text:
Why is carbon dioxide gas used in soda?
Why is carbon dioxide used in soda? What makes it preferable to other gases for this purpose?

Because it tastes better than chlorine or any other gas, I guess.

Almost a $+1$ to your comment @IvanNeretin but I personally dislike drinks with extra gasses in them ;)

You are approaching this question from the wrong angle. One does not simply ‘add a gas to a drink’ for no reason. If one wanted to add any gas, one would probably use boring old nitrogen since it is the most abundant in the atmosphere (See the Guiness example as pointed out by Ivan in a comment).
Carbon dioxide is present in fizzy drinks such as soda to give them their fizz. The reason lies in the equilibrium between $\ce{CO2 + H2O}$, $\ce{H2CO3}$, $\ce{HCO3-}$ and $\ce{CO3^2-}$. No other gas would display this behaviour and give a slight acidic taste along with the tingle on the tongue.
Note that oxygen is sometimes added to drinks to improve the assumed value. It doesn’t really do anything but sells for a higher price. I don’t think that would be the case for nitrogen- or argon-enriched water (but I am likely underestimating the powers of marketing here). Many other gases would simply make no sense to add at all because they are either toxic ($\ce{Cl2, H2S}$) or disgusting ($\ce{H2S}$) or simply way too expensive ($\ce{Xe}$).

Nitrogen is a technical challenge, for it does not dissolve that well. However, they do exactly that with Guinness beer.

Oxygen (or ozone that breaks down to give oxygen) could destroy yucky organisms or organics, although one can ask how long this effect actually lasts.

Carbon dioxide gas is comparatively more stable and better dissolves in water rather than other common gases. Moreover, addition of the gas gives the drink the fizziness which in other ways react with water to form carbonic acid which in turn produces a lemonish flavour...Over all, carbon dioxide is prefered as it is more feasible...and for a variety of other reasons.
Hope that helps!

Maybe you could expand upon what the 'other reasons' are?

Certain gases can be toxic for human consumption, CO2 also gives the tingly feeling of the drink,...


text:
Explain the enolisation process of 5,8-dihydroxynaphthalene-1,4-dione

I have no idea really how to get started...
Where does the hydrogen for the first deprotonation come from?
My working thus far:

Yep. To explain the mechanism using enolisation and protonation steps only

I just want to point out that in your mechanism there is at least one carbon with five bonds.

There is no need for any catalyst apart from acid or base. Also, the $\ce{^{13}C}$ is just there to tell you that it is not the same molecule. Also, this is not a zwitterion, it is tautomerism since there is pi bond movement involved.
Don't deprotonate any carbons. They're all $\ce{sp^2}$ and won't take kindly to deprotonation. Just push electrons from one oxygen to the other through the conjugated system.
Try doing this one first. In acid, your general tautomerism mechanism would look like this:

Protonate the carbonyl group.
Push electrons from phenol oxygen.
Deprotonate the carbonyl formed on the other side.

In base, it would be:

Deprotonate the phenol.
Push electrons from phenoxide oxygen.
Reprotonate the phenoxide formed on the other side (strictly this shouldn't happen, but whatever).

If you can do this one, the mechanism for your original question just consists of the same thing repeated. You can "fill in" the parts of the molecule that I omitted to see what I mean.

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text:
The IUPAC name of sodium nitroprusside, and back-bonding in it
In the standard test for the sulfide ion, sodium nitroprusside $(\ce{Na^_2[Fe(CN)5(NO)]})$ is used. The IUPAC name for this compound as stated by Wikipedia is sodium pentacyanidonitrosylferrate(II)
I have three queries related to this:

Does the nitrosyl ligand donate a pair of electrons (through back bonding) to form $\ce{(NO)^+}$ or is $\ce{(NO)^+}$ the ligand?
In both the cases I guess the name should be sodium pentacyanidonitrosoniumferrate(II) instead.
Now the product formed after the sulfide test is $\ce{Na^_4[Fe(CN)5(NO)S]}$ whose IUPAC name is sodium pentacyanidonitrosylsulfidoferrate(III). But the nitrosyl ligand is neutral in this case. So why in this case there is no back bonding?

Nitrosyl is the textbook example for a non-innocent ligand — that is, one that can appear in different oxidation states with different numbers of donated electrons and can make determining the bonding situation rather complicated. It can appear as the nitrosyl cation $\ce{NO+}$ (isosteric to $\ce{CO}$ or $\ce{CN-}$), the nitrous oxide radical $\ce{NO^{.}}$ or the ‘hyponitrite’ anion $\ce{NO-}$ (isosteric to either $\ce{{}^1O2}$ or $\ce{{}^3O2}$). I’m not going to start talking about possible donated electron counts yet.
Nomenclature
For nomenclature purposes, any form of $\ce{NO}$ bound to a metal is always treated as a neutral three electron donor and is always termed nitrosyl. This is because IUPAC wants rules that can be applied blindly without having to perform complicated experiments to determine how exactly the ligand is bound. And we shall see in a second that it is very non-trivial.
Considering that and checking the $\ce{[Fe(CN)5NO]^2-}$ fragment, we take away 5 $\ce{CN-}$ leaving us with $\ce{[Fe(...)NO]^3+}$ which indicates that the correct IUPAC name for sodium nitroprusside is sodium pentacyanidonitrosylferrate(III). (Not (II).)
The product formed would be, by the same approach, sodium pentacyanidonitrosylsulfidoferrate(III). But we’ll come back to that.
But what is the actual ligand?
As I already stated, nitrosyl is a non-innocent ligand. It can be $\ce{NO+}$ and allow significant π-backbonding (metal to ligand) as exemplified in the tricarbonylnitrosylferrate($\mathrm{-I}$) system $\ce{[Fe(CO)3NO]-}$. By formality, this is considered a neutral donor giving iron a formal oxidation state of $\mathrm{-I}$. But in reality the ligand is $\ce{NO+}$ meaning that we are dealing with iron($\mathrm{-II}$) — analogous and isoelectronic to the tetracarbonylferrate($\mathrm{-II}$) complex.
On the other side of the spectrum we have the complex formed in one nitrate detection method: $\ce{[Fe(H2O)5NO]^2+}$ — pentaaquanitrosyliron(II) according to IUPAC. Approaching this complex is difficult, however most evidence points towards some structure between $\ce{[Fe^{II}NO^{.}]^2+}$ and $\ce{[Fe^{III}NO- ]^2+}$. The $\ce{Fe-N-O}$ unit is still linear (VSEPR would predict $120^\circ$ for  a hyponitrite). This can be explained by considering the $\ce{NO-}$ (or $\ce{NO^{.}}$) in a triplet state much like $\ce{^3O2}$ (or a corresponding dublet state in case of $\ce{NO^{.}}$). There is a slight tendency for the unpaired electrons in iron and $\ce{NO}$ to couple; this is usually understood as antiferromagnetic coupling and results in an overall observed spin of $+\frac{3}{2}$.
In the $\ce{NO+}$ case, the bonding picture is rather clear: $\ce{NO+}$ behaves exactly like $\ce{CO}$. We have a σ donating (forwards) bond and significant π back bonding. The other case is rather hazy except for the σ bonding, which is of course ligand to metal. As I wrote, the spin coupling is considered antiferromagnetic coupling and not actual bonding usually, and if anything, it would be a covalent type of bond (each fragment donating one electron). Thus it would be wrong to speak about back-bonding — especially since metals in high oxidation states usually don’t back-bond significantly.
Removing the five cyanido ligands in nitroprusside leaves us with a $+3$ charge as shown above. Thus we are either dealing with $\ce{[Fe^{II}NO+]}$ or $\ce{[Fe^{III}NO^{.}]}$. The extremely short $\ce{N#O}$ distance points strongly towards an $\ce{NO+}$ cation. $\ce{Fe^{II}}$ is again not good at back-bonding and no π forward bonding is to be expected due to the mismatched energy of the π electrons of $\ce{NO}$. Therefore, the $\ce{NO}$ group is rather labile and can be displaced at ambient temperature and pH. This is the mode of action of nitroprusside as a drug: It releases $\ce{NO}$.
And what about the sulfide test?
I didn’t dig into it deeply, but it seems like you are mistaken in the product formed. What I found on the web[1] implies (I couldn’t read the actual article, only the abstract) that sulfur does not coordinate additionally to the iron centre (there would not be much space anyway) but rather attacs the $\ce{N#O}$ triple bond nucleophilicly on the nitrogen atom. Therefore, the final complex is better written as $\ce{[Fe(CN)5(SNO{-}\kappa N)]^4-}$ and better named pentacyanidothionitrito-κN-ferrate(III).

References
[1] S. L. Quiroga, A. E. Almaraz, V. T. Amorebieta, L. L. Perissinotti, J. A. Olabe, Chem. Eur. J. 2011, 17, 4145. DOI: 10.1002/chem.201002322
Background from Prof. Klüfers’ internet scriptum to his coordination chemistry II course.

A forum discussion in the depths of the interwebs (no source because just no) seems to confirm my assumption on thionitrite.

According to "Red Book- IUPAC Recommendations 2005 ", at page 315, $\ce{NO+}$ is oxidonitrogen.

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text:
Why is there a need of polar coordinates to solve the Schrödinger wave equation for the hydrogen atom?
Why is there a need of polar coordinate to solve the Schrödinger wave equation for the hydrogen atom?
I went through some standard text books but I am feeling rather confused about the explanation.

It's just natural because the potential energy term varies as $1/r$. That makes it much easier to separate and solve (as opposed to trying to use $1/\sqrt{x^2 + y^2 + z^2}$).

In short, that's because the problem has spherical symmetry, and you want to use the coordinates which retain as much as possible of that symmetry. Not a chemical question, really.

I'm voting to close this question as off-topic because it's purely about the mathematical methods used to solve equations and not about conceptual chemistry.

I am not sure if this would be eligible for migration though.

