bond energy is another form of chemical energy.

On the other hand, kinetic energy is due to the motion of particles. In this case, that's the radiation you talk of -- both electromagnetic (light, X-rays, infrared) radiation and from other particles like electrons (and the radiation you get from radioactive materials -- but usually this kind of radiation is different from radiation from a chemical reaction).

Electromagnetic radiation has a wave nature, but you can look at it as a bunch of moving particles, too.

Usually it's radiation. Depending on how exotermic reaction was, in which medium was it done, result is light staring with infra-red up to v.short ultra-violet. You can't generate particles like electrons  because  the low of Charge conservation must be obeyed (you transferring the electrons from one substance to the other).

The energy given of by chemical reactions is not usually directly radiation, it is usually heat, and the heat is often converted/observed as infrared radiation.  Here is a paper that models a simple chemical reaction using only rotational and vibrational motion, no direct radiation emission to accurately predict the chemical rate constant:  https://doi.org/10.1063/1.461407  The exception of course as mentioned is chemiluminescence

Actually chemiluminescence is emission from the reaction products, not directly emitted during the reaction

Let's start off on common footing by making sure we understand the definition of energy:
Energy is the capacity to do work.
An object may have many types of energy, we'll get to those in a second, but the total energy of an object will be the sum of its kinetic energy and its potential energy.  Kinetic energy is the energy associated with the movement of an object and potential energy is the energy associated with an object's position.
To take a physical example, hold out a book (preferably your heavy, expensive general chemistry textbook) and drop it on the floor (preferably, not on your little brother's toes).  The book falls to the ground and a loud noise is heard.  As the book moved, the potential energy stored in its original position was converted to kinetic energy as it fell to the ground.  Upon hitting the ground the individual atoms that make up the book crashed in to the individual atoms that make up the ground, causing them all to move.  This movement of atoms is thermal energy.
Let's focus on thermal energy for a moment, the majority of chemical reactions, including the combustion reactions you refer to, involve the transfer of thermal energy.  Hold a cup of coffee in your hand; what happens to your hand?  Assuming the coffee was hot, and your hand was not, your hand starts to feel warm.  We might say your hand is heating up.  This thing we refer to as *heat is the transfer of thermal energy; the molecules in the cup of coffee were bouncing around, crashing in to the cup, which made the molecules in the cup crash around hitting your hand.
So now let's turn to chemical energy which is the energy associated with the position of nuclei and electrons in atoms and molecules. Since we are talking about (relative) positions of objects, this energy is a form of potential energy.  Some molecules, such as methane, have nuclei and electrons that are positioned in such a way that they store a large amount of potential energy.  When the positions of these nuclei and electrons are moved, potential energy is converted into kinetic energy.
So in summary, the energy of chemical reactions is best described in terms of potential and kinetic energy, or in other words the relative position and movement of nuclei and electrons.  Often, the energy we experience in chemical reactions can be adequately described by the transfer of heat, and therefore thermochemistry and thermodynamics take up a large portion of our instruction in Chemistry.  Other types of energy, such as electromagnetic radiation, electrochemical, and nuclear, can all be described in terms of potential and kinetic energy, namely the position and movement of objects.

Thanks.  It was actually light energy that made me want to ask the question.  I knew that it was made of photons and couldn't figure out why or how it came from chemical reactions.  But I did under the term 'chemiluminescence' on Wikipedia.

@Stopforgettingmyaccounts... you might want to edit your question a bit to reflect where it came from.  Your example of a combustion reaction causes a bit of confusion.


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Photosynthesis, science fair
So I was thinking about something to do for a science fair and photosynthesis sounded like a good idea.  I was wondering what the exact process of photosynthesis was on a chemical level.  I know the formula, but how exactly is water and carbon dioxide converted into glucose and oxygen?  And where does sunlight come in, not in the equation but the chemical process.  Finally, any ideas for artificial photosynthesis?

Photosynthesis is a very broad topic for a science fair project.  You might want to explore how to design a project and also a Google search on artificial photosynthesis might be an important step.

I am assuming you are in grade school.
The problem is, as was mentioned before, that photosynthesis is very complicated.
The formula you know of is extremely simplistic and skips over 100 reactions on the way. you can find plenty of information in Wikipedia, or if that is too much you can use the Simple Wikipedia.
In high school we took some kind of water plant or algae and put it in an upside down test tube filled with water, as oxygen was formed it filled the end of the test tube. This is a fairly simple way to monitor photosynthesis and you can use it to study photosynthesis. for example: how does the temperature effect photosynthesis, how does the color of light affect it etc.

This paper provides a very good overview of the state of knowledge about the mechanism of water oxidation (which is pretty much the heart of artificial photosynthesis). So if you want to understand the process of oxygen formation from water on a molecular level this is a very good and fairly recent review of the field, which treats the heterogeneous, homogeneous and biological catalytic pathways proposed in the literature. It is solely concerned with oxygen formation and doesn't say anything about the formation of glucose from carbon dioxide - for details on that process you should look into a biology textbook. Furthermore, I want to add that the mechanistic details are still very much a matter of debate.