text:
producing ozone through electrolysis of water
I would like to produce low concentrations of ozone through electrolysis of water. The standard setup to do this uses a graphite electrode and a platinum electrode in an $\ce{H2SO4}$ solution.  However, I'd like to produce the ozone using only water and preferably using inexpensive electrodes.  I don't seem to understand the exact chemistry of this, though.  I'd like to know:
If I use different electrodes, what changes?  I'm currently using zinc and graphite and, other than the fact that the zinc degrades, is there any downside to using it?
The $\ce{H2SO4}$ solution, I expect, increases the conductivity of the water, is that right?  Why is it used?  What if I don't use it?
How can I go about detecting the ozone in the water?  When I run my system I seem to smell something ozone-y, but I'd like a much less subjective method than that.

First of all $\ce{H2SO4}$ produces $\ce{H2SO5}$ which is a peroxyacid during electrolysis  which most electrolytes will not do, so I suspect that $\ce{H2SO4}$ will be mandatory unless phosphoric acid ($\ce{H3PO4}$) can.  This peroxyacid give you a potential for a mechanism of ozone creation.
$$\ce{H2SO5 + O2 -> H2SO4 + O3}$$
Note: I do not claim this to be the mechanism but only an explaination for why the $\ce{H2SO4}$ may be important.
The carbon and platinum electrodes are not only chemically resistant, but may likely have a catalytic effect in producing ozone.  You can get platinum wire at a reasonable price if you look around.
In regard to testing if you have ozone, you can use paper strips that have potassium iodide impregnated in them.  The ozone will oxidize the iodide ion and turn the paper strip brown.

Great, thanks.  I'll try the new reaction and will try to hunt down some KI.


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How can I have a concerted addition of bromines?
I know how to add Bromines in the antiperiplanar position, but I have no idea how to get concerted addition:

Perhaps I am not seeing something obvious, but I would greatly appreciate any hint.

As briefly mentioned in the answer, concerted does not mean what you think it means. It means that a reaction proceeds in one step, without any intermediates. Examples of concerted reactions are the $\mathrm{S_N2}$ reaction, or the Diels-Alder reaction - see Wikipedia.

The proper term for concerted addition is syn addition.  I can't find any way of a direct way of producing a syn-addition product.  The only way I can figure you would have to do it is if you did the halogenation with water as the solvent you would get the halo-hydrin product with an anti-addition.  You could then follow this up with a bromination using $\ce{P,Br2}$ or $\ce{PBr3}$ which would invert the chirality and give you a syn-product.

Thank you for your answer! P is for "products", right?

I don't how a bromination would work since I don't have a double bond... Sorry. :)

P is for phosphrous I will update the answer

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pH of solution after adding SiCl4
Why is it that adding $\ce{SiCl4}$ will cause the pH of a solution to become acidic? I thought that only salts (ionic bonds break in water) can change the pH once dissolved in water. $\ce{SiCl4}$ has covalent bonds, so why does the pH change?

Silicon is very oxophilic and $\ce{SiCl4}$ irreversibly hydrolyzes in water as follows:
$$\ce{SiCl4_{(l)} + 4 H2O_{(l)} -> 4 HCl_{(aq)} + Si(OH)4_{(s)}} $$
The release of $\ce{HCl}$ creates hydrochloric acid and lowers the pH.

I wonder if the reaction in water would be: $$\ce{SiCl4 + 2 H2O -> 4 HCl + SiO_{2(s)}} $$

Depends on concentration I think

$\ce{Si(OH)4}$ will reversibly dehydrate and coagulate to some extend forming a sol over time. But when you use aqueous processes to form $\ce{SiO2}$ you get a substantial amount of $\ce{Si(OH)4}$  first.

$\ce{Si(OH)4}$ exists as $\ce{SiO2.2H2O}$ called Silica Gel. $\ce{Si(OH)4}$ as such does not stand for long, probably due to overcrowding of electrons around Si atom by $\ce{4 OH-}$ groups. Hence, $\ce{Si(OH)4}$ stabilises itself as $\ce{SiO2.2H2O}$

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Why are Pd and Pt completely soluble in Cu, but Ag only partly in Cu when forming a substitutional solid solution?
Why are Pd and Pt completely soluble in Cu, but Ag only partly in Cu when forming a substitutional solid solution? If we look at the Hume-Rothery rules, they all are within the 15% atomic radius factor with respect to Cu and all have the same crystal structure:
+----+---------------+-----------------+-------------------+---------+
|    | Atomic Radius | Crystal Struct. | Electronegativity | Valence |
+----+---------------+-----------------+-------------------+---------+
| Cu | 0.1278        | FCC             | 1.9               | +2      |
| Pd | 0.1376        | FCC             | 2.2               | +2      |
| Pt | 0.1387        | FCC             | 2.2               | +2      |
| Ag | 0.1445        | FCC             | 1.9               | +1      |
+----+---------------+-----------------+-------------------+---------+

The last two rules are:

The solvent and solute should have similar electronegativity. Intermetallics tend to form when the difference is large.
Other factors being equal, a metal will have more of a tendency to dissolve another metal of higher valency than of a lower valency.

Then Pd and Pt have an electronegativity difference of 0.3 while Ag has 0.
Pd and Pt have same valency while Ag has one less.
I would have expected that Ag would form a complete solid solution while Pd and Pt would only partly form a solid solution with Cu because difference of 0.3 in EN is quite large, because the valence difference between Ag and Cu is only 1 electron and because of the words: 'Other factors being equal,' it sounds like the other rules are more important.
To sum up, my thought process was:
Pd, Pt and Ag meet rule 1 and 2. Pd and Pt don't meet rule 3, so only form partly solid solutions. Then Ag meets rule 3 and rule 4 is less important and yet it is still close, so Ag will completely dissolve in Cu.
Why is this thought process wrong? If a difference in EN of 0.3 isn't big, what is then the border? Like 15% is the border for rule 1?

Hume-Rothery's rules aren't hard-and-fast rules, they are guidelines. Not every binary metallic mixture (more than 2200 possible) has been deeply investigated, and the ability to do so via computational modeling is only just now becoming possible. Any deeper understanding goes beyond the scope of materials engineering into chemistry or, more likely, physics. To understand fully why Hume-Rothery doesn't always work as expected would require detailed analysis or modeling of electron orbital interactions, and for transition metals it can get fairly complicated. Or so I understand.

And let's not throw Ni in there as well.  The answer is: it is complex, and simple rules can't address it with any precision.

I think it's apparent from this that the difference in valence state is a more significant attribute than electronegativity.  The electronegativity does narrowly fall within the 15% which is only an approximation ($\mathrm{0.3/2.2 = 0.136}$; $\mathrm{0.3/1.9 = 0.158}$).

Although a question on liquid phases, the point remains: http://chemistry.stackexchange.com/questions/35014/miscibility-of-molten-metals/35151#35151 - if you want I could summarize the fcc noble / near noble metal situation, but it is similarly complex. For example, Cu-Ni shows a broad range of full miscibility, with a solid fcc miscibility gap starting to form below 400C. In Au-Ni that miscibility gap starts around 600C.  For Ag-Ni, there is no temperature where there is no miscibility gap in the solid, and the miscibility gap extends even in to the liquid!


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Why are certain substances clear/invisible? Can we create invisible objects?
This is more so a question of how light interacts with matter. For example, why are substances like oxygen generally invisible? If the light hitting the molecules were absorbed, it would appear black, right?
Also, could we create invisible materials?

Well, that just means that oxygen does not absorb in the visible region of the spectrum.

http://chemistry.stackexchange.com/questions/16633/why-is-gold-golden

Mankind has done so since ancient times. Glass is effectively invisible, hence why silly insects keep flying against it.

Invisible to all wavelengths of electromagnetic radiation, and all sorts of particle beams? - No. // Invisible to a certain  range of the electromagnetic spectrum (like human eye range)? - Yes.


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Derivation of mean speed from Maxwell–Boltzmann distribution
I found that if a velocity of a gas follows the Maxwell–Boltzmann distribution, the mean velocity is given by
$$\langle v \rangle = \sqrt{\frac{8RT}{\pi M}}$$
where $R$ is the gas constant, $T$ is temperature, and $M$ is molar mass. How would one obtain this result from the Maxwell–Boltzmann distribution?

If one has a probability density function $P(x)$, then the expectation value of a quantity $f(x)$ is given by
$$\langle f \rangle = \int f(x)P(x)\,\mathrm{d}x$$
evaluated over the limits of the probability density function, i.e. if your PDF runs from $-\infty$ to $\infty$, then those are your limits of integration.
In our case, the PDF is the Maxwell–Boltzmann distribution (denoted $P(v)$ here) and the quantity we want to find the expectation value of is simply the velocity, $v$. Since velocity can only be positive,* the limits of integration are from $0$ to $\infty$.
Therefore, the mean velocity $\langle v \rangle$ is given by the integral
$$\langle v \rangle = \int_0^\infty v P(v)\,\mathrm{d}v$$
where $P(v)$ is the Maxwell-Boltzmann distribution
$$P(v) = \left(\frac{m}{2\pi kT}\right)^{3/2}4\pi v^2 \exp{\left(-\frac{mv^2}{2kT}\right)}$$
So:
$$\begin{align}
\langle v \rangle &= \int_0^\infty v P(v)\,\mathrm{d}v \\
&= 4\pi\left(\frac{m}{2\pi kT}\right)^{3/2} \int_0^\infty v^3 \exp{\left(-\frac{m}{2kT}v^2\right)}\,\mathrm{d}v
\end{align}$$
The integral can be evaluated using integration by parts repeatedly. The process is not interesting and you could consult a table of standard integrals to find the result:
$$\int_0^\infty v^3 \exp{(-\alpha v^2)}\,\mathrm{d}v = \frac{1}{2\alpha^2}$$
Setting $\alpha = m/2kT$,
$$\begin{align}
\langle v \rangle &= 4\pi\left(\frac{m}{2\pi kT}\right)^{3/2} \frac{4k^2T^2}{2m^2} \\
&= \sqrt{\frac{8kT}{\pi m}} \\
\end{align}$$
$m$ here refers to the mass of one molecule, whereas $M$ in your question refers to the molar mass of the compound. They are related by $M = N_\mathrm{A}m$, where $N_\mathrm{A}$ is the Avogadro constant. Since $R = N_\mathrm{A}k$, you can multiply top and bottom by $N_\mathrm{A}$ to obtain the desired result
$$\langle v \rangle = \sqrt{\frac{8RT}{\pi M}}$$

Footnote
* If the assertion that velocity is non-negative is confusing, it's worth noting that the $v$ in the Maxwell–Boltzmann distribution $P(v)$ is not the vector quantity $\vec{v} = (v_x, v_y, v_z)$, but rather the magnitude of the velocity $v = |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$. Part of the derivation of the Maxwell–Boltzmann distribution involves converting from the vector quantity to its magnitude: that's where the $3/2$ exponent comes from (it's actually $1/2$ per dimension in three dimensions), and also where the factor of $4\pi v^2$ comes from ($4\pi v^2$ is the formula for the surface area of a sphere: loosely speaking, this factor represents the "collection" of all points $(v_x, v_y, v_z)$ which have the same magnitude $v$). A fuller explanation may be found in any typical physical chemistry textbook.

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Why do octahedral metal ligand complexes have greater splitting than tetrahedral complexes?
Octahedral complexes have greater splitting in the d orbitals. Is it because octahedral complexes have more atoms and thus more interactions?

The answer is, as @orthocresol mentioned, due to two parts.
Initially, if we are assuming the ligands are point charges (crystal field theory) then in octahedral complexes, the ligands are more closely associated with the d orbitals, as shown in this diagram:
But, in tetrahedral complexes, the ligands are not so closely associated with the d orbitals, as shown in this diagram:
By associating the d orbitals seen in the first diagram, with the tetrahedral point charges in the second diagram, you can see how close the point charges are to the d orbitals in the octahedral case compared to tetrahedral.
Also as otho mentioned, octahedral has 6 ligands interacting with 5 d orbitals, whereas tetrahedral has only 4 ligands interacting with 5 d orbitals. Octahedral has the more negative charge due to more electrons, therefore more repulsion.