I’m going to vote to leave open even though it is a maths/physics question technically. Reasoning being that as far as I know all the basic quantum stuff is usually handled in chemistry courses.

Since the problem exhibits spherical symmetry, it certainly is easier to solve it in spherical coordinates, with spherical harmonics being the straightforward answer. Use math to your advantage, not to your detriment.

The question is really just wrong. There is no "need" to use polar coordinates, it is just simpler to do the math in polar coordinates.

I was of the opinion that using polar coordinates would simplify everything but I'm questioning it now I see the solutions to the spherical harmonics are actually simpler expressed in cartesian coordinates than polar (see https://en.wikipedia.org/wiki/Table_of_spherical_harmonics).

Tl;DR
When solving the Schrödinger equation for the hydrogen atom, its potential energy is dependent on $r$. Using Cartesian co-ordinates $r$ is expressed as $\sqrt{x^2+y^2+z^2}$ which greatly complicates the process of solving the equation. Therefore we use polar co-ordinates as the variables are$~r,\theta,\psi$ which means that there is no nasty square root sign as we don't have to split $r$ into terms of $x,y,z$.
Long Version
This is because to solve the Schrödinger wave equation for the hydrogen atom you require to use the technique known as, separation of variables. Separation of variables is a method for solving partial differential equations. For a partial differential equation in a function $\Phi~(x, y~...)$ and variables $x, y,~...$ we assume that it can be factored into functions which is function of only one variable:
$$\Phi~(x, y,\dots) \equiv X(x)~Y(y)~\dots$$
By breaking the resulting equation into a set of independent ordinary differential equations, we are able to easily solve for $X~(x), Y~(y), ~...$ and then plug them back into the equation to get our original function, $\Phi~(x, y~...)$. For example, lets consider this differential equation of
$$\frac{\partial^2 U(x,y)}{\partial x^2} - \frac{\partial U(x,y)}{\partial y} = 0$$
To solve this partial differential equation we use the separation of variables method. We assume that $U(x,y)$ can be written as $X(x)~Y(y)$. Therefore the equation becomes:
$$\frac{\partial^2 (XY)}{\partial x^2} = \frac{\partial (XY)}{\partial y} $$
Now, the beauty of this is that $X$ and $Y$ are independent of each other meaning that $X$ is not a function of $Y$ and $Y$ is not a function of $X$. Therefore we can rearrange that equation to give:
$$\frac{Y~\partial^2 X}{\partial x^2} = \frac{X~\partial Y}{\partial y} $$
The equation can be rearranged to give:
$$\frac{X''}{X} = \frac{Y'}{Y}$$
Since each side of the equation is a constant with respect to the other since one is a function of x and the other is a function of y, we can write:
\begin{align}\frac{X''}{X} &= C, &\frac{Y'}{Y} &= C\end{align}
Now each differential equation can be solved independently of other to obtain $X(x)$ and $Y(y)$. From that, we can easily find $U(x,y)$.
So as you can see, for the method of separation of variables, it is crucial that you are able to split the variables apart so that you can solve each differential equation independently of other variables. This same technique can be applied to solve the Schrödinger equation for the particle in the $3$D box. However when we try to do this for the hydrogen atom we get a problem. The Schrödinger equation for the hydrogen atom using Cartesian co-ordinates looks like this:
$$\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right)\psi + \frac{2m}{\hbar^2}\left(\frac{e^2}{r}+E\right)\psi = 0 $$
This is a partial differential equation in three variables and we suspect that we need use separation of variables technique to solve it. So lets assume that $\psi(x,y,z) \equiv X(x)~Y(y)~Z(z)$. Now all we have to do is substitute $\psi$ with that and then separate the $x$, $y$ and $z$ variables so that they are together then solve them independently.
But there is one problem. In the equation, $r$ which is the radius is equal to $\sqrt{x^2+y^2+z^2}$ That is a problem because due to the square root sign there is no way we can separate the $x$, $y$ and $z$ variables.
Therefore to solve this equation more easily, we use polar co-ordinates. By using polar co-ordinates, we take advantage of its spherical symmetry which is convenient as the entire system also has a spherical symmetry. So we now transform the Cartesian Schrödinger equation to polar co-ordinates, so that it is dependent on the variables, $r, \theta, \phi$. The Schrödinger equation using polar co-ordinates is:
$$
  \frac{1}{r^2}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial}{\partial r}\psi\right)
   + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial}{\partial \theta}\psi\right)
   + \frac{1}{r^2\sin^2\theta} \frac{\partial^2}{\partial \phi^2}\psi
   + \frac{2m}{\hbar^2}\left(\frac{e^2}{r} + E\right)\psi = 0
$$
This may seem really complicated at first, however with some manipulation, we will be able to separate out all three variables, $r, \theta, \phi$.
We assume that $\psi ~(r, \theta, \phi) \equiv R(r)~\Theta(\theta)~\Phi(\phi)$. For simplicity we will write them as $R, \Theta, \Phi$ without showing the functionality. The Schrödinger equation becomes:
$$
  \frac{1}{r^2}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ \Theta\ \Phi}{\partial r}\right)
   + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial R\ \Theta\ \Phi}{\partial \theta}\right)
   + \frac{1}{r^2\sin^2\theta} \frac{\partial^2 R\ \Theta\ \Phi}{\partial \phi^2}
   + \frac{2m}{\hbar^2}\left(\frac{e^2}{r} + E\right)R\ \Theta\ \Phi = 0
$$
This simplifies to:
$$
  \frac{\Theta\ \Phi}{r^2}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ }{\partial r}\right)
   + \frac{R\ \Phi}{r^2 \sin\theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right)
   + \frac{R\ \Theta}{r^2\sin^2\theta} \frac{\partial^2  \Phi}{\partial \phi^2}
   + \frac{2m}{\hbar^2}\left(\frac{e^2}{r} + E\right)R\ \Theta\ \Phi = 0
$$
Divide both sides by $R\ \Theta\ \Phi$ and multiply by $r^2\sin^2\theta$:
$$
  \frac{\sin^2\theta}{R}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ }{\partial r}\right)
   + \frac{\sin\theta}{\Theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right)
   + \frac{1}{\Phi} \frac{\partial^2  \Phi}{\partial \phi^2}
   + \frac{2mr^2\sin^2\theta}{\hbar^2}\left(\frac{e^2}{r} + E\right) = 0
$$
Inspection shows that no other terms except for the $3$ terms has a functional dependence on $\phi$. Therefore with respect to the other variables, the third term can be treated as a constant, which we will call $-n^2$ for convenience. So we obtain: $$\frac{1}{\Phi} \frac{\partial^2  \Phi}{\partial \phi^2} = -n^2$$ The general equation to this can be easily solved. So we have separated one of the three variables. Now the equation can be written in the form:
$$
  \frac{\sin^2\theta}{R}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ }{\partial r}\right)
   + \frac{\sin\theta}{\Theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right)
   + -n^2
   + \frac{2mr^2\sin^2\theta}{\hbar^2}\left(\frac{e^2}{r} + E\right) = 0
$$
Dividing by $\sin^2\theta$ and rearranging:
$$
  \frac{1}{R}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ }{\partial r}\right)
   + \frac{2mr^2}{\hbar^2}\left(\frac{e^2}{r} + E\right)
   + \frac{1}{\Theta\sin\theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right)
   + \frac{-n^2}{\sin^2\theta}
    = 0
$$
Now by inspection, the first two terms depend on $r$ while the last two terms depend on $\theta$. As before, we will set the two terms equal to a constant, $\beta$, so that:
$$
  \frac{1}{R}\frac{\partial}{\partial r}
     \left(r^2 \frac{\partial R\ }{\partial r}\right)
   + \frac{2mr^2}{\hbar^2}\left(\frac{e^2}{r} + E\right) = -\beta
$$
$$
     \frac{1}{\Theta\sin\theta} \frac{\partial}{\partial \theta}
     \left(\sin\theta \frac{\partial \Theta\ }{\partial \theta}\right)
   + \frac{-n^2}{\sin^2\theta} = -\beta
$$
Now we have successfully separated all 3 variables into 3 separate single variable differential equations that can be solved relatively easily.

But there is still an $r$ in the expression (in $\tfrac{e^2}{r}$), and that's the same $\sqrt{x^2+y^2+z^2}$ you mentioned as a problem for the Cartesian version, right?

@AndersMB Yes, there is still the same $r$ (and it still equals $\sqrt{x^2+y^2+z^2}$)  in the expression,. However the beauty of polar co-ordinates is that the variables depend on $r, ~\theta, ~\phi$ not $x, ~y,~ z$. Therefore we just have to group all the $r$'s together then solve the equation independently. There is no need to split $r$ into terms of $x,~y,~z$ as they aren't variables in the polar equation.

For an extensive answer, check out Nanoputian’s one.
In a nutshell it boils down to the following:

Mathematically, it does not matter which coordinate system you use.
Since your entire system is of spherical symmetry (the effective charge decreases proportionally to $\frac{1}{r}$) you benefit from choosing a coordinate system which has the same inherent symmetry.
Cartesian coordinates do not have this symmetry why polar coordinates do.
Thus, polar coordinates allow for easier calculations.


text:
Rusting without water?
I was under the assumption that corrosion (rust specifically) required water + oxygen to occur. But the word equation below states that only oxygen is necessary. I know this is true but I do not know the explanation behind it. Could someone help me understand? Apparently water magically appears from somewhere.
A middle to high school level of explanation would be ideal, not too advanced.
metal + oxygen ->  metal oxide
metal + oxygen + water -> metal oxide

What's your question really, 'cause I don't see much sense here? Water is a often a problem if you want to get oxide - that's why rust isn't pure oxide.

Quite simply I want to know how does rusting occur without the notable presence of liquid water? I thought it was essential to rust formation. But as I stated apparently rust can form by reacting iron with only oxygen? I'd like to understand how this works if you'd be so kind?

Why, many metals would react with oxygen directly (just like they teach in middle school, metal + oxygen $\to$ metal oxide ). Iron, for one, certainly would. It may be going slower than with water, and produce different resulting compounds (because common rust is not really oxide, it is hydroxide), but anyway.

Try an experiment: Using tongs, hold a piece of steel wool in a propane torch flame. Let us know the results. Hopefully you will have answered your own question.

It seems the rust reaction was not well explained.  The following is the formula for iron rusting with water.
$$\ce{4Fe + 3O2 -> 4Fe^3+_(aq) + 6O^2- _(aq) -> 2Fe2O3}$$
Water will dissolve the iron because often carbon dioxide is available in the atmosphere to make the water slightly acidic. With the help of acidity electrons can be transported much easier from from dissolved iron to dissolved oxygen. Where oxygen, iron, and water meet is most often where rust pits start to form.
Without water the formula is the following:
$$\ce{4Fe + 3O2 -> 2Fe2O3}$$
This reaction is very slow.  More energy is needed to initiate the reaction as in the case with burning steel wool, or another means is needed to transport electrons.  Dirt, grime, residues, or any number of things can transport electrons resulting in rust.

Rusting is a more sophisticated reaction than what I have described. For the benefit of the person who asked the question a simple metal and oxidizing component was chosen. The real world is likely to have metallic impurities, hydroxide, a mix of iron II and iron III, condensation, coatings, and a whole host of other things specific to a particular circumstance. Going into detail is not necessary to answer the question.


text:
How to calculate the mass of magnesium necessary to produce a certain volume of hydrogen gas?

How many grams of magnesium metal are required to produce 44.8 liters of hydrogen gas at STP according to the chemical equation shown below? (Answer: $\pu{48.6 g}$)
$$\ce{Mg(s) + 2 HCl(aq) -> MgCl2(aq) + H2(g)}$$

What I have done:
Amount of hydrogen to be produced= 44.8/22.4 = 2 moles of $\ce{H}$ = 2.02 grams of $\ce{H}$.
Amount of magnesium metal needed $$=\frac{\pu{2.02 g} (\text{of}~\ce{H})\cdot \pu{1 mol} (\text{of}~\ce{H2})\cdot \pu{1 mol} (\text{of}~\ce{Mg})\cdot \pu{24.31 g} (\text{of}~\ce{Mg})}{\pu{2.02 g} (\text{of}~\ce{H2})\cdot \pu{1 mol} (\text{of}~\ce{H2})}$$
I'm not getting the correct answer, can someone please explain how to do this using dimensional analysis? Please just tell me only how to do this, do not show it.

You don't have moles of $\ce H$. You have moles of $\ce{H2}$.

The molecular weight of Mg is 24.31 g/mol. Moles multiplied by MW gives the mass (in grams).

I have updated your post with some chemistry markup (Your attempt confused me a bit too much, so I skipped that). If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details.

Note that, apparently, the question still uses the old definition of STP with a pressure of $p=1\ \mathrm{atm}$ and a molar volume of an ideal gas of $V_\mathrm m\approx22.4\ \mathrm{l/mol}$.
Since 1982, according to IUPAC recommendations, STP corresponds to a temperature of $T=273.15\ \mathrm K$ and a pressure of $p=100,000\ \mathrm{Pa}$. At this temperature and pressure, the molar volume of an ideal gas actually is $V_\mathrm m=22.710,947(13)\ \mathrm{l/mol}$.

It is really simple to solve this problem. You want to have 2 moles of H2. So you'll require twice the amount of Mg required to produce one mole of H2.
Since 1 mole of Mg produces 1 mole of H2. 2 moles of Mg will produce 2 moles of H2.
We dont need to go to the basics to solve this problem.

It is very helpful when you start with using proper notation, i.e. $V(\ce{H2})=44.8~\mathrm{L}$. Proper notation like $M$ for molecular mass, $m$ for mass, and $n$ for amount of substance will help you follow through your thinking.
When do the calculations, always carry the units with every quantity you use.
It is necessary to know a few principle formulae, for example that molecular mass and mass are connected through the amount of substance: $$M = \frac{m}{n}$$
There are a few more you will need and with their repeated use you will get to know them easily.
When attacking this problem, find out how the quantities relate to each other. In this example it is: If I want to produce one mole of hydrogen gas, how many moles of magnesium do I need?

 one, because $n(\ce{Mg})\ce{->}n(\ce{H2})$

A key principle (and asumption) in this exercise is the ideal gas law,
$$pV=n\mathcal{R}T.$$
You need it to calculate the amount of substance of hydrogen that should be produced.
(You have indirectly used it by dividing the given value by the molar volume of an ideal gas at STP.)

 In this case about two moles of hydrogen gas shall be produced, i.e.
 $\displaystyle n(\ce{H2})=\frac{pV(\ce{H2})}{\mathcal{R}T}\approx2~\mathrm{mol}$
 (Old definition of STP: $p=1~\mathrm{atm}$ and $T=298.15~\mathrm{K}$)

Now you also know the amount of substance of magnesium you need.

 two moles, i.e. $n(\ce{Mg}=2~\mathrm{mol}$

You just have to use the molecular mass of magnesium to find the mass of magnesium you need with the above connection.

I'm sorry but how did you get 2 moles of magnesium? Also I'm mostly confused on how to solve this using dimensional analysis, no matter what I do I can't seem to find the correct answer.

@user3882522 Can I stress Martin’s points of 1) using proper notations, and 2) properly using units?


text:
Why is the smell of car exhaust much stronger in winter?
I noticed that in winter, the smell of cars (to be precise the smell of the exhaust gases) is much stronger in winter. I just can't figure out why. Has this got something to do with the air temperature or with the humidity?

While car engines might look simple there is a vast amount of technology installed to ensure as complete a combustion as possible. Complete combustion according to the equation
$$\ce{C_{$m$}H_{$n$} + \frac{2$m$ + $n$}{2} O2 -> $m$~CO2 + \frac{$n$}{2} H2O}$$
(not accounting for oxygen, nitrogen or other elements present in the fuel) requires a precise amount of oxygen added. Too little oxygen will result in uncombusted hydrocarbons while too much oxygen will result in $\ce{NO_{$x$}}$. A small oxygen sensor is thus installed to improve the oxygen regulation. (It is a very interesing compound because it functions as an oxide ion conductor.†)
The thing about this oxygen sensor is that its function is temperature dependent: It needs to be heated to $\approx 500~\mathrm{^\circ C}$ to function properly. This is achieved quickly due to the heat of the exhaust, but does take a few minutes from starting the car. So throughout the first few minutes, the oxygen sensor does not function and incomplete combustion results in $\ce{NO_{$x$}}$ and uncombusted hydrocarbons that cause the odour. Once working temperature has been reached, the smell disappears mostly.
In cold weather it will take longer for the oxygen sensor to function properly and thus more incomplete combustion is observed. This becomes more pronunced the colder it gets.

†: Its German name is Lambdasonde from the Greek letter $\lambda$ which is used as a variable to determine if the correct amount is present or not. At the correct oxygen concentration, $\lambda = 1$.

A couple of points- oxygen sensors are usually heated in modern cars (if the sensor has 4 wires running to it, it's heated) so the warmup time is much shorter. However, engine management systems will not enter closed loop control (where the oxygen sensor reading is used to control the fuel mixture) until the engine has reached a certain temperature anyway.

@squigbobble I didn’t know that. (But all of what I do know is from a bachelor’s introductory lecture to solid state chemistry which was long ago, so no surprises there ;).)

It's the air temperature. Petrol engines need a richer (a larger proportion of fuel to air) mixture in order to run properly when the cylinder head and intakes are cold due to the fuel condensing onto the inside of the intake and ports before reaching the cylinder. The richness is roughly proportional to the temperature of the engine so the colder the engine is, the richer the mixture will be. The excess fuel does not burn completely and ends up as a variety of smelly chemicals in the exhaust.
For modern vehicles with a catalytic converter, there's also a secondary effect where the cat takes longer to reach its operating temperature so more uncatalysed emissions escape out of the exhaust, they smell bad too.


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What happen to the Salt in Water Electrolysis
I haven't found yet what happen to the Salt Ions in Water during Electrolysis and is there an Electron Flow or Ion Flow.
I have expierienced strong magnetic Forces between the Electrodes but couldnt measure an outside Magnetic Field.

When a battery is connected to an electrolytic cell one electrode becomes positively charged the other negatively charged. The ions in the liquid are attracted to the electrode with the opposite charge and move through solution. Ions with a negative charge move to the positive anode (anions), Ions with a positive charge migrate to the negative cathode (cations).
Ionic solids produce electrolytes. When an ionic solid is dissolved in water the aqueous ions are free to move and they carry the current. The anions of the salt and water give up electrons which flow through the circuit. The anions are oxidized at the anode while the cations of the salt and water recieve the electrons and are reduced at the cathode.
Cations will cluster around the cathode and anions will cluster at the anode. It is usually easier to displace hydrogen atoms from water molecules in the aqua ion than to displace the cation from the system  so electrons from the cathode react with water molecules , and hydrogen is displaced.
The result is decomposition of water molecules:
2H-OH(l) + 2e- → H2(g) + 2OH-(aq)
The solution becomes alkaline around the cathode due to the hydroxide ions being formed. If a metal is above hydrogen in the electrochemical series, hydrogen will always be produced at the cathode during electrolysis of solutions of its salts. If the ionic salt consists of a metal below hydrogen in the electrochemical series, electrolysis of its salt solution will always produce metal at the cathode.
Both ion flow and electron flow are occurring.
The intensity of an electromagnetic field reduces significantly the further away from the source.

You're mixing two different concepts.
First a magnetic field has nothing to do with an electrochemical cell. A magnetic field is created when a direct current flows in coil. Think of wrapping a wire around a nail.
As far as the ions in solution, the cations have a mass flow in one direction and the anions have a mass flow in the other. Think of having a theoretical "force field" that can gate anions and cations. As many cations would flow through as anions while current flows in the system.