The crystal field theory given in Benzene’s answer is a nice simple model, but we can get a deeper, maybe more logical explanation if we check out molecular orbital theory. For a general octahedric complex, the MO scheme looks like depicted in figure 1 (only σ-donors, π effects not included because I was too lazy to draw another image).

Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.
We realise that the $\mathrm{e_g^*}$ orbitals are raised in energy because they are antibonding to the metal-ligand bonds. $\mathrm{t_{2g}}$ orbitals are not affected. Including π basic effects on the ligands, they would be raised slightly because they again would be antibonding. These interactions always stem from 6 ligands. Also, $\mathrm{e_g^*}$ are actually directly pointing at the ligands as Benzene nicely showed.
A typical tetrahedral complex, this time including π interactions to a π basic ligand is shown in figure 2.

Figure 2: Tetrahedral $\ce{[ML4]}$ complex including ligand to metal π donating orbitals. Image copied from this answer.
Here, the ligand orbitals that interact in a σ manner transform as $\mathrm{a_1 + t_2}$ so the stronger destabilisation is present on $\mathrm{t_2^*}$. $\mathrm{e^*}$ is also destabilised due to being antibonding but lesser so. It is a little bit hard to see in Benzene’s image, but all of the orbitals are now pointing away from the ligands they are interacting with. Thus, the interaction must be weaker than for the octahedral case above. Finally, we also need to consider that we are dealing with the mixing of less orbitals here: there are only four ligands this time further reducing the energy difference.


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Technecium and Technetium
Was technecium "recently" renamed technetium in English? I know when I formally studied chemistry (in the USA) that earth, air, water and fire were on the periodic table, but I really remember the element as being technecium.

After German chemists claimed to have discovered element 43 in 1925, the name masurium was used in the first publications. However, the German experiments could not be reproduced.
The official discovery of element 43 was accomplished by C. Perrier and E. Segrè in 1936. They proposed the name technetium (see Perrier, C.; Segrè, E. Technetium : The Element of Atomic Number 43, Nature, 1947, 159, 24.).
In 1949, the 15th IUPAC Conference adopted this name.
The name technetium was also used in the early American literature, for example in Radioactivity Applied to Chemistry; Wahl, A. C., Bonner, N. A., Eds.; John Wiley & Sons: New York, 1951.

There are plenty of English publications where it is spelled "technecium" https://books.google.com/books?id=J6k4-N5lPX8C&pg=PA275&dq=%22technecium%22&hl=en&sa=X&ved=0ahUKEwjLj7WequjLAhVIRyYKHZpRAkgQ6AEIIjAB#v=onepage&q=%22technecium%22&f=false It's either a common misspelling or an alternative spelling.

And may with "Technicium" https://books.google.com/books?id=WuFFAAAAYAAJ&q=%22Technicium%22&dq=%22Technicium%22&hl=en&sa=X&ved=0ahUKEwi7_d6irOjLAhXCMSYKHZUwAYoQ6AEIJjAC https://books.google.com/books?id=8ztrAAAAMAAJ&q=%22Technicium%22&dq=%22Technicium%22&hl=en&sa=X&ved=0ahUKEwi7_d6irOjLAhXCMSYKHZUwAYoQ6AEIQzAI

LOL - That "early American literature" was what I cut my teeth on. My first high school chemistry class was in 1967.

It's always been technetium in English.
In fact, I can't find an English entry for technecium on the internet at all.  I'd venture to guess you're Czech based on the search results, which did yield other language pages.

No, I'm American not Czech.

Here is a black swan, one of many actually...  https://books.google.com/books?id=dQ0neMLQNskC&pg=PA1194&dq=%22technecium%22&hl=en&sa=X&ved=0ahUKEwjXr9-y85HPAhVE7SYKHetkClMQ6AEIMjAE#v=onepage&q=%22technecium%22&f=false


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Do polycarbonate and/or polyurethane have latex?
I have a severe latex allergy which hinders my health every day via hives, inflammation, and breathing trouble. I am trying to find a cell phone that is not going to cause anaphylaxis or other reactions. Neither smartphones nor their companies are smart enough to know what their products are made of.
I know from wheelchair experience that polyurethane bothers me. I believe it has latex or a latex-like component in its make-up, but I would like to clarify this.
It appears by touch that polycarbonate is not as bad, but it just depends on the item in question...and usually I have already reacted to something else before trying to touch this.
Could you verify/explain if polycarbonate or polyurethane have latex? For that matter does polyethylene?

You should perhaps consider looking for a silicone case.  Silicone is completely synthetic and inert.  This should not carry risk of an allergic reaction.

@A.K. Thank you very kindly. That was another material I was unsure of. I will look into finding and trying one.

@GeoffHutchison Thank you very kindly for answering and providing this information.
You are correct, I am and was not asking a medical question nor medical advice. I am, in all honesty, working to learn more about materials and to educate myself about what to avoid and watch for with my severe allergy. The quest is daunting, to say the least, and help is often far away or, though filled concern and compassion, lacks the knowledge required...

@GeoffHutchison Polymers are so much a part of our daily lives these days that it is difficult if not impossible to avoid them. And, the manufacturer who strives to be helpful, caring, considerate, and honest about how and what their product is made of is a rare gem.
Speaking of a welcomed treasure... this site must be one. It is wonderful to have received an answer and help in gaining more of an understanding. May I add that it is also greatly appreciated that a chemist uses common sense and puts things plainly for the rest of us to understand and educate ourselves.
Thank you very Kindly.

@Lisa If you feel that way I hope you will join some of the other communities and ask and answer questions in the future.  Here you don't need to know everything about everything, just something about something, and you will typically relearn things you have forgotten.  Enjoy!

We have a general rule against "medical questions," but I think yours is different.
First off, latex is a form of poly-isoprene. The "isoprene" part here is the key chemical identity - the letter (so to speak) that repeats in the polymer.
So latex is chemically quite distinct from polyurethane and polycarbonate. Similarly, polyethylene and polypropylene are also distinct.
In short, from a chemical perspective, all of these materials are all different.
Now, that's not to say that:

You may have allergies to multiple polymers / plastics.
Manufacturers may blend multiple chemical components into one object.
Some latex residue may end up on various objects.

For example, many people in healthcare (as you know) use latex gloves, often with latex powder. Small amounts of the latex powder can remain on a wheelchair - causing problems for sensitive individuals like yourself.
If you're looking for a smartphone, you should be able to find one with a metal / glass case (e.g., iPhone, but also others). In that case, the amount of plastic (polymer) should be minimal and you (or a friend) can thoroughly wipe down the device before you use it (e.g, the wrapping plastic might cause sensitivity).

I believe with the mistake wheelchair hand rims, that the polyurethane had been blended with rubber/latex. That's why I was curious if polyurethane is always this way or just may be depending on the product. Thank you for explaining. I gather, now, that this is a possibility not the rule.

I have a severe latex allergy, what you're reacting to in the wheelchair armrest is the FOAM inside. Most foam contains latex. So if the foam is compressed, it puts some particles air borne that you're inhaling. I think you can get latex free wheelchairs/armrests, you have to look for it though.

I have a latex allergy and reacted to a charging cable - the manufacturer told me afterwards (when issuing my refund) that their PU was mixed with latex but that this isn’t always the case.
I would also agree with another comment that foam, especially low density foam, has latex in it (think make up sponges), the glue in some plasters does and also the shine on gloss paper is made by adding latex to the coating.
Good luck negotiating all the ways it’s in life, Nina xx

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Why should a system try to reverse any change that has been done to it?
I understood Le Chatelier's principle perfectly but my question is: Why should a  system try to reverse any change that has been done to it?
I understand equilibrium constant, mole concept and partial pressures but I do not understand quantum mechanics yet. If you use any other technical terms in your answer, please also explain them. Answers with intuitive examples will be highly appreciated.

At an abstract level, Le Chatelier's principle is a (not so obvious) consequence of the second law of thermodynamics. The second law, along with the fact that a small part of the system should be in equilibrium with the rest, leads to thermodynamic stability conditions, which can be summarized as the entropy being a concave function (or, equivalently, the internal energy being a convex function). Le Chatelier's principle follows from the concavity/convexity requirements.

Further reading: Section 22 of L. D. Landau and E. M. Lifshitz, "Statistital Physics Part I",  3rd revised and enlarged ed. (1980) and Section 8.5 of H. B. Callen, "Thermodynamics and an Introduction to Thermostatistics", 2nd ed. (1985).

Would be great if I could write a detailed answer exemplifying this point, but  it seems difficult to produce something more readable than what is in the books I  recommended.  I might try when I have enough time and energy.

Well, lets first write out the general equation of a chemical reaction:
$$\ce{aA + bB <=> cC + dD}$$
As you obviously know, since this is at equilibrium, not all the reactants react to form the products. Hence in a solution which is at equilibrium, all species of $\ce{A, B, C, D}$ will be present. However the special thing is that the ratio of the concentration of these species will usually be constant, no matter how much of each species you start with. This ratio is known as the equilibrium constant. The only thing that effects this ratio is temperature. The equilibrium constant is given by:
$$K_\mathrm{c} = \ce{\frac{[A]^a[B]^b}{[C]^c[D]^d}}$$
One more equation you need to know is the reaction quotient which gives the ratio of products to reactants at a certain time during a reaction and is given by:
$$Q_\mathrm{c} = \ce{\frac{[A]^a[B]^b}{[C]^c[D]^d}}$$ Note how this is similar to the equilibrium constant, but here, $\ce{[A]}$, $\ce{[B]}$, $\ce{[C]}$ and $\ce{[D]}$ are the concentration of the substance at a specific time during the reaction and doesn't necessary have to be when the system has reached equilibrium. This means that if $Q$ is smaller than $K$, the reactants will react to form more products so that $Q$ will increase so that it will eventually equal $K$ and when $Q$ is larger than $K$, vice versa. When $Q$ equals $K$, the system is at equilibrium.
So now lets apply this concept to help us explain Le Chatelier's Principle by considering different scenarios. Lets consider a mixture of hydrogen and nitrogen gas in a container. They react to form ammonia gas. The equation of this reaction is:
$$\ce{N2(g) + 3H2(g) <=> 2NH3(g) + heat}$$
Also the equilbrium constant is:
$$K_\mathrm{c} = \ce{\frac{[NH3]^2}{[N2][H2]^3}}$$
Scenario 1: Taking or adding more reactants or products
Lets say that this mixture of gases have finally reached equilibrium, so $K = Q$. Now I suddenly take some hydrogen has out.This would decrease the concentration of hydrogen gas and hence increase $Q$. So now $Q>K$. Therefore for the system to adjust itself, it needs to decrease the value of $Q$. It does this by creating more products and decreasing the amount of reactants. So in short, the equilibrium shifts to the left. This is what Le Chatelier's Principle predicts. Now by using the exact same logic, we can predict what happens if I were to change the concentration of ammonia, nitrogen, etc.
Scenario 2: Changing pressure or volume
Pressure and volume are inversely proportional to each other. Meaning that an increase is pressure will result in a decrease in volume and vice versa. This is due to the ideal gas law:
$$pV = nRT$$
Therefore, increasing the pressure of a system has the effect on the equilibrium  as decreasing the volume.
Now, back to example of hydrogen, nitrogen and ammonia gas in a container. If I were to suddenly decrease the volume of the container by two (equivalent to doubling the pressure) this would mean that the concentration of each species would double. Now, how would this effect our $Q$ value, would it be larger or smaller than $K$? Well the numerator would increase by fourfold while the denominator would increase by sixteenfold. Hence $Q<K$. Therefore as mentioned above, the equilibrium will shift to the right.
This can be applied to if the pressure was decreased, volume was decreased, etc.
Scenario 3: Temperature is changed
Now this one isn't really explained by using equilibrium constant. Instead will use a more intuitive approach.
For endothermic reactions, heat is required for the reaction to go forward. So by increasing the temperature, there is an increase in heat energy available to the system. Therefore the forward reaction will occur more readily. Hence it is expected that the equilibrium will shift to the right. Meanwhile, a decrease in temperature means less heat energy, hence equilibrium will shift to the left. The same logic can be applied to exothermic reactions.