An electromagnetic coil creates an electromagnetic field when a current is applied, energy radiates away from the source,  a regular magnet is static  and creates a Near magnetic or magnetic field where energy is stored...

Sorry I meant a electrochemical cell creates an electromagnetic field which is an electric field and a magnetic field at the same time.

@Joel - Well you're right any current flowing will create a magnetic field, but in a typical straight wire the field is very weak. An electromagnetic coil creates a much stronger field which is very easily detected. // A wire concentrates the current flow in a very small cross-section.  To detect the magnetic field in the electrochemical cell itself would probably require some sort of very special cell design and some sort of very special detector (e.g. a SQUID).


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Distinguishing Test for Organic Compounds
Given the following pairs of samples, how do I distinguish between them?

benzamide and ethanamide
ethanoyl chloride and ethanoic anhydride

My Approach:

We could easily convert the amides to amines by reducing them with $\ce{LiAlH4}$. Now, the question reduces to identifying the methyl group against the benzene group.

I am really lost here but my guess would be that we are looking for some characteristic reaction for anhydrides.

Since this sounds like an assignment, I am not going to give it all away.
Does you have to use qualitative chemical tests? I don't even teach those anymore. They were supplanted by spectroscopic techniques in most professional labs decades ago.
IR spectroscopy would answer both questions quickly.

Benzamide would have peaks characterizing an aromatic ring, for example:  $\ce{C}_{sp^2}\ce{-H}$ stretches around $3050-3100\ \mathrm{cm^{-1}}$. Ethanamide will not have these peaks, but will instead have $\ce{C}_{sp^3}\ce{-H}$ peaks $2800-3000\ \mathrm{cm^{-1}}$. There are other differences as well.
There are differences in the position (and sometimes number!) of carbonyl peaks between an anhydride and an acid halide. What are they?

So would mass spectrometry. Benzamide and ethanamide have different molar masses and different fragmentation patterns. Ethanoyl chloride... chlorine has two major isotopes. How can that help?
NMR spectroscopy would also be useful. However, IR is cheap, fast, and easy.

While I understand the answer, and agree that there are better ways to distinguish between chemicals, this unfortunately does not answer the question. I'm sure there are situations, where the OP's approach has value, too.

For the first case, we need to compare benzamide and ethanamide, One of the ways I can think of is comparing aniline and methylamine.
But we need to get there first, so we first react both our substrates with $\ce{Br2/NaOH}$ (Hoffman Bromamide degradation). As taken from Chem Libretexts/Hoffman Degradation:

Hofmann rearrangement, also known as Hofmann degradation and not to be confused with Hofmann elimination, is the reaction of a primary amide with a halogen (chlorine or bromine) in strongly basic (sodium or potassium hydroxide) aqueous medium, which converts the amide to a primary amine. eg:

Therefore, we now have aniline and methylamine. A way to compare these two is via diazotization, wherein we treat the compound with $\ce{cold NaNO2/HCl}$. After which we add β-napthol. Now, this reaction is known as the azo-dye test, where we see aromatic amines give an orange dye which is insoluble in water. Methylamine doesn't give this test and hence, we have distinguished the two compounds.
Case two, we have ethanoyl chloride and ethanoic anhydride, this is a much easier test, you need to add water to both substrates, one produces white fumes of HCl (ethanoyl chloride) during hydrolysis and the other substrate (ethanoic anhydride) shows no reaction but the water becomes acidic due to the formation of the carboxylic acid.

Hydrolysis of the amides produces in the first case acetic acid which is a liquid miscible with water. In the second case it makes benzoic acid which is solid and insoluble in water.
Let's consider now the second group of compounds. Acetic anhydride does not react quickly with water, and it produces only acetic acid. Acetyl chloride reacts quickly with water, producing acetic acid and hydrochloric acid.


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Reaction of Sulfur dioxide with Potassium
What are the products of the reaction: $$\ce{K + SO2->}$$
Here, $\ce{S}$ is in $+4$ oxidation state. Hence, $\ce{SO2}$ has both oxidizing and reducing character. $\ce{K}$ is a stronger reducing agent and hence, it will reduce $\ce{SO2}$.
So, $$\ce{K + SO2->KO2 + S}$$ is a possibility.
But the answer given is $$\ce{K + SO2->K2SO3 + K2S2O3}$$

$\ce{KO2}$ is definitely out of question. True, it does form when $\ce{K}$ reacts with $\ce{O2}$, but that's another story. It is a peroxide (in a broader sense), so it contains oxygen in oxidation states other than -2. $\ce{SO2}$ contains oxygen in -2. Who is going to oxidize oxygen?
Turning down peroxides, we resort to the more realistic option of $\ce{K2O}$. Now, that's a basic oxide, and $\ce{SO2}$ is an acidic oxide. Surely some chemistry is going to happen between them.
As to where the reduced sulfur would go, it is not quite clear and depends on many things. With excess of potassium one might expect ending up with $\ce{K2O}$ and $\ce{K2S}$. Obviously that's not our case. Then we must consider a reaction between $\ce S$ and $\ce{K2SO3}$. Such reaction is known (though probably under different conditions than we have here) and indeed produces thiosulfate.
That being said, partially reduced compounds of sulfur are numerous and I see no reason to favor one above the other without some deep investigation. Have you heard about dithionites, for example? They are known to appear when reducing $\ce{SO2}$ with certain metals; why not here?

I am hearing it for the first time. I have learnt reactions of sulphur dioxide with metals iron and magnesium. In both case, oxide and sulphide of the metal is formed just like you have specified.

It is much more complicated than that. $\ce{Zn}$ in water solutions gives dithionite. Without water (and with some heating), it probably would end up in oxide and sulphide as well.

Where can I read about this? (I am using JD Lee and Shriver,Atkins book for inorganic chemistry)

Are you writing "peroxide (in a broader sense)" instead of superoxide?

2 Aditya Dev: I'm no expert when it comes to literature. I learned this all many years ago, in a different language.
2 Mithoron: yes, exactly.


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What exactly happens at the critical solution temperature of a phenol water system?
What makes phenol that was turbid go clear beyond the critical solution temperature? We say beyond the critical solution temperature the constituents become soluble. What is the reaction or the exact change involved in the phenol water system (say with water in excess) beyond the critical solution temperature.

There is no reaction. As for the exact change... well, it's like you said: the components become soluble. I mean, they were soluble to some extent even before that point, but here they become soluble infinitely.

Critical Solution Temperature (CST)

In chemistry a system a two liquid system is said to have two phases. The system is turbid below the CST because one phase is droplets suspended in the other phase.


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electronegativity and how it relates to bond strength
my question is, if the strength of the bond between two atoms increases as the difference in electronegativity increases as well.
so say, a H-F bond would be stronger than a O-H bond.
I'm a little confused because I know that acidity increases from left to right across the periodic table. So H-F is a stronger acid than H-CH3. Wouldn't this mean that H-F has weaker bonds than H-CH3?

No it doesn't, those two things are independent. H-F bond is stronger but it's only one of factors influencing acidity.

Bond strength usually refers to the energy required to break a bond homolytically--with one electron going to one of the atoms and the other electron going to the other atom. In the case of HF this requires more energy than in CH4. Acidity is the tendency for the bond to break heterolytically--with two electrons going to one atom and none going to the other atom. In the case of HF, we get the two ions H+ and F-. This still requires a lot of energy, but in water solution, the ions are stabilized by hydrogen bonding, so ionization is much more favorable. Methane on the other hand would ionize to give a CH3- anion, which is a lot worse than F- due to carbon being less electronegative.

First off, you're comparing an ionically bonded $\ce{HF}$ with a covalently bonded $\ce{CH4}$.
HF is a stronger acid than $\ce{CH4}$ because its able to furnish out $\ce{H+}$ ions due to its ionic bonding.
So I think your question boils down to: Why is ionic bonding in $\ce{HX}$  (X=Halogen) stronger than covalent bonding?
This is because in covalent bonding there is only a mutual agreement to share electrons whereas electrostatic attraction in ionic bonding.

Bond In HF isn't ionic!


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Moisture Absorbancy & Odor Elimination
What would be the typical DRY ingredients in moisture absorbing and odor eliminating materials that work especially well on perspiration/organic odors?

Why type of clothing/application do you have in mind? You can't really absorb much moisture, the body makes a lot. It is more of a matter of wicking it away to increase  surface area and speed evaporation.

Thanks for question --  Application will be for removing moisture and odor from athletic shoes, gloves, gear.  I appreciate your thoughts!

Silica gel is often used to absorb moisture; activated carbon is effective at odor removal; and zeolites (also called molecular sieves) come in different varieties and can be effective for both purposes.


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Understanding the bond dissociation energy of poly-atomic molecules
Now I have a breaker which is containing $\ce{CH4}$ of some amount at standard conditions and then I put it on a burner.