For a simple reaction, we can derive the free energy change as
$$\Delta G (T,p)=\Delta G^\circ(T)+k_\mathrm{B}T\ln K_\text{eq}$$
Here, $\Delta G (T,p)$ is the free energy change of the reaction, $\Delta G^\circ (T)$ is the free energy change at that temperature but 1 bar pressure or you can use any scale of absolute pressure as long as you are using $K_\text{eq}$ expression correctly. Now $K_\text{eq}$ can be expressed as (for a simple $\ce{A <=> C}$ reaction; here $p_\ce{A}$ and $p_\ce{C}$ are partial pressures of $\ce{A}$ and $\ce{C}$)
$$K_\text{eq}=\frac{p_\ce{C}}{p_\ce{A}}$$
So, at equilibrium $\Delta G(T,p)=0$ yields
$$K_\text{eq}=\exp\left[-\frac{\Delta G^\circ(T)}{k_\mathrm{B}T}\right]$$
So far everything is fine and also the effect of temperature, pressure on equilibrium has been explained nicely in the above answer but still the question is why $Q$ goes to $K$ at equilibrium is unexplained.
At any particular temperature $\Delta G(T,p)$ would be zero if and only if ratio of $p_\ce{C}/p_\ce{A}$ is such that value of $\Delta G^\circ(T)$ is cancelled out by
$k_\mathrm{B}T\ln K_\text{eq}$ or in mathematical term
$$\Delta G^\circ(T)=-k_\mathrm{B}T\ln K_\text{eq}$$
So it's a single point on thermodynamic surface. Now if you change one pressure the other pressure must change itself to keep the ratio constant. On the hand if you change temperature your pressure ratio will also change to keep $\Delta G(T,p)$ zero.
Now the final question is, why is $\Delta G(T,p)$ is zero at equilibrium. This is something like gravitational surface. In gravitational surface every object will try to go as down as possible to make it's center of mass as close as possible to the gravitational field. On a macroscopic scale it can be seen as equilibriation of chemical potential. If you keep two different temperature body together they will exchange heat between each other and eventually end up being at the same temperature. In the same way if you mix two chemical compound they will try to exchange chemical potential between themselves (The currency here is conversion of one molecule into another) as long as their chemical potential is not equal. The above formulation can be written as
For $\ce{A}$, chemical potential is
$$\mu_\ce{A} (T,p)=\mu_\ce{A}^\circ(T)+k_\mathrm{B}T\ln\frac{p_\ce{A}}{p^\circ}$$
similarly for $\ce{C}$
$$\mu_\ce{C} (T,p)=\mu_\ce{C}^\circ(T)+k_\mathrm{B}T\ln\frac{p_\ce{C}}{p^\circ}$$
At equilibrium
$$\mu_\mathrm{A}(T,p)=\mu_\mathrm{C}(T,p)$$
And you will get
$$K_\text{eq}= \exp\left[-\frac{\Delta G^\circ(T)}{k_\mathrm{B}T}\right]$$
I hope it will help you to clarify the concept.


text:
Usage of ammine vs amine in nomenclature
In the nomenclature of complex salts we use ammine for NH3 instead of amine.
I thought this was to differentiate between ammine ligand and amine in organic ligand (like en).
However, Wikipedia cites the usage has historical reasons. So why exactly do we use ammine instead of amine?

I always remember it because it directly derives from ammonia (with a double-m). Whereas the amines are only modified versions of ammonia, they no longer can dissociate to liberate it.

(Will do more research into ammines come February.)
Premise
Ammonia is spelled with two 'm's. The more natural derivative of the word in a linguistic sense would also have the same number 'm's. Thus whichever predates the other, ammine complexes or amines, would claim the throne.
In a structural sense, when awknowledgement of the first ammine complex came it was literally in a solution of ammonia (see more below).
Etymology of ammonia and amine

Our travels begin in sunny Ancient Egypt. One important deity of the time was Amun, the god of wind and air along with his wife-in-mythology Amaunet. At the beginning of the Middle Kingdom of Egypt (c. $2055-$$1650$ BC), he was annointed patron saint of Thebes.  Later Amon become part of a trinity, along with gods Re and Ptah. This coincided with Thebes restored as capital city due to pharaoh Ahmose I (c. $1539-$$1514$ BC).
Eventually Amon would become the god of gods. This is when he also goes by Amon-Re (or Ra), after uniting with sun god Re.$^{\text{[a]}}$
Egypt's glory shone far beyond its borders. Ancient Greeks adopted their own version of Amon. There he would be called Ammon, with two 'm's.

As it so happens, near one of Amun's temples in then-Egypt is a depository of $\ce{NH4Cl}.$ From this háls ammôniakós (Greek for 'salt of Amun') we can easily extract $\ce{NH3}.$
$$\ce{NH4Cl(s) ->[t^\circ] NH3(g) +HCl(g)}.$$
It makes sense for the option with two 'm's to outdate amine. Etymonline agrees (in ~ English):

ammonia (n.) $-$ $1799$, modern Latin, coined $1782$ by Swedish chemist Torbern Bergman ($1735$$-$$1784$) for gas obtained from sal ammoniac [----];
amine (n.) $-$ "compound in which one of the hydrogen atoms of ammonia is replaced by a hydrocarbon radical," $1863$, from ammonia + chemical suffix -ine (2).

Note: has been expanded upon under 'Amine in chemical literature' further down.
The why
Why ammine for coordination compounds and not amine? It is probably because the study of metal ammines predates any knowledge of amines. Indeed, in $1798$ Tassaert$^{[1]}$ reported on 'ammoniacal solutions of cobalt(II) chloride [which when] allowed to stand overnight yield an orange-colored crystalline product containing six molecules of ammonia'$^{[2]}$. The compound in question was $\ce{CoCl3*6NH3}$. These and similar compounds were dubbed metal ammines$^{\text{[b]}}$.
Note that at the end of  $18^{\text{th}}$ century organic chemistry had not yet developed. The crucial and accidental evidence against vitalism$^{\text{[c]}}$ came in $1828$ when the German chemist Wöhler artificially synthesised an organic compound $-$ urea $-$ an amide with two amine qroups!
$$\ce{AgNCO + NH4Cl-> AgCl + NH4NCO \\ NH4NCO ->[t^\circ]NH_3 + HNCO <=>\underset{urea}{CO(NH2)2}}$$
First pivotal trace of amines pops up in $1842$, $44$ years after Tassaert's discovery of coordination chemistry. Russian chemist N. N. Zinin uses dihydrogen sulfide to reduce nitrobenzene$^{[3]}$$^{\text{[d]}}$:
$$\ce{C_6H5NO2 + 3H2S ->[NH3]\underset{aniline}{C6H5NH2} + 3S + 2H2O}.$$
On Zinin's research, co-founder of amine chemistry A. W. von Hofmann declares$^{[4]}$,

If Zinin had done nothing more than to convert nitrobenzene into aniline, even then his name should be inscribed in golden letters in the history of chemistry.

Aliphatic amines were simultaneously and independently synthesised in $1849$ by chemists Wurtz of France and Hofmann of Germany. Both could be credited with the discovery of ethylamine. Hofmann's method allowed for the additional preparation of secondary and tertiary amines.$^{[3]}$
Wurtz treated ethyl isocyanate with potassium hydroxide; Hofmann applied heat to solutions of alkyl halogenides and ammonia.$^{[3]}$
$$\tag{Wurtz}\ce{C2H5N=CO + H2O->[KOH]C2H5NH2 + CO2}$$
$$\tag{Hofmann, I}\ce{R-I + NH3->[t^\circ]R-NH2*HI}$$
$$\tag{Hofmann, II}\ce{2R-I + NH3->[t^\circ]R2-NH*HI + HI}$$
$$\tag{Hofmann, III}\ce{3R-I + NH3->[t^\circ]R-N*HI + 2HI}$$
For quaternary amines, Hofmann used the following scheme:
$$\tag{Hofmann, IV}\ce{R3N + RI->R4-N*I}.^{[3]}$$
But because ammine was already reserved, they went with amine.$^{[5]}$ Furthermore, if you think about it, the naming is rather successful. Well, at least for primary amines.$^{\text{[e]}\ \text{[f]}}$
$$\text{metal ammines contain the complete}\ \ce{NH3->} \text{keep the}\ m\\ \text{pr. amines have one less hydrogen or}\ \ce{-NH2->} \text{lose the}\ m$$
Amine in chemical literature
In $1834$, German chemist Liebig optained a dirty white or buff-coloured amorphous substance by heating ammonium thiocyanate.$^{[5]\ [6]\ [7]\ [8]\ [9]\ [10]\ [11]}$
$$\tag{$\alpha$}\ce{KSCN + NH4Cl ->[t^\circ] NH4SCN + KCl}$$
$$\tag{$\beta$}\ce{NH4SCN <=> \underset{thiourea as thione}{S=C(NH2)2} <=>[t^\circ] \underset{thiourea as thiol}{HS-C(NH)NH2}}$$
$$\tag{$\gamma$}\ce{\underset{thiourea as thiol}{HS-C(NH)NH2} ->[t^\circ] NH3 + HNCS \\ \underset{thiourea as thiol}{HS-C(NH)NH2} ->[t^\circ] H2S\ \ \ + \underset{cyanamide as aminonitrile}{N#C-NH2}}$$
$$\tag{$\delta$}\ce{\underset{cyanamide as aminonitrile}{N#C-NH2} <=> \underset{cyanamide as diimine}{HN=C=NH}}$$

He named the product melam without any admitted etymology other than his own fantasy. Indeed, Liebig mentions in his original paper$^{[5]\ [6]}$,

[These designations] are, if you like, grasped from the air (i.e. pure inventions), and serve the purpose just as well as if they were derived from the colour or one of the properties.

Liebig proceeded to boil melam with a solution of potassium hydroxide and ammonia. This yielded melamine.$^{[5]\ [6]\ [7]}$

Neither the structure of melam nor melamine were known to Liebig at the time.$^{[5]}$ Melamin(e) was most likely derived by Liebig as melam $+$ -in(e).$^{[5]}$$^\text{[g]}$
A few years before $1849$, Liebig had predicted the existance of amines.$^{[5]}$$^\text{[h]}$ Liebig called them compound-ammonias$^{[5]}$. The word amine itself was used already in $1853$$^{[12]}$ in the works of French chemist Gerhardt. Often times this was put in parentheses as an alternative name. For example$^{[12]}$,

méthyl-ammoniaque (méthylamine),
amyl-ammoniaque (amyl-amine).

Acceptance of the word amine was not immediate. Besides having to distinguish between amines and metal ammines, amides were too an obstacle. The restriction to retain amine for ammonia [where any] hydrogen atom [has been] successively replaced by hydrocarbon radicals was generally accepted in $1863$.$^{[5]}$
TL; DR
$$\text{Amun in Greek}+\text{nearby }\ce{NH4Cl}\ \text{depository}\ce{->[Tassaert dibs on ammine in 1798][\text{vitalism until a.l. 1828}]}\text{amine in 1863}$$

Extra information

$\text{[a]}$ According to some texts, the unification as Amon-Ra goes back to the reign of Mentuhotep II ($2055-$$2004$ BC). Amon's (or Amon-Ra's) importance declined, then increased again during Amenemhet I ($1985-$$1956$ BC), fell and rose to national prominence with Ahmose I.$^{[13]}$
$\text{[b]}$ Most of the early work was done with ammonia, and thus the resulting class of complexes were already then known as metal ammines$^{[14]}$. Developments in naming specific compounds are summarised in this diagram:
$$\text{after discoverer}\ce{->[\text{m.c.l.}]}\text{basis on colour}\ce{->[\text{m.c.l.}]}\text{count number of ammonias only}\ce{->[\text{m.c.l.}][Werner's coordination theory 1893]}\text{premodern}$$
where $\text{m.c.l.}$ is short for many compounds later. Some examples of such practice is included the table$^{[14]}$ below.
\begin{array}{|c|c|c|c|}
\hline
\mathbf{Complex} & \mathbf{Colour} & \mathbf{Name} & \mathbf{Present}\ \mathbf{formulation}\\
\hline
\ce{CoCl3*6NH3} & \text{yellow} & \color{gold}{\text{luteo}}\text{cobaltic chloride} & \ce{[Co(NH3)6]Cl3} \\
\ce{CoCl3*5NH3} & \text{purple} & \color{purple}{\text{purpure}}\text{ocobaltic chloride} & \ce{[CoCl(NH3)5]Cl2} \\
\ce{CoCl3*4NH3} & \text{green} & \color{green}{\text{praseo}}\text{cobaltic chloride} & trans\text{-}\ce{[CoCl2(NH3)4]Cl} \\
\ce{CoCl3*4NH3} & \text{violet} & \color{violet}{\text{violeo}}\text{cobaltic chloride} & cis\text{-}\ce{[CoCl2(NH3)4]Cl}  \\
\ce{CoCl3*5NH3*H2O} & \text{red} & \color{pink}{\text{roseo}}\text{cobaltic chloride} & \ce{[Co(NH3)5(OH2)]Cl3} \\
\ce{IrCl3*6NH3} & \text{white} & \color{gold}{\text{luteo}}\text{luteoiridium chloride} & \ce{[Ir(NH3)6]Cl3} \\
\hline
\end{array}
An attentful reader ought to notice that $\ce{CoCl3*6NH3}$ is yellow and $\ce{IrCl3*6NH3}$ is white. But they both have the same prefix -luteo. At first it did indeed mean yellow but later the same scheme was adopted to notate an equal number of ammonias. Crazy but it is true!$^{[14]}$ You can probably see that proper nomenclature could not get there soon enough.
$\text[c]$ In chemistry, vitalism was the idea that compounds found inside living things are fundamentally different from inorganic chemistry. Such organic molecules were believed to be impossible to synthesise outside an organism.
To captivate how little of organic chemistry was understood in the first half of the 18th century, Wöhler stated in $1835$$^{[5]}$,

[Organic chemistry is] like a primaeval forest of the tropics, full of the most remarkable things.

Proponents of vitalism had a hard time accepting the evidence. They      argued that the process was achievable since carbamide is simply a waste product. However, luckily notable progress followed.$^{[15]}$

M. Berthelot $-$ natural fat ($1854$ doctoral thesis)
H. Kolbe, E. Frankland $-$ acedic acid ($1861$)
A. Butlerov $-$ sugars ($1861$, see Formose reaction)

${\text{[d]}}$ Aniline had been successfully separated in $1826$ by German merchant and chemist O. Underderben. Note, however, that this was not a synthesis $-$ he used dry destillation of indigo.$^{[5]\ [16]}$
${[\text{e}]}$ This was very much intentional. Wurtz recognised in $1849$ that methylamine and ethylamine could be regarded as ammonia in which one equivalent of hydrogen is replaced by methyl or ethyl.$^{[5]}$
$\text{[f]}$ Classification of primary, secondary, and tertiary amines was introduced by Gerhardt in $1856$.$^{[17]}$
$\text{[g]}$ Some authorities$^{[18]}$ claim melamine as having come from mel(am) $+$ amine. This is wrong since amines were unknown at that time.$^{[5]}$
$\text{[h]}$ From Zinin to Liebig, amines were probably grouped as alcaloïdes artificiels (French for artificial alcaloids) because the structure had not yet been proposed.$^{[19]}$

Refrences and bibliography

$[1]$ B. M. Tassaert, Ann chim. phys., 28, 92 (1798).
$[2]$ Fred Basolo, Ralph G. Pearson. Mechanisms of Inorganic Reactions. (1958). (page 2)
$[3]$ Michele Giua. Storia della Chimica. (1962) (pages 342$-$344)
$[4]$ A. W. von Hofmann. Ber. Deut. chem. Ges. (1880). 13,. 449$-$450.
$[5]$ W. E. Flood. The Origins of Chemical Names (1963) (pages xxii, 33$-$35, 37$-$38, 137$-$138)
$[6]$ J. von Liebig. 'Uber einige Stickstoff $-$ Verbindungen'. Justus Liebigs Annalen der Chemie, 10, 1, 1$-$47 (1834).
$[7]$ Bernard Bann, Samuel A. Miller. 'Melamine And Derivatives Of Melamine'. Chemical Reviews, 58 (1), 131$–$172. (1958)
$[8]$ Klaus Bretterbauer, Clemens Schwarzinger. 'Melamine Derivatives – A Review on Synthesis and Application'. Current Organic Synthesis, 9, 342$-$356 (2012).
$[9]$ Michihiro Ohta, Shinji Hirai, Hisanaga Kato, Vladimir V. Sokolov, Vladimir V. Bakovets. 'Thermal Decomposition of $\ce{NH4SCN}$ for Preparation of $\ce{Ln2S3}$ ($\ce{Ln}$ = $\ce{La}$ and $\ce{Gd}$) by Sulfurization'. Materials Transactions, vol. 50, No. 7, 1885$-$1889 (2009).
$[10]$ Zerong Daniel Wang, Motoko Yoshida, Ben George. 'Theoretical study on the thermal decomposition of thiourea'. Computational and Theoretical Chemistry, vol. 1017, 91$–$98 (2013).
$[11]$ A. Kawasaki, Y. Ogata. 'Kinetics of the formation of melamine from dicyandiamide'. Tetrahedron, vol. 22, 1267$-$1274 (1965).
$[12]$ M. Charles Gerhardt. Traité de chimie organique. First edition. (1853) (pages 8, 134, 210$-$211, 277, 396$-$397, 427, 463, 545, 551, 611, 613, 616$-$619)
$[13]$ Seppo Zetterberg. Maailma ajalugu. (2015). (pages 41$-$43)
$[14]$ Fred Basolo, Ronald Johnson. Coordination Chemistry. (1964) (pages 4$-$13)
$[15]$ H. Karik. Üldine keemia. Handbook for students. (1987) (page 223)
$[16]$ Alex Nickon, Ernest F. Silversmith. Organic Chemistry: The Name Game. (1987) (page 305)
$[17]$ M. Charles Gerhardt. Traité de chimie organique. Fourth edition. (1856) (page 592)
$[18]$ The American Heritage Dictionary of the English Language: Fourth Edition. 2000. https://web.archive.org/web/20081201105219/http://www.bartleby.com:80/61/24/M0202400.html (Jan 3, 2017)
$[19]$ Encyclopædia Universalis. 'Amines'. http://www.universalis.fr/encyclopedie/amines/ (Jan 2, 2017)

Please by all means criticise, correct and add as you see fit! This is by no means complete but shouldn't be far off from the probable answer. If someone has access and can understand A. Werner, Neuere Anschauungen auf den Gebieten der Anorganischen Chemie, Third edn., Vieweg, Braunschweig, 1913, pp. 92–95 it might provide extra insight.

HISTORY

AMINE

PRIMARY, SECONDARY, AND TERTIARY
The qualifiers primary, secondary, and tertiary were first applied to the classification of the organic amines in 1856 by the French chemist, Charles Gerhardt, in volume four of his famous Traité de chimie organique in order to distinguish between amines resulting from the first (primary), second (secondary), and third (tertiary) stages in the progressive substitution of the three hydrogen atoms of the ammonia molecule (NH3) by various alkyl radicals:1 One might call the nitrogen compounds primary, secondary or tertiary according as they represent the ammonia type with substitution of one, of two, or of three atoms of hydrogen [1]

AMMINE

Werner published his coordination theory in 1893, which postulated that single atoms or molecules could be joined and grouped around a central atom. It is said that in the middle of night in late 1892, Werner woke up suddenly after visualizing the solution for the structure of coordination complexes in a dream. Throughout the remaining night and the next day he wrote down the details of his coordination chemistry, which he published in his legendary paper: On the constitution of inorganic complexes" ... "To validate his theory, Werner had to work for nearly 25 years and in the process, he prepared more that 8000 compounds. in 1907, Werner prepared a compound, an ammonia-violeo salt, predicted by his coordination theory. [2]
Werner assigned the violeo salt the cis configuration by preparing them by
treating the carbonato complex with conc. HCl. $\ce{[Co(NH3)4(NO2)2]X}$ ..." [3]

BIBLIOGRAPHY
[1] Jenssen, W.B. J. Chem. Educ., 2012, 89 (7), pp 953–954.
[2] Mahanti, S. "Alfred Werner. Founder of coordination chemistry" http://www.vigyanprasar.gov.in/scientists/alfred_werner.pdf Last accesed 3 Dec., 2016.
[3] Panda, B.K. Indian Journal of Science, 2013, 3(6), 25-31.


text:
At which temperature does carbon monoxide explode?
I had doubt whether carbon monoxide explodes or not. I checked out this resource and found out that carbon monoxide explodes when heated. But they didn't mention the temperature at which it explodes. So my question is that at which temperature does carbon monoxide explode and also at which pressure ?
Thanks in advance!


text:
Reaction of alcohols with bromine
In a reaction between $\ce{R-OH}$ and $\ce{Br2}$, what will the product be?
Initially I thought it would be $\ce{R-Br}$, but that doesn't seem right.
I'm oscillating between the product being either $\ce{R-OBr}$ or no reaction occurring at all. I've searched a bit on the web too, but couldn't get an answer.
What is the product?

$\ce{R-OBr}$ does not quite exist. Another reaction may be possible, though.

ROBr is actually  an reaction intermediate. See this: https://pubs.acs.org/doi/pdf/10.1021/ja00990a036

Also see this: https://pubs.acs.org/doi/pdf/10.1021/ja01176a069

No reaction occurs with the reagents you have stated at room temperature. Maybe with different conditions (eg. basic conditions) the haloform reaction that Ivan commented would occur.


text:
At which temperature does carbon monoxide explode?
I had doubt whether carbon monoxide explodes or not. I checked out this resource and found out that carbon monoxide explodes when heated. But they didn't mention the temperature at which it explodes. So my question is that at which temperature does carbon monoxide explode and also at which pressure?

The key explosive risk is from flammability with oxygen in air, described in your linked table as "explosive limits, vol% in air: 12.5%-74.2%".  More chemistry than physics.

At which temperature does carbon monoxide explode?

At a wide range of temperatures.
If mixed with air in a proportion between 12 and 75% it is explosive (assuming a there is an ignition source) - I think this is at standard temperature and pressure - 273K, 0.1 MPa.
Note that 1% of CO in air will cause death within 3 minutes due to its toxic effects.

Does that mean that if there is no ignition source and we supply lots of heat it will not explode?

I'm assuming you mean autoignition. Carbon monoxide autoignites at $609~\mathrm{^\circ C}$, as stated in this website.

Well, this is a perfectly fine answer. The question is only asking for the number and you provided it, so don't feel bad about posting it as an answer.

On it's own Carbon Monoxide will not explode no matter how much you heat it, unless it is mixed with an Oxidizer. Which is more than can be said for a lot of other gases such as Acetylene.


text:
How to identify the ratio of oxidation states in terbium ions in its mixed oxide?

Estimate the ratio of $\ce{Tb^3+}$ to $\ce{Tb^4+}$ in  $\ce{Tb_6O_11}$.

For my solution I knew that I had to get a sum of 22 positive charges from 6 $\ce{Tb^3+}$ and $\ce{Tb^4+}$ ions, so I just added different combinations until I found that $4(\ce{Tb^4+})+2(\ce{Tb^3+})$ gave me the desired value to have an overall neutral oxidation state on the molecule. This gives a $2:1$ ratio also. Is this the correct way to do this, or is there a more formal method?

your way looks good to me

Well, trying different combinations is definitely a lot better than just sit there and stare at it. You could go with a linear equation, but with numbers like these it is hardly any faster.