In Thermodynamics we are taught about how to find the bond enthalpy of poly-atomic atoms. They say that such molecules (here in this case methane) all four $C-H$ bonds are identical. However, the energies required to break the individual $C-H$ bond in each successive step differs
$$\begin{alignat}{2}
\ce{CH4(g) &-> CH3(g) + H(g)}\qquad &&\Delta_\text{bond}H^\circ=+427\ \mathrm{kJ\cdot mol^{-1}}\\
\ce{CH3(g) &-> CH2(g) + H(g)}\qquad &&\Delta_\text{bond}H^\circ=+439\ \mathrm{kJ\cdot mol^{-1}}\\
\ce{CH4(g) &-> CH(g) + H(g)}\qquad &&\Delta_\text{bond}H^\circ=+452\ \mathrm{kJ\cdot mol^{-1}}\\
\ce{CH(g) &-> C(g) + H(g)}\qquad &&\Delta_\text{bond}H^\circ=+347\ \mathrm{kJ\cdot mol^{-1}}
\end{alignat}$$
and now their bond enthalpy is the mean of all those numbers.
now going back to the diagram we heat up the beaker containing methane then the energy that is absorbed by any one of the $\ce{C-H}$ bond should be equal to the energy absorbed by the other $\ce{C-H}$ bond because all the bond in a molecule should absorb equal amount of energy at the same time because at time $t$ the temperature is same all over the beaker and  all bonds are equal. so it should be like that while calculating the bond dissociation energy of a poly-atomic molecule (take the example of methane) calculate it like $4\times 427\ \mathrm{kJ\cdot mol^{-1}}$. (is this a common sense?)

all bonds Should broke up at once!

but we know that this does not happen, which means that all the bonds in a molecule does not absorb heat of equal amount at the same time or are they unequal? And fairly speaking it is painful to me to visualize this thing maybe because I am not able to understand this reality.
How do I comprehend this thing?
In Short

all bonds of $CH_4$ are equal
The temperature of beaker is same all over its volume
energy that is absorbed by any one of the $\ce{C-H}$ bond should be equal to the energy absorbed by the other $\ce{C-H}$ bond
bonds should absorb equal energy at time t
the should all break at once at same time t

this do not happen so how to understand this thing?

You are making several assumptions that assume equilibrium conditions. In this sense all the bonds are equal: they are on average over time.
However, at a molecular level (one molecule at a time) this is not true especially if enough energy is being added to break some bonds. The energy distribution for each bond will be Boltzmann-Like so some bonds will have enough energy to break. Once broken, of course, the other bonds no longer are CH4 bonds but are CH3 bonds of some sort. Hence we would not expect the next dissociation energy to be the same as the first.
The critical idea here is that the thermodynamic quantities of bond dissociation are averages and at a molecular level the actual energy is distributed according to statistical rules: some atoms or some of their bonds have more energy than others. The reactions occur not when the average energy input is equal to the dissociation energy, but when a small number of bonds at the extremes of the distribution have enough energy.
This is a very important idea (called statistical thermodynamics) in understanding chemical reactions.

Whoa!!!
The C-H bonds in $\ce{CH4(g)}$ are not the same as the C-H bonds in $\ce{CH3(g)}$. The $\ce{CH3(g)}$ "molecule" is electrically neutral but it is a radical with an unpaired electron in one orbit. In fact the physical orientation of the molecule changes to a planar configuration.

You can use \ce{} to format chemical expressions. See this meta post or, if you have lots of time, the mhchem package documentation.

okay, i got that, that bonds of $CH_4$ not same as that of $CH_3$, i pointed out that at time t all the bonds should break up at once but it is not that, so i want to know that why this doesn't  happen

If guess you're missing the point that C(g) doesn't exist at STP. A gaseous carbon atom would only exist in a very rarefied gas. The same goes for the rest of the "weird"  molecules. // A chemical reaction only happens when two "things" interact "physically." How would you pluck 4 atoms of H from methane at the same time (four directions at once)?

To look at it another way, let's imagine two very high velocity methane molecules (thus each molecule has a high "temperature") colliding in deep space. I can calculate the kinetic energy necessary to break all 8 bonds and yield 10 free atoms. That chemical reaction just wouldn't  happen. You'd get some weird assortment of polyatomic  species depending on the orientations of methane molecules as they collided.


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How come ancient Egyptian ink didn't fade during thousands of years?
This ink's main ingredients were charcoal, honey and gum arabica.
I understand charcoal is carbon and it would remain black for long time or forever just like coal. But gum arabica is organic and I would assume it would decay just like any organic material thus breaking the carbon bond and then carbon would've just turned into dust and disappear from papyrus or clay vessels, but apparently that's not the case.

You mostly seemed to have answered your own question. I could only add that ink is a liquid so that it wicks into the nooks and crannies of the substrate (papyrus or clay). So the bond between the dry ink and the substrate is a mechanical one.

I was trying to say that ink should've disappeared by now, and we shouldn't be seeing any writings at all, but you may have answered the question, I haven't considered the fact that it bonds mechanically with media. Then I'm curious why paints once covering most Roman statues didn't survive.


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Why are noble gases used in "neon" lamps
Neon lamps are lamps that contain noble gases... They light due to the presence of energetic levels for electrons (according to the definition of it in books). But I don't understand yet why noble gases are used in such lamps. Any help please?

there isn't much gases to begin with, and many complex ones may dissociate under electric discharge, producing solid deposits and degrading the color (carbon derivitives) or even react with parts of the tube even if producing a nice color (halogens)

A) to get the desired color (possibly in conjunction with a phosphor coating), B) stability, C) cost.

Strictly speaking "neon" lamps contain neon but the term is often used colloquially for a whole range of coloured lighting probably because the red neon tubes are one of the commonest.
But it isn't just noble gases that are used, though they are the most common. Carbon dioxide is sometimes used. And there are many discharge lamp that add metals or metal salts, though these are more common in other uses such as high intensity street lighting.
The noble gases and mixtures of them are commonly used in coloured advertising lighting because they offer a range of colours in a simple low-pressure discharge lamp without the added complexity of fluorescent coatings (as in common fluorescent tubes) or high-pressure systems (as is required in high intensity systems used in street or industrial lighting). Advertising signage often consists of long glass tubes in custom shapes and the simplicity of simple gas discharge makes the results cheaper than more sophisticated alternatives.
The colours are derived from the energy levels of the electrons in the gases and it just so happens that the range of those energy levels in noble gases correspond to a useful variety of colours.
There is a good summary of the colours (and other types of discharge lamps) in this Wiki article.

You sort of missed the question. It is a combination of factors. Yes noble gases have excited states which fluoresce with colors. But also noble gases are relatively chemically  inert. So the ions convert back to uncharged atoms without chemically reacting with the electrodes or the container (glass tube).

@MaxW I'm not sure this is major factor at all. There are plenty of discharge lamps that use non-noble gases, metal vapours or salts but most are not as simple or convenient to operate.

I understood the cause of using noble gases, but I got confused because their justification was because they are inert gases. Really, I don't know what does it mean to have an inert gas. I just got to study that the family of column 18 is inert or noble gases and nothing else.

@user225430 - Noble gases for practical purposes don't chemically react with anything. "practical purposes" means barring exotic laboratory conditions. Thus the atom is the "molecule".  Thus my sentence "So the ions convert back to uncharged atoms without chemically reacting with the electrodes or the container (glass tube)." You could get a gas discharge through oxygen, but the oxygen atoms and ions would react with the metal electrodes to form metal oxides.


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I found instructions on how to make nerve gas. Is it possible to be real?
I just happened to find instructions how to make nerve gas. I found it on the Deep web. And since I know next to nothing about chemistry, I'm gonna ask here:
Is it possible to make nerve gas according to the instructions below? And if it is, how hard is it and how hard it is to acquire ingredients?
It just boggles me that something like this would be available, even on the Deep web. It's like complete instructions to make nuclear bomb, except plutonium is harder to come by, right?
Thanks for answering and here's the link to the image, because it's pretty big and comprehensive.

Well, if presence of such receipts disturbs you, than why are you asking if they are ok, using a link? Many explosives, drugs or gases aren't difficult to make. On the other hand access to firearms is so easy in many places that, in comparison, it can be considered minor problem.

Well, you certainly can produce lots of stuff that kills. But, can you do them on an industrial scale that would actually pose a threat? Hmm, not if you have better things to do with your money.

Once you have mastered all skills needed to follow that recipe, you are probably old and wise enough not to do that. Just by guessing it is at the edge of what a PhD in synthetic organic chemistry can survive, literally.

Mithoron: It does not disturbs me, it boggles me. I always thought that knowledge like this is kept in secrecy in some highly guarded labs. I am full aware that full country armed with guns is much more horrible. But this is somehow offtopic. inɒzɘmɒЯ.A.M: I asked if that recipe is real, not "create nerve gas in industrial quantities". It is somehow apparent even to me that this thing is out of reach for normal criminals/humans. ssavec: So it's real, no buzzwords put together to be intimidating - "hey guise i creatud nerve gus"...

I don't think many would confirm or deny if it's real, for lots of obvious reasons. Like the problem you have with the release of such a receipt in the first place...

I am genuinely scared.That SUCH RECIPES are available on net is INERADICABLE.If you are BOGGLED by this WHY SHARE AND SPREAD? why?

Some nerve gases are very simple chemicals so it is likely that the instructions are real.
This, however, isn't the point. They are very hard to make without killing yourself in the process and acquiring the ingredients is often very hard as governments restrict the supply and monitor uses very carefully. Even research scientists in approved facilities may not be allowed access to the key ingredients (I've had this for legitimate uses while doing real research).
Even if they ingredients are obtainable, the expertise to handle the products safely are not and almost nobody is likely to be able to apply the recipes safely. A similar situation exists for conventional and nuclear explosives. The recipes are available (e.g. for nitroglycerine or a simple uranium nuclear bomb). Both require both large, expensive facilities to avoid self-destruction or death by radiation. So the apparently simple recipes are useless unless you are major industrial enterprise or a government.
We're safe, despite the recipes.

Chlorine gas was used in WWI. Just mix bleach and ammonia which are both common household chemicals. Read about someone getting killed doing this every so often. http://articles.philly.com/1995-01-09/news/25712055_1_household-chemicals-bleach-ammonia


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No buffer region in the titration of strong acids or bases?
So recently I have conducted a acid-base titration experiment. I encountered that there is no buffer region in strong acid-strong base reaction. However, when take a weak acid or base buffer region appears.
WHY THIS IS LIKE THAT? What I mean is there any rational explanation for this?

Yes there is a rational explanation for this. Do you actually know what a buffer is?

Yes, it is how the solution resists the change in pH

But do you know how to make one?

Strong acids and strong bases essentially completely dissociate. So if you're adding the strong base to the strong acid, the solution stays very acidic, then neutralizes, then goes basic with just a small amount of the strong base.
With a weak-acid at the pK value you have as much $\ce{H+}$ as $\ce{A-}$. Thus a small amount of strong base or strong acid won't change the pH value much. That is the very notion of a "buffer solution."
So yes there is a rational chemistry explanation. :-)


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Mechanism for synthesis of 4-nitrophenylhydrazine from the diazonium salt of 4-nitroaniline
I am trying to figure out the mechanism to create 4-nitrophenylhydrazine from the diazonium salt of 4-nitroaniline. I've found a couple of explanations for what reactants are used, but I don't understand the mechanism behind them.
The mechanisms overall seem to suggest that $\ce{HCl}$ is used, and the $\ce{Cl-}$ bonds to the $\ce{N2+}$. That makes sense, but then it is suggested that $\ce{Na2SO3}$ and $\ce{SnCl2}$ are used to somehow change this into 4-nitrophenylhydrazine. If anyone could explain how this works, I would be very grateful!
Here is the article showing which reactants are used.

In the 1st step you also need $\ce{NaNO2}$ and is a standard textbook mechanism. The second step is a reduction. You should correct your reagent to $\ce{Na2SO3}$


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Is water a heterogeneous molecule?
My question is based upon the statement found here:

an element may be homogeneous on a larger scale, compared to being heterogeneous on a smaller scale. This is known as an effective medium approach, or effective medium approximations.

So, is it acceptable to say that water molecule is a heterogeneous molecule?

I have never heard of heterogeneous molecule or any similar terminology, but it seems you mean a heterogeneous mixture.

I think you are confused. Heterogenous and homogeneous are words that are used to describe mixtures (of substances). In homogenous mixtures, the substances exist only in one phase, while heterogeneous mixtures the substances exist in two or more phases.
A molecule is homonuclear if it consists of atoms of a single chemical element ($\ce{O2}$). A molecule is heteronuclear if it is composed of more than one element ($\ce{H2O}$). Note that any pure molecule is considered a substance, and it is the combinations of these substances that form mixtures, either homogeneous, heterogeneous, or colloidal.

No. "Element" in the article cited doesn't mean a chemical element or molecule but the it is considering the "volume of stuff" under consideration. The "element" could be a solution, the ocean, or a galaxy sized masses. Also heterogeneity/homogeneity are typically applied on a "macroscale" sample meaning large enough that the sample isn't quantum sized.
Thus for water (or some solution) the heterogeneity is really a concept of the volume of solution under consideration not the water molecule itself. So a liter of salt water would be homogeneous, but clam chowder would be heterogeneous.
If we were considering homogenized milk then yes/no gets a bit tricky. We'd have to consider how we define homogeneous since 1 ml sized samples would be the same but the homogenized milk really has two phases mixed together. (I'd guess that you could centrifuge homogenized milk and get it to separate.)
As another example the oceans are heterogeneous. The surface water will have a different chemical composition (pH, salt concentration) as compared to the water in the deep parts.


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Polyethylene biodegradation, how?
Worm guts have bacteria that apparently can degrade polyethylene:
The only agents that I have heard of that can degrade a fully-saturated hydrocarbon at room temperature are fluorine (and related nasties), other halogens (with UV light), magic acid (which is extremely reactive with almost anything). None of these seem biochemically feasible.
How is the polyethylene being attacked? Are the ends of the molecules more reactive double bonds that can be be broken off one carbon at a time like beta oxidation of fats?

How is the polyethylene being attacked?

Polyethylene is attacked by microbes that contain oxygenase or oxidase enzymes.  These enzymes can attack C-H bonds even at moderate temperatures.

The only agents that I have heard of that can degrade a fully-saturated hydrocarbon at room temperature are fluorine (and related nasties), other halogens (with UV light), magic acid (which is extremely reactive with almost anything).

One prominent example of a hydrocarbon degrading enzyme is methane monooxygenase, which is capable of catalyzing methane oxidation to methanol even at temperatures as low as 0 °C.  The C-H bonds of polyethylene are about as reactive as methane's.  (Actually, they're slightly more reactive.)  Thus enzymes can do the job.  They don't need elemental fluorine or magic acid.
However, your point about halogens reacting with hydrocarbons is very interesting.  Several research groups have reported that adding halide salts to aqueous (and aerobic) suspensions of bacteria accelerates the rate of oxidation.
It's possible that this finding can be explained by the involvement of haloperoxidase enzymes in the degradation.  However, the mechanism of microbial polyethylene degradation is a very new, and very active, area of ongoing research.

Very good answer.

from Wikipedia

Polythene or polyethylene film will naturally fragment and biodegrade, but it can take many decades to do this. There are two methods to resolve this problem. One is to modify the carbon chain of polyethylene with an additive to improve its degradability and then its biodegradability; the other is to make a film with similar properties to polyethylene from a biodegradable substance such as starch. The latter are however much more expensive.

So the cheap way is with a filler so that the film/part disintegrates not necessarily the polyethylene molecules of the polymer itself.


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How is Molecular mass is the molecular weight?
Wikipedia says molecular weight is the mass!! I can't get it the weight is the mass times 10 so we cannot neglect that, because 10 amu is not equal to 1 amu

I don't understand your statement that "the weight is the mass times 10". If my answer doesn't help, please explain.

Weight is not simply "10 times mass". The weight in Newtons is 10 times the mass in kg.

Or more precisely, the weight in Newtons is the mass in kg times the acceleration due to gravity in $\mathrm{m~s^{-2}}$ which is approximately 10 at the earth's surface.

ah  yes, 9.81 m s^-2 rounds to 10. It was 9.82  in my 1970 physics book... ;-)

http://physics.stackexchange.com/questions/43195/what-is-the-difference-between-weight-and-mass

Please do not change the premise of your question once it has recieved an answer with non-negligible score that would be invalidated. You can ask a new question any time.

I'd absolutely consider molecular mass as the preferred term. Wikipedia even has the article with that title and the sloppily used term molecular weight redirects to molecular mass.
The point being that mass of some sample would be the same on earth as the Moon or Mars, but the weight would change. Since we're stuck on earth we really weigh samples in Earth's gravitational field of $9.806\mathrm{~m~s^{-2}}$. But all laboratory balances use mass units. The weight in Newtons is 9.806 (rounds to 10) times the mass in kg. In science we don't like standard units that depend on arbitrary standards, and the gravity of the earth is arbitrary value on a galactic scale. Gravity isn't even constant over the surface of the earth. Read about how the meter has been standardized. Fascinating to me.

He doesn't get the difference between weight and mass. He thinks that an object with a mass of 1 kg has a weight of 10 kg.

I very much refuse to accept that the term molecular weight is the same as molecular mass. This is not a preference, the first expression is just plain wrong. Mass and weight are two completely different things. In sloppy spoken words we might understand what is meant; it might be used very often, even by teachers and professors and people who should know better; but it is incorrect nevertheless. If you refer to a mass of any kind, don't call it weight. Sorry, but I have to give it a down vote.

@Martin - Ok, you're right this is supposed to answers should be educating so I reworded the first part.

@orthocresol An object with a mass of 1 kg has a weight of 1 kg-wt but in my question I converted it to Newton that doesn't alter the point of the question! Why is atomic weight is referred to as atomic mass!? despite they both have different dimensions one has a mass dimension the other has a force dimension

The correct and internationally standardized name for this quantity is “relative molecular mass”. For historical reasons, IUPAC accepts the use of the special name “molecular weight” for the quantity relative molecular mass. However, the use of this traditional name is deprecated. Also note that the quantity relative molecular mass is dimensionless and thus it is not to be confused with the quantity molar mass $M$ (with the usually used unit $\mathrm{g/mol}$).

@MaxW The current Wikipedia article on “molecular mass” contains several severe mistakes. In particular, the article confuses various quantities and units.

@Loong - Thanks! I looked at the discussion on the wikipedia article and it shows quite a bit of bantering about all of this. IUPAC is certainly the dog and wikipedia the tail on this.

@AhmedElsawy well, if you were clear about the difference between weight and mass, you would not have written that "weight is the mass times 10", and you would not have referred to a "weight" of 10 amu, because amu is actually a unit of mass.

@Ahmed-Elsawy - the confusion in the wikipedia article is about one atom or a mole of atoms is being considered. The Wikipedia article is inconsistent with IUPAC who is the definitive source on terminology.

@orthocresol then tell me why atomic weight is measured in amu?

@AhmedElsawy If you took the trouble to read everyone else's comments, you would see that "atomic weight" is just an old, imprecise synonym for "atomic mass".

@orthocresol this answer is not satisfying

Wikipedia is inconsistent. They sometimes use the term "molecular mass" to be what IUPAC calls the “relative molecular mass” (i.e. grams/mole). and sometimes Wikipedia is using "molecular mass" to mean the weight of one molecule which is measured in amu.

@Ahmed your comment is not satisfying. I would ask another question only after I've read what I should've.


text:
Why is styrene an aromatic compound?
Why is styrene aromatic? Doesn't it have 8 pi electrons, thereby violating Hückel's rule? Can someone offer an alternate, more through explanation?

A compound is aromatic if it has 6 carbon atoms in a ring with 3 "double bonds", e.g. a benzene type ring. The rings can be joined. It doesn't matter what else is attached to the rings. Typically the hexagonal ring is drawn with one line between the hexagon vertices and a circle inside the hexagon to indicate the aromatic character of the bonding.

@MaxW That is a very limited definition of aromaticity which isn't really useful. There are many aromatic compounds which do not have six membered rings (cyclopentadienide anion for example) and have rings containing elements other than carbon (pyridine, furan etc.).  Also most people still draw benzene with alternating double and single bonds because it is more useful for drawing mechanisms.

As Wikpedia Article says ,the criteria for a molecule to be aromatic are --

the molecule must follow Hückel's rule, having 4n+2 electrons in the delocalized, conjugated p-orbital cloud;
the molecule must be able to be planar;
the molecule must be cyclic; and,
every atom in the ring must be able to participate in delocalizing the electrons by having a p-orbital or an unshared pair of electrons.