There is certainly a formal way. Let $x$ and $y$ be the number of $\ce{Tb^3+}$ and $\ce{Tb^4+}$ ions respectively. Then:
$$\begin{align}
3x + 4y &= 22 \\
x + y &= 6
\end{align}$$
This is a simple system of two linear equations in two unknowns, and I am sure you know of many ways to solve it. Let's just solve it using matrices for the fun of it:
$$\begin{align}
\begin{pmatrix}
3 & 4 \\
1 & 1
\end{pmatrix}
\begin{pmatrix}
x \\
y
\end{pmatrix} &=
\begin{pmatrix}
22 \\
6
\end{pmatrix} \\
\begin{pmatrix}
x \\
y
\end{pmatrix} &= \frac{1}{3\cdot 1 - 1\cdot 4}
\begin{pmatrix}
1 & -4 \\
-1 & 3
\end{pmatrix}
\begin{pmatrix}
22 \\
6
\end{pmatrix} \\
&= -
\begin{pmatrix}
1\cdot 22 - 4\cdot 6 \\
-1 \cdot 22 + 3\cdot 6
\end{pmatrix} \\
&= \begin{pmatrix}
2 \\
4
\end{pmatrix}
\end{align}$$
as you found. It seems that trial and error may well be faster!


text:
Why are fats glossy?
I am talking about the standard fats we see in everyday use (triglycerides). It seems that no matter what the composition, saturated or unsaturated, all fats are glossy when liquid. I don't know if this extends to other esters, as I have seen only few of them.
What gives fats this shiny look?
Sorry if the question appears unresearched, but all my attempts to combine "fat" and "glossy" in a search query resulted in nutrition advice.

Come to think of it, aren't all liquids glossy?

If they all are, then I want to know why all liquids are glossy :) But that would be a very hard claim to make. If you are looking at anything liquid as defined in layman terms, then many suspensions and runny emulsions count, and they are not glossy. If you restrict yourself to a single compound in a liquid phase (as long as it has a liquid phase at room temperature - I wouldn't describe molten iron as glossy) then everyday fats are excluded, because they are mixtures. So, the two questions may not have that much of an overlap.

OK, let's refine my claim. Aren't all transparent liquids at room temperature glossy? (That would exclude emulsions, suspensions, and even some individual compounds, but leave many mixtures which are true solutions.)

It will depend on your angle of incidence. From the Fresnel equations the reflectivity depends greatly on angle of incidence so that even an ordinary flat piece of paper appears reflective if the angle of incidence is almost 90 degrees, i.e. at glancing angle whereas it appears matt at zero degrees.

Reflected light (gloss) is a maximum when the angle of observation is equal to the angle of incidence:  https://en.wikipedia.org/wiki/Gloss_(optics). The image shows reflection from a plane surface with some roughness that gives diffuse reflection.

At Brewster's angle, the reflected ray is strongly polarized. The image is intended to indicate a depth of penetration in transparent materials; the polarization is not relevant here.

In conductors, like metals, and strong absorbers, like iodine, there is essentially no transmission, and almost all the light is reflected from a plane surface, because the penetration depth is exceedingly small.
In a dielectric (i.e., a nonconductor like glass or fats), some light will be transmitted, and reflection occurs after a penetration of some depth - on the order of a wavelength of the light. Glass is atomically uniform on this scale, and  a fat molecule (e.g., glyceryl trioleate) would be a glob about 2 or 3 millimicrons in diameter, therefore essentially homogeneous on this scale. Any liquid volume larger than that would have no discernible inhomogeneity and reflection would be a uniform wavefront, specular (mirror-like) rather than diffuse.
But suspensions and runny emulsions and milk are not homogeneous on that scale. They have particles on the order of a wavelength of light (~500 millimicrons) in the liquid, so reflection occurs from sites of different composition within this penetration depth. These materials reflect similar to reflection from a roughened metal surface and give a diffuse reflection rather than a gloss.

The equation for reflection ($\mathrm R_0$) of interfaces of clear mediums is given by:
$$\mathrm R_0 = \left[ \frac{n_1 - n_2}{n_1 + n_2} \right] ^2$$
where $n_i$ is the refractive index of its respective medium. If air is $n_1 = 1.00$ and olive oil is $n_2 = 1.47$ (air-oil interface) then
$$\mathrm R_0 = \left[ \frac{1.00 - 1.47}{1.00 + 1.47} \right]^2 = \left[ \frac{-0.47}{2.47} \right]^2 = 0.1903^2 = 3.62\%$$
Therefore $3.62\%$ of light will be reflected off of the smooth surface of the oil giving it the glossy appearance.


text:
Pauli's Exclusion Principle
While reading Pauli's Exclusion principal from here, they said that

The Pauli Exclusion Principle states that, in an atom or molecule, no two electrons can have the same four electronic quantum numbers." Now again in we know that the The orbital angular momentum quantum number $l= 0,1,2... (n-1)$ – Source

So from here we can say that $l\neq n$. So here automatically no two electrons can have the same four quantum numbers.  So what's new in Pauli's principle?

I think you've misunderstood, Pauli's exclusion principle doesn't mean that the four numbers are different it means that no electron can have the same respective values for $n$, $l$, $m_s$, $m_l$ as another electron (in some case some of the numbers themselves can be the same). i.e. electrons in the same suborbital must have different spins.

$n$ indicates the orbital shell
$l$ indicates the type of suborbital, s, p etc
$m_l$ indicates the orientation of the suborbital, px, py etc
$m_s$ indicates the electron spin

Pauli's exclusion principle means that no electron around an atom can occupy identical orbitals, i.e two electron with up spin in a 2s1, as all their electron numbers would be the same as their respective equivalents ($n_1 =n_2 = 2$, $l_1 = l_2 = 0$, $m_{l,1} = m_{l,2} = 0$,  $m_{s,1} = m_{s,2} = -1/2$).


text:
Curing of polyester resin that did not reach top exothermic temperature
When working with polyester resins for fiberglass or geloat, there are times where the thickness of the resin on the substrate is not sufficient to get a strong exothermic reaction and achieving a high temperature to "kick". I have been told in the industry to use artificial heat source to help it get to a high enough temperature for a good kick.
My question is: If you are working with a thin layer of resin or the ambient temperature is too low for the resin to kick on its own, and it was not artificially brought up to an appropriate temperature of say 250 degF and the resin stayed at a low ambient temperature of about 45 degF for several days before finally brought up to about 70 degrees, would the resin still cure eventually or will the overall cure be significantly compromised?
This type of scenario can happen a lot when doing minor gelcoat repair outdoors in colder climates.


text:
Why is the ionic product for water $10^{-14}$?
Why is the reference value of $K_w=10^{-14}\ \mathrm{M}^2$, such that $[\ce{H+}]= 10^{-7}\ \mathrm{M}$ and $[\ce{OH^-}]= 10^{-7}\ \mathrm{M}$ for pure water at $20\ ^\circ \mathrm{C}$?
For example, why isn't $K_w=10^{-18}\ \mathrm{M}^2$, and thus $[\ce{H+}]= 10^{-9}\ \mathrm{M}$ and $[\ce{OH-}]= 10^{-9}\ \mathrm{M}$?

Why would it be? Do you know what that number means?

Questions of the "why" of specific numerical values of physical constants exit the realm of chemistry and enter the sphere of scientific theology. One might well ask in the same vein why Planck's constant is $6.626\times 10^{−34}\ \mathrm{J\cdot s}$ instead of $7.358 \dots$, or why  the universal gravitational constant is $6.674\times 10^{−11}\ \frac{\mathrm{N\cdot m^2}}{\mathrm{kg^2}}$ instead of $5.123\dots$.

That being said, the mechanistic (if not explanatory) answer is that the quantum and statistical mechanics of the interactions of $\ce{H2O}$, $\ce{H3O+}$, and $\ce{OH-}$ in the condensed phase just work out that way.

@Brian see section 15.3.6 "Computation of the Water Dissociation Constant" in Bridging the Time Scales: Molecular Simulations for the Next Decade and see if still think it is the same vein as Planck's constant https://books.google.com/books?id=7H4U0lwnbtYC&pg=PA425&lpg=PA425&dq=Autodissociation+of+a+water+molecule+in+liquid+water&source=bl&ots=--0OXMoJ5I&sig=ZqTQtUQKmuINFxoHNC32WNJ9HSI&hl=en&sa=X&ved=0ahUKEwjDn_eErd7JAhVN62MKHQ55BeMQ6AEIQjAH#v=onepage&q=Autodissociation%20of%20a%20water%20molecule%20in%20liquid%20water&f=false

@DavePhD It's not fundamental in the same way that Planck's constant is, I agree, but I'd argue the fact that it bears the value it is observed to have at a given set of conditions is scientifically unexplainable in the same fashion as is the fact of the value of $h$.

$K_w$  is actually measured for pure water. The value of $K_w$ does change with temperature. The value is $1 \times 10^{-14}$ at room temperature.

$$\ce{2H2O <=> H3O+ + OH-}$$
$$K_w = \frac{a(\ce{H3O+})a(\ce{OH-})}{[a(\ce{H2O})]^2}$$
where $a$ is "activity of".
Then you can approximate activity of water as 1 if appropriate and activity of the ions as concentration of the ions.
Just like any other equilibrium constant, the relative chemical potential (Gibbs energy) of the products and reactant determine the constant.
Consider the relative strengths of the bonds of $\ce{H2O}$, compare to those of $\ce{OH-}$ and $\ce{H3O+}$, as well as the number and strength of intermolecular hydrogen bonds to each of these.
For $\ce{D2O}$, the equilibrium constant is 8 times small.

But why does the $\Delta G$ have the value such that $K_w=10^{-14}$?

@Brian see Ab Initio QM/MM Simulation with Proper Sampling:  “First Principle” Calculations of the Free Energy of the Autodissociation of Water in Aqueous Solution . Phys. Chem. B, 2002, 106 (51), pp 13333–13343 http://pubs.acs.org/doi/abs/10.1021/jp021625h

But why does this highly accurate QM/MM simulation with proper sampling provide this particular $\Delta G$ value? Why not a different one? Why do the quantum and statistical mechanics give this numerical value and not another?

@Brian yes, you can keep asking way, but we should get to a reasonable set of constants that QED and QCD are based upon.

Agreed! And then we reach the parallel to Planck's constant: the values of the QED and QCD constants, of which $h$ is one, are "scientifically arbitrary" -- science cannot explain why they hold the values they do. Thus, values derived from these constants are similarly "arbitrary," in this sense.  Depending on how fundamental an explanation Mehdi is looking for, your answer is either spot-on, or no answer can be given.


text:
Standard reduction potential of MnO4-/MnO2 couple
This is the data given:
$$\ce{MnO4- /Mn^{2+}} = 1.5~\mathrm{V}$$
$$\ce{MnO2 /Mn^{2+}}= 1.23~\mathrm{V}$$
I know that :
$$\ce{MnO4- +5e- + 8H+ ->Mn^{2+}}$$
$$\ce{MnO2 + 4H+ +2e- -> Mn^{2+}}$$
After that here's what's done which I absolutely don't get. After using $\Delta G^\circ = -nFE^\circ$, won't we have to multiply these energy values by the coefficients of the balanced equations before we can get $E_{\mathrm{cell}}$?
In particular, shouldn't the $\Delta G^\circ$ of the first equation be multiplied by two and that of the second by three?

The important thing to note with electrochemistry questions like these is that, if you add two equations together to get a third, their $\Delta G^\circ$ can be added, but their $E^\circ$ cannot.