When we are talking about aromaticity , we consider the electrons inside the ring. You are right that the styrene has 8 pi electrons but when checking for Huckel's rule, we only consider the electrons inside the ring. The alkenyl group is just an attachment to the stable aromatic compound which might take part in resonance but wouldn't affect the compound's aromaticity.

It's not quite that simple. Consider benzoquinone; isn't it a planar cyclic molecule with 6 electrons in a delocalized, conjugated p-orbital cloud and some side attachments (oxygens)? True, it is. But is it aromatic?

@Ivan Neretin Benzoquinone is a case where double bonded oxygen atom is directly attached to the molecule which can undergo resonance to produce a resonance stabilised compound. And therefore it exists in aromatic state. A compound which has an option to enter either aromatic or antiaromatic state, will always go for aromatic state because it is more stable.  And that is why we call it to be aromatic.

That's where we disagree. I would never call quinone aromatic.

@IvanNeretin You confused me.  Why is Quinone aromatic or anti aromatic?

Quinones are a class of organic compounds where two single bonds outside the hexagonal carbon ring have been replaced by a double bonded O atom. Since  Quinones only have 2 carbon double bonds in the planar hexagonal ring they are not aromatic, but they do follow Huckel's 4n+2 rule.

As to me, quinone is not aromatic because the C-C bond lengths in its cycle are distinctly different, and it does not manifest chemical reactions typical of all aromatic compounds. The number of electrons is of secondary importance here. As for the Hückel's rule, it is not really applicable to the compounds with conjugated side chains.

@MaxW But why would it exist in such an unstable form when it can become aromatic?

Because it can't. These oxygens insist on having double bonds, and they aren't going anywhere.

@Ivan beware of saying "all aromatic compounds" undergo a typical set of reactions.  Benzenoid aromatic compounds do, but some other systems such as cyclopropenyl cations (which are electron deficient and so more amenable to nucleophilic rather than electrophilic substitution) and [18]annulene (which undergoes polyene type reactions) are different.


text:
What does Ae mean when used in a chemical formula?
In this video on slide 38 for example, the 35 minute mark, the formula appears
HAeBe

With the explaination "some chemestry has taken place..." and an infrared spectra.
What is Ae? I tried looking it up thinking it might be a common shorthand, but the only hits are answers that say "no, did you mean Al, Ag, etc.".

I'm voting to close this question as off-topic because HAeBe is not a chemical substance, rather it is a star type.

Gee, downvoting a months-old question that’s already closed? Pointless and not nice.

HAeBe stands for Herbig Ae/Be star, which is a young star before the main sequence of its formation.
The spectra of HAeBe stars show hydrogen and calcium emission lines, which are shown in the video.
The cloud could be referring to the gas-dust envelope that surrounds the star, but I'm not an expert in this field and so I'm unsure.


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NMR spectrum of cyclohexanol - hints on integration
I've been given this NMR along with the IR and Mass Spec and assigned the task of figuring out the unknown compound. I know that it is cyclohexanol and by analysing the NMR I know that the peak at around 2.8 is the OH and the peak around 3.6 is the deshielded CH. However I can't pair up the other protons and the peaks between 1-2 ppm. I know that the difference between the peaks/protons is their vicinity to the OH group.
My questions is:
Is there any way to do the integral of the points lying between 1-2 ppm. This will be useful as I think that the peak at around 1.3 ppm represents more protons than the others. Also could the peaks be different due axial and equatorial protons resonating at different frequencies, or would the temperature
Or is there a better way to pair up the peaks between 1-2 ppm and the remaining protons in the cyclic.

Yes you can measure integrals; resort to the NMR visualising software of your choice. (No, the problem is non-trivial so it cannot be precisely done by hand unless you want to cut out slabs of paper and weigh them.) Also it is most likely that axial and equatorial protons give different signals.

(Unfortunately, I cannot link you to the exact page, but note that SDBS has an entry for cyclohexanol which contains the relevant assignments. Check the $400~\mathrm{MHz}$ spectrum presented there.)

To answer your first question about measuring the integrals for the peaks between 1 and 2 ppm: If you seriously need to do a very accurate integral of these peaks without having access to NMR processing software (ie you've just been given the spectrum as a handout), you can resort to one of the methods of choice from the good 'ol days.

Print your spectrum out on a nice big piece of paper, get a nice sharp pair of scissors and cut out your peaks and weight them (a 4 or 5 figure  balance may be required). These method gives incredibly accurate integrals
Print out your spectrum on graph/grid paper and count the grids. Again, very accurate, depending on the scale of grid you choose to print on.

As for 'pairing up' the remainder of the ring protons, this is not a trivial exercise, made even less so with the dispersion of the spectrum you are working with. You are quite right in your assessment. The protons closer to the -OH group will be at higher chemical shifts; equatorial and axial protons will be non-equivalent (equatorial at higher chemical shifts); and the whole fluxional ring dynamic will be temperature, concentration, solvent and field dependent. So, probably it is not worth trying to pair up any of these protons. You could (and hopefully would, if you are a thorough researcher) assign all of your peaks with the assistance of 2D COSY, HSQC/HMBC methods.

Geeze you all are spoiled by high tech. Going really old school. Look at the peak heights. Crude but good enough for 1-2 ppm region. Key is that bonds are similar enough that peaks will be the same width. So height is proportional to area. From left to right I see 2, 2, 1, 4, 1. You know it has to add to 10.

I must’ve stopped looking at peak heights sometime when my NMR software gave me a different integral by factor two, but I’ll agree it can be helpful @MaxW.

Considering that the $400~\mathrm{MHz}$ spectrum on SDBS looks exactly like the image in the question, I’ll assume some TAs were lazy and copied. Assuming SDBS is correct, we get distinct peaks for axial and equatorial protons. Otherwise I agree; $+1$.

I don't understand the "otherwise". I assure you it is 2,2,1,4,1. You have to understand that the molecule isn't planar, nor is it static relative to the time that the NMR scan takes. // The Diophantine solution with peak heights only works for small numbers of atoms with fairly well separated lines.

I just did the integrals, it seems to work by just printing it on graph paper and measuring the heights. Thanks for the help.

Calculating the spectrum (APFD cc-pVDZ NMR GIAO) with -OH in equatorial position gives the following signals:

H-1 axial 3.3
H-2 axial 1.1, equatorial 1.8
H-3 axial 1.1, equatorial 1.6
H-4 axial 1.0, equatorial 1.4

That would fit the spectrum decently but it also would mean that flipping through the conformations is surprisingly slow.

Welcome. It might help to format the calculated shifts into a list to make it easier to read. Did you perform the calculations yourself?

On its own, I don't think these shifts provide any evidence for slow ring flipping. At RT ring flips should be fast, and you haven't evaluated this possibility to see if it fits better...? I think to be thorough, one would need to calculate shifts for the other conformer, then simulate the resulting spectrum for a variety of exchange rate constants (not done with a typical QC programme, you need specific software which deal with NMR / spin dynamics Hamiltonians), and see which rate constant (or range thereof) best matches the experimental spectrum.


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Allylic bromination with NBS
I know that NBS is a selective brominating agent especially on the allylic position. And its reaction mechanism is of free radical.
But I can't figure out the reason for the change in position of the double bond in the products.
It would be helpful if someone could give the mechanism along with its reason.

This is what I'm thinking, that might be happening there but still I don't understand how and why something like this would happen.

My mistake, number 3 and the product which I have marked as not formed are same.
2)Which one among these three products (disregarding stereochemistry) is the major product?

You're going through an allylic radical intermediate, right? How is the allylic radical stabilised?

The stabilization process for the free radical is the same as the carbocation, right.Because both are electron deficient. But I really can't understand the displacement of double bond in this case.

Can you draw a diagram to indicate the stabilisation?

I have added it to my answer,and I don't understand why that is not formed when acting upon allylic site (B)

The triradical structures look horrifying, rather you would draw appropriate fish-hooks to move the double bond in one step. 2) Yes it is formed. Look closely. Look harder.

Please give some more hint or explanation, I'm trying but I don't seem to get anything

Actually the answer contains only those 3 as shown in the first image. Also, I want to know ,if this mechanism is correct, which one would be the major product? Andwhy would a double bond split like that, because even before the free radical was on secondary carbon, and its the same even after bond displacement, what I really mean is how would that double bond displacement help increase the free radical stability?

What Jan is saying is, that one that you said was "not formed" is in fact one of the three products. Look closely. As for how the radical is stabilised - it's exactly the same way that a carbocation is stabilised. You can think of both the radical and the carbocation as being electron-deficient species.

Yes i see itin now ,thank you. But what about my other question, I.e., if my mechanism is correct then, I want to know what is the reason for shifting of the bond like I have shown, I don't know how it gets more stable. Please tell me a bit more in detail, I might be over looking few points ,which I don't seem to realise at this moment.

What are is the reason for splitting of the bond and forming bond like that? And which one would be the major product ?

SciFinder doesn’t find any reaction of 4,4-dimethylcyclohexene with NBS, for your reference(s). So the entire exercise is academic.

For the product mixture in allylic bromination, see here: https://youtu.be/BV_9wkYTeyA?si=kuTU6e4SGp09QhYv&t=290

For a textbook chapter that uses this example, see worked example 10.1.

[OP] I can't figure out the reason for the change in position of the double bond in the products.

This problem is a worked example in McMurry.
The first step is abstraction of a hydrogen atom (radical), leading to an allylic radical. These show resonance:

As a consequence, a bromine atom (radical) can form a bond in two positions for the top, and in two position for the bottom resonance structure (for the bottom, there is symmetry, so you end up with the same molecule). All of the products are chiral, so products come in pairs of enantiomers, in a 50:50 ratio because everything else is achiral.

[OP] Which one among these three products (disregarding stereochemistry) is the major product?