Your half-equations are incomplete, which might not be the worst thing, but it certainly doesn't help your case. Using $\Delta G^\circ = -nFE^\circ$, we have the two equations
$$\begin{align}
\ce{MnO4- + 8H+ + 5e-} &\ce{-> Mn^2+ + 4H2O} & \Delta G_1^\circ &= -728.46~\mathrm{kJ~mol^{-1}} \tag{1} \\
\ce{MnO2 + 4H+ + 2e-} &\ce{-> Mn^2+ + 2H2O} & \Delta G_2^\circ &= -237.35~\mathrm{kJ~mol^{-1}} \tag{2}
\end{align}$$
The trick is to find the appropriate combination of equations $(1)$ and $(2)$ that give the equation $(3)$, which you are interested in:
$$\ce{MnO4- + 4H+ + 3e- -> MnO2 + 2H2O} \qquad \qquad \Delta G_3^\circ = \,\,? \qquad \qquad \tag{3}$$
If you just look at the manganese-containing species, it's actually not too hard to see that $(1) - (2) = (3)$. That's why I said it's probably not the worst thing that you omitted $\ce{H2O}$ in your half-equations - since it doesn't really make a difference in terms of finding the answer - but it's just not good practice to write something that's incorrect. Anyway, that means that:
$$\begin{align}
\Delta G_3^\circ &= \Delta G_1^\circ - \Delta G_2^\circ \\
&= -728.46~\mathrm{kJ~mol^{-1}} + 237.35~\mathrm{kJ~mol^{-1}} \\
&= 491.11~\mathrm{kJ~mol^{-1}} \\
E_3^\circ &= \frac{-\Delta G_3^\circ}{3F} \\
&= +1.697~\mathrm{V}
\end{align}$$
If you obtain a slightly different answer, it may be due to rounding inconsistencies, or perhaps because I used more accurate values for $F$ and the initially given reduction potentials. In any case, you should get an answer around $+1.70~\mathrm{V}$.

Well I still have a question - The balanced cell reaction would $2$ times $(1) - $ $5$ times of $(2)$ and thus in total 10 $e^{-1}$ are used, So shouldn't we multiply the gibb's energy by them and after subtracting divide by 10? Why would that be wrong?

Can you calculate what the resultant chemical equation is if you take $2\times (1) - 5 \times (2)$? Is that what you are trying to find $\Delta G^\circ$ for?


text:
Why does freezing point decrease on adding impurities?
It is a well known fact that freezing point decreases on adding impurities. I feel that it should increase on adding impurities.
Reason: I have watched this video on supercooled water. It says that when ice forms, it needs some nucleation points. If water is pure and left undisturbed, then there would be no nucleation points and water won't freeze even below $0^o$C.
So, my reasoning is that when impurities are added, they provide additional nucleation points where water crystals can gather. This would cause water to freeze faster.
But it doesn't happen. Why?

It depends on the "impurity." If the impurity dissolves then that is one kind of impurity. If the impurity doesn't dissolve, then it could provide nucleation sites at the freezing point of water.

The kinetics (nucleation sites) is very different from the thermodynamics (Gibbs free energy) of the situation. Whether nucleation occurs quickly or slowly does not impact whether a particular phase is thermodynamically stable. Thermodynamically, adding impurities will always reduce the melting point (enthalpy change is linear/polynomially influenced by impurity concentration, while entropy is logarithmically related).

In a hypothetical case that the impurity is more soluble in solid ice than in liquid, then would the freezing point increase?

No, but good question. Freezing is a thermodynamic process. Nucleation is a kinetic process.
Kinetic Explanation: At the freezing point of water, water wants to become ice (thermodynamically driven). In order for freezing to initiate and ice to form a temperature dependent minimum number of water molecules must arrange them selves in a crystal or else the particle will dissolve and no freezing will occur.  Impurities or containers with rough surfaces can help arrange and hold molecules which allows freezing to occur with a fewer number of molecules.
Thermodynamic Explanation:  Freezing is a thermodynamic process dictated by $\Delta G_f = 0 = \Delta H_f -T\Delta S_f$.  When water freezes it must dispel the impurities to form an ice crystal (reverse mixing). Since mixing increases entropy, this means that for water to freeze the negative change in entropy ($\Delta S_f$) is larger than that of pure water.  This must be compensated by a lower temperature ($T$) neglecting and differences in ethalphy. The enthalpy ($\Delta H_f$) of mixing will contribute as well to the freezing temperature of water and must be accounted for.  In the case of Sodium chloride though the enthalpy mixing is positive which would favor a higher melting point, but the entropy contribution is large and magnified by the temperature such that entropy dominates and reduces the temperature of freezing.
If you look at the phase diagram of water and salt you will see that the mixing of the salt and water depresses the melting temperature due to the increased change in entropy from mixing water and salt to make brine from ice and salt.

Nice answer. This makes sense. But is the kinetic part less dominant or is it completely irrelevant?

There is a molecular process for water to become ice and the kinetic part perimts the water to undergo that process.  Even if the water were to be at -30C, without the kinetic process to form ice, the water can not freeze.

So, if I were to add some salt to this water at $-30^oC$ it would freeze above that temperature, right?

no it would freeze immediately as the salt surface would provide nucleation sites for freezing.

No. What I meant was that if I were to take that sample of water, heat it to normal temperature and then add salt, then would it freeze below or above $-30^0C$?

If you added ~$23 wt$% salt, it would start to freeze at  $0^oC$ , and would finish freezing at ~$-22^oC$.

Think in terms of needing to remove the dissolved substance: the water that crystallizes is (fairly) pure $\ce{H2O}$, and the remainder is a concentrated solution (e.g. brine, if the solute is $\ce{NaCl}$. Work is done to separate the water from brine, so it takes more "effort" to freeze than does pure water.
There are many good explanation for the phenomenon on the web, e.g. Physics Animations.
BTW, some bacteria nucleate water freezing, such as Pseudomonas syringae. The protein from these bacteria is used in artificial snow-making. The water still has to be below 0 C to freeze, but supercools less.


text:
why are simple organic molecules usually white?
I have heard that it is because they absorb electronically in the ultraviolet range, if this is true, why is it true. What is it about the nature of organic molecules that make them absorb in the ultraviolet range.

Possible duplicate of Why is snow white?

nah, i don't think this is a duplicate. I'm pretty sure that snow is white for a different reason.

Everything is white when it doesn't absorb or disperse visible light but is scattering it.

IMHO the question is not a duplicate, but its answer is. Indeed, snow is white for pretty much the same reason.

If it's already answered, it is a dupe, but there is another mechanism, indeed involving absorption of UV - fluorescence, you could modify your quesrion, so it wouldn't get closed

@Mithoron It is only a duplicate if the question is the same.

@Martin-マーチン Yes, but if answer is appropriate on many questions, it determines level of similarity needed to treat them as dupes.

@Mithoron It really does not. Closing a question as duplicate is pretty straight forward and it only concerns the question. If the answer on two different questions would be the same, then the questions are still not, they are related. Even if the answer goes as far as explaining the question, too, then the questions are still different. On first sight, small organic molecules and snow don't have much in common. I would be really confused if one question leads to the answer of another that seems completely unrelated.

@Martin-マーチン Looks like another theory vs practice problem...

Most substances are white in powder form if they have no strong absorption of visible light. They whiteness is a consequence of light scattering by small particles and the fact that there is no absorption within the visible range.
The lack of visible absorption is another issue. Light is absorbed by molecules when the energy difference between a filled orbital and an unfilled one in the molecule matches the energy of a photon of light. For light photons the energy is the equivalent of wavelength and therefore colour.
In most small, simple organic molecules, the lowest unfilled orbitals are a long way from the highest filled ones. So they need a lot of energy to cause them to jump to an unfilled orbital. So they absorb only high energy photons (UV light is higher energy than visible light so this implies the absorption bands will be in the UV and we will not notice them as our eyes can't see UV).
The characteristics that create colour in molecules are often long runs of unsaturated bonds where the electrons can spread along the chain of unsaturation. In these delocalised systems the gap between occupied and unoccupied orbitals is smaller than in simpler molecules and often falls into the visible spectrum. These molecules are perceived as coloured.
The molecule below is a simple example of a dye (most are bigger and more complex than this):

The electrons in this molecule can delocalise over all three rings via the NN bridge and this creates orbitals with the right energy gap to absorb blue light.
In simpler molecules with little delocalisation the energy gap is far into the UV. Even in polycyclic aromatic systems the energy gap is often still in the UV.
Of course, this is a simplification, but the essence of what sort of molecules you need to cause visible colour is a relatively large unsaturated system and by definition this doesn't happen in very simple organic molecules. Wikipedia lists many more examples of dyes so you can get a feel for what sorts of structures make strongly coloured compounds.


text:
How to create different levels of hydrophobicity on glass beads surface?
I have some glass beads and I need to create different levels of hydrophobicity on their surfaces (various contact angles). I know that this is possible by 'silanization'. However, I need to know about the following questions:

Whether there is any other reasonable method to do that.
In silanization, how can I change the level of hydrophobicity that my glass beads achieve? (e.g. by changing coating time, using different solutions or concentrations, etc.)

PS. I have the following solutions for silan. Please let me know if anybody has any comments on them or knows a more suitable solution for this purpose.
- CHLOROTRIMETHYLSILANE
- DICHLORODIMETHYLSILANE
- TETRAETHYL ORTHOSILICATE
- OCTADECYLTRICHLOROSILANE

Sure, there are plenty of methods. For example, you may cover them in cooking oil and carefully heat in a stove. However, I'd go with silanization, for it is clean and controllable.

Lets assume first that the surface $\ce{Si-O}$ bonds of teh glass are terminated by $\ce{Si-OH}$ groups. Hydrophilicity will depend greatly on the ration of $\ce{Si-OH}$ to $\ce{Si-R}$ groups on the surface of the glass. Chlorotrimethylsilane is great for a completely hydrophobic surface as it will cap all of the $\ce{Si-OH}$ groups to $\ce{Si-O-Si(CH3)3}$ groups allowing no polar interaction of the surface with the water. dichlorodimethylsilane is not as good leaving you with $\ce{-Si-O-(Si(CH3)2)-O-Si{-}}$ and $\ce{-Si-O-Si(CH3)2OH}$ groups at the surface which affords a bit more interaction.  Octadecyltrichlorosilane does provide even more polar groups than the last two, but because of the large octadecyl group this will likely make a more hydrophobic surface than the chlortrimethylsilane.  The tetraethyl orthosilicate or rather tetraethoxy silane (TEOS) will hydrolze to $\ce{Si(OH)4}$ in water which you can use to help make the surface more hydrophilic.
If I had to do this I would make various binary mixtures of TEOS, chlorotrimetyhlsilane, and dichlorodimethylsilane and treat the surface with them under constant conditions and see what spectrum of hydrophilicity you achieve. For the most hydrophobic surfaces I would use mixtures of octadecyltrichlorosilane and chlorotrimetyhlsilane.

There is a very nice comparison in answer above in terms of how well each solution can hydrophobize the glass; and that's exactly what I want to do as I need one solution to give me the highest hydrophobicity and then choose a few more solutions such that each one gives me a lower level of hydrophobicity so that in general I can create glass surfaces with different contact angles (e.g. varying from let say 140 deg to 40 deg.). Do you think, the 4 solutions that I've chosen are able to do this for me? or I can have a better selection (especially for lower hydrophobicities). Thanks

I believe that the chemicals you have are sufficient.  The real devil is in the that tenant of materials science no one talks about-processing.  You can do the varying mixtures that I suggested, also you can dilute your chemicals is a non-protic solvent like kerosene to vary the extent of reaction, and lastly you can vary the time of the treatments.  I believe you have everything you need, but some experimentation will be required.

I have two more questions and thanks in advance for your answers! 1) my solutions are mostly 98% or 99% concentrated. Should I deep coat particles into dense solution or should I dilute them and how much? 2) For varying the time, could you please give me an order of magnitude for the time required for a full treatment and the time steps that I should decrease from that in each new experiment (I need first treat the flat glass slides, measure the CA on them and use the same treatment for glass beads, supposing that it will give me the same CA at the same treatment time)

I would Dilute in kerosene.  Your solutions will react quickly so time on the order of 10's of seconds should suffice.


text:
Determining the nature of common pH indicators (methyl orange) using Le Chatlier's principle
I got some good information about this topic yesterday, and I understood what I was told, but I have come up with a specific way of phrasing the problem I'm having. It is slightly different than before, and I'm looking for someone to tell me if what I'm doing makes any sense or not, rather than just rote learning facts like 'methyl orange indicator is a weak acid/base'. I want to be able to use this process for any weakly acidic/basic indicator I come across.
(Note: Methyl orange is taken to be orange when it has a moderate pH (HIn, not very acidic or basic, and red between a pH of 4.4 and 3.1 - i.e. when it is In- or In+ depending on the scenario below)
Assume methyl orange, an indicator, is a weak acid. It dissociates as follows in aqueous solution;

We know that if an acid is added to this system at equilibrium, the concentration of hydrogen ions will increase. Using Le Chatlier's principle; to oppose this stress, the reaction shifts to the right, and the colour of the initial indicator (in this case methyl orange, so orange) will dominate and become more intense.
The solution will not go red, like it would if a base was added and proton concentration was decreased, as the equilibrium would shift in the other direction.
The problem is, that if you add an acid to this system in real life, a colour change occurs at a pH of 4.4, from orange to red. This means that methyl orange indicator cannot be an acid. Right?
If you instead assume that methyl orange is a weak base, it dissociates like so in aqueous solution;

So here, if you add an acid, the concentration of hydroxide ions will decrease (due to bonding between the hydrogen ions in the acid and these hydroxide ions to form water), so the equilibrium shifts to the right, to compensate.
This means that the red colour will dominate.
This result is in line with observation, as methyl indicator will change colour from orange to red at a pH of 3.1-4.4, which is acidic.
Does my reasoning therefore demonstrate that methyl indicator must be a weak base?

It's too broad, you could read many books acid/base properties of organic compounds

Ok. but surely for indicators where their properties (in terms of acid/base) are very relevant, there is some answer on this? For some reason phenolphthalein is defined as a 'weak acid' but Methyl Red/Orange can't be defined? How am I supposed to know how they dissociate in solution?

I've seen mechanisms for at least two first and imo it would be good to know them as they are interesting.

Methyl orange is weak base, their properties are well known. And if you want to know them you can check pKa or pKb+ values. You should edit your question - narrow it down.

According to my textbook and according to my state exam papers, methyl orange is a weak acid. According to another textbook I have, it is a weak base. There is no info either way I could find online. This is my point!

I don't think it is too broad. Did you have a look at any of the structures of the indicators you have mentioned? If you find controversial statements, include them in the post so that they can be cleared up. And since you have not done it yet, I invite you to take the [tour].

Methyl Orange is the common name for Sodium 4-[(4-dimethylamino)phenyldiazenyl]benzenesulfonate. So the compound is a salt or a conjugate base. It is the sodium salt for the indicator since that form is more soluble than the acid form.

The pKa is 3.47 in aqueous solution, so at pH values of 3.1 or lower the molecule is protonated.

At pH values greater than 4.4 all the molecules in solution are deprotonated as the sulfonate anion.

Linked two images from ChemWiki.

The confusion is caused by rather poor wording in a question.

I'd prefer the wording to be:

... In aqueous solution methyl orange indicator behaves like a weak acid. Letting HX represent the protonated form of the indicator, it dissociates as follows:

It is splitting hairs, but methyl orange is the name of the sodium salt, not the protonated form of the indicator.

So... if it is a conjugate base that means it dissociates like a base would in aqueous solution? (XOH ---> X+ and OH-)? I apologise for not using the correct terminology - I'm just beginning to understand what is involved in acid/base titrations

No, when the acid form loses a proton, then the anion is a conjugate base - it needs a proton to become neutrally charged. But the anion isn't an anion of "methyl orange" per se since methyl orange itself is a salt. // So this is analogous to acetic acid and sodium acetate.  In solution acetic acid is the protonated acid form, and acetate is the deprotonated anion form. The acetate anion is a conjugate base.

It doesn't get protonated on sulfonic group, which is least basic, but on N atom connected to the same ring as sulfonic group. This way most stable cation is created.

So, the acid is the one used as the indicator? I'm essentially trying to find out the properties of the indicator itself - not the conjugate base - so that I can know how it dissociates in aqueous solution. If indeed it is an acid, how can I explain what I've done above when I assume methyl orange to be an acid, then use Le Chatlier's Principle to effectively show that not to be true? Did I make a mistake?

@Mithoron - Thanks for the correction. I'll do some more research and fix the answer. I must have been brain dead. The amine group would be protonated in acid too. I'm surprised that the double bonded nitrogen group reacts.

@Malnurtured Gee, it's the salt, which has basic anion

@Malnurtured - Methyl orange itself can't be an acid since there isn't any proton to donate. Methyl orange is the name for the sodium salt, not the acid form.

@MaxW Great! Makes sense, that's what I thought except my exam papers (state level standardised tests, mind you) have this question where they call Methyl Orange an acid and have it dissociate like an acid: https://gyazo.com/086545c1acb65e7a2dd019c6d9c92984 What do you make of this? Just wrong?

@Malnurtured - Someone made the same mistake I did, and didn't really check the chemistry. The wording seems a bit sloppy. I think "methyl orange indicator" would be better wording than "indicator methyl orange." Given the aqueous chemistry I doubt that 4-[(4-dimethylamino)phenyldiazenyl]benzenesulfonic acid exists as a solid. So the protonated form isn't really "methyl orange" per se.

@MaxW Gah, so there isn't really any way to tell whether they are talking about the conjugate base or the acid form unless they tell you (in terms of the exams I'm going to get at least). Sucks. Thanks man.

?!? I have to wonder if the terminal amine group is protonated in weakly acidic solutions. The pKb for N,N-Dimethylaniline is  8.85. When the double bonded nitrogen protonates, then the hydrogen on the amine group would leave.

No there's quick equilibrating and azo protonated is strongly stabilised major form.

@ Mithoron - I guess a better way to put this is - I wonder what the ratio of the azo protonation to the amine protonation is as a function of pH. I looked and couldn't find anything on that. I'd guess that I just didn't know how to formulate the query. The ration has surely been evaluated.

let's continue in chat


text:
2 modes of radioactive decay for hydrogen-5?
While looking at a table of helium isotopes it said that helium-8 turns into Tritium(hydrogen-3) and helium-5 through fission and beta decay.
But the pathway to helium-5 isn't so direct. It is more like this
helium-8 -> hydrogen-3 and hydrogen-5 (fission) -> helium-5 (beta decay)
But the table of hydrogen isotopes said nothing about hydrogen-5 going through beta decay. All it had there was double neutron emission producing tritium which after 12 years turns into helium-3 via beta decay.
Does this mean that I discovered that hydrogen-5 does go through beta decay but not nearly as often as double neutron emission?

@A.K. http://meta.chemistry.stackexchange.com/q/2988/7475

Sorry, you haven't.  I don't know where your information is from but I think there is an error in your decay chain.  For the 0.9% of $\ce{^8_2He}$ decays that proceed via spontaneous fission your equation should look more like this:
$$\ce{^8_2He ->[fission] ~^3_1T + ^5_1He + \beta ^{-}}$$
$$\ce{ ^5_1He  ->[neutron~emission]  n + ^4_2He} $$

Looking at the A=8 energy level diagrams (http://www.tunl.duke.edu/nucldata/figures/08figs/menu08.shtml) gives a slightly different picture. 0.9% of the time you get He-8 undergoing beta decay to Li-8, which than can fission in to He-5 and T. 16% of the time the beta decay instead lands on Li-8 levels that decay to Li-7 + neutron. 84% of the time it decays to an even lower Li-8 energy level, drops to the ground state, and then will beta decay to Be-8. Nuclear energy level diagrams are quite useful.

I am aware of the other decay modes.  I chose to focus on the one pertinent to the question. I have updated the question to be more explicit about this.

For clarity it seem that there should be two separate equations. One for $\ce{^8_2He}$ and one for $\ce{^5_1He}$.

But fission of helium produces 2 hydrogens. So the larger hydrogen(hydrogen-5 in this case) goes through beta decay.

This is because fission not only splits the nucleus but it also splits the electrons. So to get helium-5 from helium-8 it first has to go through fission into tritium and hydrogen-5. Then it has to go through beta decay and capture that electron in the 1s orbital to turn into helium-5


text:
How to calculate air tank capacity from tank displacement, tank fill (gauge) pressure, and specific type of gas?
I am trying to calculate the amount of a specific compressed gas that will fit in an air tank of specific displacement, at a specific pressure. This is in reference to compressed air guns.
The inputs are:

Tank Displacement (units: cubic inches or cubic centimeters)
Tank Fill (Gauge) Pressure (units: psi or bar)
Type of Gas:

Dry Air
Carbon Dioxide
Nitrogen
Helium

*For this scenario, Temperature is assumed to be constant @ 20C / 68F / 293.15K.
The output is:

Tank Capacity (units: cubic feet or cubic meters)

In other words... an example of a possible question that could be asked pertaining to this would be: "How many cubic feet of helium will fit in a 122cc tank when filled at 3000psi?"

My question is: What amount of specific gas will fit in an air tank of specific displacement when filled at a specific pressure? (Please provide equation along with an explanation)
I am having trouble coming up with the proper equation to use with the given inputs. I have tried the Ideal Gas Law, Boyle's Law, and the Combined Gas Law. I have no way to verify if the answer I am getting is correct because I'm not sure if I'm using the correct equation.
Note: this is not homework. I am a PHP developer. This is functionality I am implementing for a client's website.
Thanks!

Welcome to Chemistry SE! You should still put some of your calculations, so we can see what was wrong.

Well for Carbon Dioxide, Nitrogen, and Helium you can assume that the ideal gas law works and it will be "reasonably close" $$PV = nRT$$ Dry air is a bit funny in that it is a mixture of gases. You should be able to find the mass of a liter of dry air at some combination of T,V and P.  22.4 liters would be a "mole" of air.

The gist is that the "cubic feet" is measured at standard temperature and pressure and then pumped into the smaller volume under pressure.

@MaxW
How do the inputs above plug into the Ideal Gas Law?
P = Tank Fill Pressure
V = Tank Displacement
n = # moles = mass of specific gas / molar mass
R = Specific gas constant
T = Temperature
Where I'm getting confused is how there is only one Volume in the equation. If I'm given Tank Displacement (as volume), how can I solve for Tank Capacity (as volume)?

Is air rifle a Air Arms S200? // Helium would give much greater velocity to pellet, but it is a small molecule and leaks would be much more likely. In other words a rifle that holds air fine might leak helium.

@Mithoron Thank you for the tip. Next time I'll be sure to show exactly what I've tried in order to provide a better picture of where things went wrong.

The compressibility factor (z factor) for carbon dioxide under the conditions in the tank is about 0.35. So the ideal gas law will not be accurate enough for CO2.  You should use PV=znRT, with z = 0.35 for inside the tank.

Changing the question after there have be answers given is a big NO-NO. You can always ask a new question on this site. I am currently researching if a rollback is necessary and would not invalidate the answers given further. Note: This still is a homework question. You can inform yourself about this site when you take the [tour].

"Quick and Dirty" for gases other than carbon dioxide
You're trying to figure out what volume of gas as STP (standard temperature and pressure) would fill a 122 cc tank at 3000 psi. From ideal gas laws:
$PV = nRT$
but since nRT is constant
$P_1V_1 = P_2V_2$
so
$V_{14.7psi} = \dfrac{(3000\text{ psi})V_{3000psi}}{14.7\text{ psi}} = 204 \times V_{3000psi}$
It takes one tank of gas to fill the "empty" tank, so 203 more tanks of gas can be pumped into the tank full of gas at 14.7 psi to get to $3000 \text{ psi}$.
$ 203 \times 122 \text{ cubic centimeters} = 24.7 \text{ liters}$
1 liter is 0.0353147 cubic feet. So:
$24.7 * 0.0353147 = 0.87$ cubic feet of whatever gas.
I lifted the following figure from this webpage. In basically shows that for a given temperature and amount of gas (number of moles) then the product of pressure and volume, PV, ought to be a constant which isn't observed with "real" gases. The lines show the real behavior and the data points estimates from the Van der Waals equations. The gist here that for most normal gases at 3000 psi (200 atmospheres), you'll only be off by about 20%.

More exact solution using Van der Waals equation for gases other than carbon dioxide
Starting from the Van der Waals equation

$(P + \dfrac{an^2}{V^2})(V - nb) = nRT$

Where:

P - pressure
V - volume
n - number of moles
T - absolute temperature
R - ideal gas constant = 0.08206    L atm/(K mol)
a - constant
b - constant

T must be an absolute temperature. Typically in science the Kelvin scale is used.
P is conveniently converted to atmospheres where 1 atmosphere = 14 psi.
a,b are looked up in tables of gas constants
In the following table the constant $a$ has units $L^2 \cdot \text{atm} \cdot mole^{-2}$ (assuming 1 bar = 0.987 atm )and $b$ has units $L \cdot mole^{-1}$
   Gas                a             b
   Nitrogen          1.352          0.0387
   Helium            0.0342         0.0238
   Dry Air(a)        1.33           0.0366
   Carbon dioxide    3.593          0.04267

(a) reference
Steps:
(1) To calculate the number of moles of gas at high pressure, you solve the following cubic equation for $n$. (Knowing $P, V, T, a,$ and $b$). At high pressure (hundreds of atmospheres), and well above the critical temperature there will only be one root for $n$ for all the gases listed except carbon dioxide.

$(\dfrac{ab}{V^2})n^3 - (\dfrac{a}{v})n^2 + (Pb+RT)n -PV = 0 $

(2) To calculate the volume of gas at room temperature and pressure, you solve the following cubic equation for volume, V, using $n$ from step 1. (Knowing $P, n, T, a$, and $b$). At room temperatures and pressure there will only be one root for $V$ for all gases including carbon dioxide.

$(P)V^3 -(nbP+nRT)V^2 +(an^2)V -abn^3 = 0$

Solution for carbon dioxide
At room temperature and pressure gaseous carbon dioxide gas would follow Van der Waals equation as noted in step 2 above. However with "moderate" pressure carbon dioxide liquefies at room temperature. Thus a carbon dioxide tank is filled to about 85% of its volume capacity with liquid carbon dioxide.
The overall idea is that the liquid carbon dioxide expands to a gas in an airgun. This creates a relatively stable pressure as the airgun is shot. (You do have to allow the carbon dioxide reservoir to warm back to ambient temperature since vaporizing the liquid to a gas cools the reservoir.)
There are two other considerations here. First above the critical point 304.18 K (31.03 °C, 89 °F) and 72.8 atmospheres, carbon dioxide exists as a supercritical fluid. Second a carbon dioxide tank is never filled completely full of liquid. There is always some gas headspace, hence the 85% fill. Carbon dioxide tanks have been documented to explode in a hot car on a summer day for example.

(Image from Wikipedia)

Dah... I really need to change the answer. @ChesterMiller is right. CO2 is very different than the other gasses. At "high" pressure (much less than 3000psi)  it would be a liquid at room temperature. As far as the Van der Waals equation, you look up the additional constants in a table of such constants for various gases. For example: https://en.wikipedia.org/wiki/Van_der_Waals_constants_%28data_page%29

Which equation to use for each step?

?!? The Van der Waals equation of course. // I'm still working on the whole answer. I need coefficients for air and more on CO2.

Solving for V in Van der Waals yields a cubic equation. Is this intended?

To clarify, to solve for n:
n = ((P * V) / (R * T)) * (1 + ((n^2 * a) / (P * V^2))) * (1 - ((n * b) / V))?
And to solve for V:
V = (n * R^3 * T^3) / (((P * R^2 * T^2) + a * P^2) + (n * b))?

I didn't try to expand the Van der Waals equations and collect terms. Your first equation for $n$ has $n$ on both sides. Having $R^3$ in the second equatio0n just seems wrong.

I think I have most of this sorted out. I'm still working on how to calculate the amount of carbon dioxide that would be in a high pressure fill for a tank with a volume, V.

@MaxW I just wanted to let you know that editing your post very often triggers an automatic flag. It would be nice if you could avoid it in future. Thank you.

You may not be using absolute temperatures in your calculations.  It honestly should matter as all of them stem from the Ideal gas law:
$$PV = nRT \Rightarrow \frac{P_1V_1}{n_1T_1} = R = \frac{P_2V_2}{n_2T_2}$$
Where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal gas constant, and $T$ is the absolute temperature.  Boyle's, Charles, Gay Lussac's, and the combined gas Laws assume that the same number of moles (molecules) are used for each initial and final condition ($n_1 = n_2$).
The Combine gas law makes no further assumption than constant number of moles ($n_1 = n_2$) giving:
$$\frac{P_1V_1}{T_1} = nR = \frac{P_2V_2}{T_2}$$
Boyle's law, adds that the  initial and final temperatures are assumed to be the same ($n_1T_1 = n_2T_2$) thus:
$$P_1V_1 = RnT = P_2V_2$$
Charle's law, adds that the initial and final pressures are assumed to be the same ($\frac{P_1}{n_1} = \frac{P_2}{n_2}$) thus:
$$\frac{V_1}{T_1} =\frac{V_2}{T_2}$$
Gay-Lussac's law, adds that the initial and final volumes are assumed to be the same ($\frac{V_1}{n_1} = \frac{V_2}{n_2}$) thus:
$$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$
The ideal gas law is for ideal gases which gives values close enough to most real gases.  If your customer would like to be more accurate though there is the Van Der Waals equation:
$$P = \frac{nRT}{V-nb} - \frac{n^2a}{V^2}$$
Where $a$ and $b$ are constants specific to the gas.  But again your most likely malfunction is using Celsius temperatrue for $T$ in stead of Kelvin temperatue ($T_\mathrm K = T_\mathrm C+273.15$)


text:
Oxidation of alkenes with KMnO4
This is a weird question but it's finals week and it's impossible to get in contact with my prof. I have no idea what he means by the last line where he says a C-O bond is formed for every C-C and C-H bond. Can someone just clarify this for me?

The reaction mechanism is complicated. You just have to know that hot alkaline KMnO4 converts an aldehyde to a carboxylic acid. In case of formaldehyde, its converted to carbon dioxide. KMnO4 cleaves alkenes just like ozonloysis. If cold dil. KMnO4 is used, instead of forming a carbonyl compound, a syn dihydroxylation compound is formed. http://chemwiki.ucdavis.edu/Organic_Chemistry/Reactions/Oxidation_and_Reduction_Reactions/Oxidation_of_Organic_Molecules_by_KMnO4

Yeah I know that part. Just confused about this rule of  thumb stuff. If that's true CH2-CH2 would end up with 4 CO bonds??

Yes. It will end up with two molecules of carbon dioxide. First two molecules of formaldehyde which then becomes HCOOH and finally CO2.

ohhh got it. Thanks for the clarification. But if I extend the length of the alkene wouldn't this "rule of thumb" not really work anymore? eg CH3-CH2-CH2-CH-CH-CH3. Wouldn't there need to be 17 CO bonds?

In your "rule of thumb", its given "alkene". But you are talking about an alkane.

I put a CH-CH in there though, does that not make it an alkene?

I didn't read it properly. Only the hydrogen of the carbons having a double bond is oxidized. Initially its oxidised to aldehydes. Then, the alpha hydrogen is oxidized to OH to get carboxylic acid. After that, there is no remaining alpha H to be oxidized. So reaction stops with two carboxylic acids.

Your professor obviously wants you to consider the $\ce{C-C}$ bonds of the $\ce{C=C}$ double bond plus any $\ce{C-H}$ bond to either of the alkene carbons.
The first step in double bond oxidation by permanganate will always be replacing a $\ce{C=C}$ by $\ce{C=O + O=C}$. If you consider the double bond two bonds, then for each of those bonds each carbon atom gets one $\ce{C-O}$ bond to again give a double bond.
After we have generated the inital carbonyl, we can turn our attention to the other bonds the former alkene carbon has. Remember that aldehydes are easily oxidised to carboxylic acids while ketones are rather resistant to oxidation. The difference is the presence of a $\ce{C-H}$ bond in the aldehyde; permanganate can oxidise this to another $\ce{C-O(-H)}$ bond. This is the second part of your rule.
If accidentally the carboxylic acid species generated is formic acid, which has another $\ce{C-H}$ bond, this can again be oxidised to give carbonic acid. Since carbon’s oxidation state is maximal in $\ce{H2CO3}$, this is truly the end of the oxidation.


text:
How to calculate the concentration of the diluted solution given a source concentration in ppm?

A student adds $37.4~\mathrm{ml}$ of a $50.0~\mathrm{ppm}$ solution to a $100~\mathrm{ml}$ volumetric flask and dilutes to the mark. What is the concentration of the diluted solution?

$13.7~\mathrm{ppm}$
$7.5~\mathrm{ppm}$
$37.4~\mathrm{ppm}$
$23.6~\mathrm{ppm}$

I used $$M_1V_1 = M_2V_2$$ which worked for all the others and I assumed would work for this considering I had all of the variables except for the 2nd concentration, but those calculations gave me $18.7~\mathrm{ppm}$, which is not an answer choice.

Let us assume for a moment, that the parts-per-notation is actually unambiguous. We then know that we can convert the concentration given in ppm $c^\dagger$ into the concentration $c$ in SI units $\mathrm{mol\,L^{-1}}$, or $c^\dagger\propto c$. Then we know that the concentration is defined as the amount of substance per volume of solution.
$$c = \frac n V$$
When diluting a solution the amount of substance stays the same, hence we have two equations, that we can combine.
$$\begin{align}
  c_0 &= \frac{n}{V_0}  &&\text{source solution}\tag1\\
  c_\mathrm{dil} &=\frac{n}{V_\mathrm{dil}}  &&\text{diluted solution}\tag2\\
  c_\mathrm{dil} &=\frac{c_0V_0}{V_\mathrm{dil}}  &&\therefore\tag{1=2}
\end{align}$$
Which is exactly the formula you used (when I assume that $M$ stands for molarity).
Ergo your calculation result, as orthocresol already pointed out, of $c^\dagger_\mathrm{dil} = 18.7~\mathrm{ppm}$ is correct.
The error therefore has to be in the question, or in copying the same.


text:
What is the purpose of finding log k?
I understand that k alone is the equilibrium constant and know how to find this number or when given it I can take the log of this number, but what is the purpose of this number? What does the log of k tell us in relation to the reaction? Why do we find log k?

$\log K$ relates to the thermodynamic changes in entropy and enthalpy of a reaction.
$$\Delta H - T \Delta S = \Delta G = \Delta G^\circ - RT \ln Q$$
At equilibrium $Q = K$ and:
$$\Delta G = 0 = \Delta G^\circ - RT \ln K \Longrightarrow  \Delta G^\circ = RT\ln K$$
$$\Delta G^\circ = \Delta H^\circ -T\Delta S^\circ = RT \ln K$$
and:
$$\ln K = \frac{\Delta H^\circ -T\Delta S^\circ}{RT} = \frac{\Delta H^\circ}{RT} - \frac{\Delta S^\circ}{R}$$

The logarithm of k is equally effective to understand the thermodynamics of a reaction.As the logarithmic function is increasing, you can compare log k of two reactions and arrive at similar qualitative interpretations as you would for k.The most important thing is that the the standard free energy change for a reaction is directly related to the logarithm of k.
                 $\Delta G^\circ = -RT \ln K$
Which actually tells you that the logarithm of k is often more useful for thermodynamic computations.