For other product mixtures, where the degree of substitution at the double bond differs between possible products, the one with the highest degree of substitution is favored (see e.g. this video). Here, there is no such difference. There might be a slight preference for abstracting the hydrogen further from the dimethyl group. There might also be a preference for not having the bromine end up right next to the dimethyl group (as in 3-bromo-4,4-dimethylcyclohexane).
In the absence of any of these preferences, the product formed from the symmetric allylic radical (3-bromo-5,5-dimethylcyclohexane) would be favored 2:1:1, just because of the statistics.


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Types of argon gas regulator?
I have a high pressure bottle of argon gas (99.98 %) that I need to use in an experiment. The problem is that the regulator I have is a generic type that is seemingly optimized for high flow rates (welding). I need a fairly low flow rate of under 200 cm³/min. However, I do not know what kind of regulator would be suitable. Any suggestions for search description?

I'm voting to close this question as off-topic because it's not about chemistry.

@Jan Disagree. I used 200sccm flows of argon in some of the experiments in my chemical engineering Ph.D. work.  Knowledge/use of the proper gas handling equipment was important. Thought experiment: Replace $\ce{Ar}$ with a non-inert gas like $\ce{SO2}$ or $\ce{Cl2}$ -- still off-topic?

This is used as part of chem eng PhD work by a student at Imperial College London

@Brian imho yes.

So discussing equipment used in chem labs is off topic?

@Jan Interesting... under what discipline would you file it? No others immediately come to mind.

@Brian some engineering something. The question is about the pressure regulator if I read it correctly. Much like I would not consider it on-topic here to ask for how to repair vacuum pumps. It is simply lacking the direct connection to chemistry that it would require imho. Three others agreed with me.

Would how to get a gas tight seal on an Erlenmeyer flask be similarly off topic?

@Jan No, I see your point: the scope addressed by the question is much broader than just chemistry.  However, in the definition of [tag:equipment], it says right up front "This tag should be applied to questions that are about the equipment used in labs, and techniques of keeping them, their uses, or their maintenance."  That gas regulators &c. are used many other places does not negate the fact that they are also used in labs. But, I can see it going either way--in the end, I will be content with the decision of the community.

For the sake of full disclosure: I do not think it is off topic here and hence I hesitated to act on the raised flag. It should be noted, that there was a request in chat. Please also consider this reply. I stand by my word and I am now considering migration. Please make yourself heard with reasons why this should not be done. One option would be to open a discussion in [meta] or simply reply here. (cc @Brian) [OP: If you want migration please flag.]

also cc @Jan (may the 15 char threshold be with me)

@Martin-マーチン Migration to Engineering.SE is an acceptable resolution to me. Nick Alexeev's point about the possibility for additional perspective is a good one.

No cylinder regulator that I know of will provide fine control of flow rates on the order you require.  The purpose of a regulator is to control the line pressure, not the line flow rate.  In theory you could stick a flowmeter in-line and try to finesse the regulator pressure and shutoff valve (if there even is one) to reach the flow you want, but it would probably be a maddening exercise.
What you need is a mass flow controller (MFC) of some kind.  Depending on your budget and your needs for adjustability and precision of the flow rate, there is a variety of options:

If you know you will only require a specific, set flow rate (or, a small number of set flow rates) you might do okay with a flow orifice: set the upstream pressure appropriately and the flow should match the spec.
If you need to be able to vary the flow rate and an approximate value is acceptable, something like these dial-knob flow control valves would probably work okay.
If you need precise control of the flow rate, you're probably up against something like these Porter mass flow controllers. They're quite pricey (a few thousand USD) but they work extremely well.  You'd need both the in-line MFC unit and the separate control box to make it work.

IMPORTANT: One thing to be aware of is that all of these technologies are gas-specific to a greater or lesser extent.  The orifices and dial-knob controllers depend at least on the viscosity and density of the gas, and the Porter MFCs depend also on the specific heat of the gas.  The differences will be minor, so I wouldn't worry about it for the first two, but if you need highly precise flow rates, you will have to compensate for the differences if you should need to switch gases.  Porter actually calibrates each MFC to a specific gas, so if you only need argon flows, you'll be all set.

No problem. I found someone who knew about this stuff and they got some bits and pieces from around the lab and fixed it for me. Part of the original problem turned out to be a faulty regulator. When that was replaced and a flow controller put inline it did the job.

@DirkBruere Heh, yeah - if the membrane in your regulator's blown, say, that'll definitely mean a lot more flow than you want. :-)


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Rate of SN2 reaction depending on following different conditions in case of primary halides
Most important points to remember when thinking of $\mathrm{S_N2}$ reactions are that they are nucleophilic reactions, a pentavalent transition state is formed, depends on steric hindrance, strength of the bonds, nucleophilic and substrate used i.e. what kind of leaving group is attached to it, etc. I know and understand these.
I also know that in the following conditions the rate of $\mathrm{S_N2}$ increases, but I don't seem to understand the reason behind it.
1) The presence of unsaturation on beta-$\ce{C}$ in primary halide increases $\mathrm{S_N2}$ reaction rate.
I really don't seem to understand this one, I know for the alpha $\ce{C}$ it would be difficult because the halide and the carbon will acquire partial double bond character which would make it very difficult to break.
2) The presence of carbonyl group at alpha carbon is most suitable for $\mathrm{S_N2}$ reaction.
My reason: it might be because of tautomerism, i.e., due  to its conversion to enol form it will increase the reactivity.
3) Presence of hetero $\ce{O,N,S}$ atom at alpha carbon in primary halides increase $\mathrm{S_N2}$ reaction.
This too, I fail the understand.

Please don’t use MathJax in titles due to searching issues. 2) Please use \ce{...} always and exclusively for chemical notations i.e. atoms and compounds. Note that $\mathrm{S_N2}$ reactions are better typeset using $\mathrm{S_N2}$ because the 2 should not be an index. 3) Please note that all punctuation marks except for opening brackets, opening quotation marks or mid-word apostrophes are followed by a space. (Groups of quotation marks — like at the end of this phrase — have a space at the end of the group.) That makes your post easier to read.

I know this question was from awhile ago, but I'll answer anyway.
a) The key to understanding this is the concept of the transition state. Below is an image taken from Khan Academy

For an $\mathrm{S_N2}$ reaction to take place, there must be enough activation energy to reach the transition state. In this type of reaction, the transition state comprises an alpha carbon in a pseudo-$\mathrm{sp^2}$ state. The nucleophile and the leaving group are directly above and below the alpha carbon in this state, with atoms bound to the carbon in a horizontal plane.
The reason this reaction proceeds more quickly with an unsaturated beta-carbon is the pi-electrons from the unsaturated carbon stabilize the transition state. This lowers the activation energy of the reaction, making it more likely to happen with each molecular collision.
c) I'm assuming you're comparing $\ce{N, O}$ and $\ce{S to C}$. $\ce{N, O,}$ and $\ce{S}$ are more electronegative than carbon, which will impart a more positive-like characteristic on the alpha carbon, making it more susceptible to attack from the nucleophile.


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Which reaction to use for standard formation enthalpy?

What reaction illustrates $\Delta H^\circ_\mathrm f$ for $\ce{CaCO3}$?

Answer: $\ce{Ca(s) + C(s) + 3/2O2(g) -> CaCO3(s)}$.
Ok so why isn't it: $\ce{2Ca(s) + 2C(s) + 3O2(g) -> 2CaCO3}$
Or
$\ce{Ca(s) + C(s) + O3(g) -> CaCO3}$ ?
For example in another similar question:

Choose the reaction that illustrates $\Delta H^\circ_\mathrm f$ for $\ce{Ca(NO3)2}$.

Answer: $\ce{Ca(s) + N2(g) + 3 O2(g) -> Ca(NO3)2(s)}$
Can someone confirm this assumption of mine:  it's ok to use fractions in these types of equations to blance it out as long as it's in oxygen's coefficient.
As that assumption correct?

The heat of formation refers to the formation of one mole of the compound from its elements in their natural states at $1\mathrm{~atm}$ and $298\mathrm{~K}$.  Your first example forms 2 moles and your second example uses ozone, which is not the "natural state" of oxygen at $298\mathrm{~K}$ and $1\mathrm{~atm}$.
Regarding your question about oxygen, it has nothing to do specifically with oxygen. The important thing is to have 1 mole of the compound on the right hand side of the equation.

Since 1982, $p=1\ \mathrm{bar}=100,000\ \mathrm{Pa}$ (not $1\ \mathrm{atm}$) is used as the standard pressure for tabulating thermodynamic data. Furthermore, the definition of standard states make no reference to fixed temperature. It is possible to have a standard state at various temperatures. Hence, the assumed temperature should be indicated, e.g. $\Delta_\mathrm fH^\circ(298.15\ \mathrm K)$. It is true though that $T=298.15\ \mathrm K$ is the most favoured and the one recommended for use.

298.15 is typically the temperature that applies to the heats of formation you find tabulated  in books.  According to Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics, "The usual choice for this temperature is 298.15  K, or 25 C."


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Calculating heat of combustion via calorimetry
Question:

When $\pu{60 g}$ of $\ce{C60}$ is combusted in a bomb calorimeter that has a water jacket containing $\pu{300.0 g}$ of water, the temperature of the water increases by $\pu{10 ^\circ C}$. Assuming that the specific heat of water is $\pu{4.18 J g^{-1} K^{-1}}$, estimate $\Delta E$ of combustion per mole of $\ce{C60}$.

Answer: $\pu{-150.5 kJ/mol}$
What I have tried:
$$
\begin{align}
Q &= m \times c \times \Delta T \\
Q &= 300 \times 4.18 \times 10
\end{align}$$
That's the only equation I know and it doesn't seem to work for this problem.

chemistry help question and more chemistry help? are not titles that would be in any way likely to help future visitors of the site. You have a rather specific problem in both your questions, you should have titled them appropriately. 2) I don’t believe that that be the only equation you know.

@Jan Ok, I changed the title and obliviously I know more equations but in this regard I don't. I don't know any other equation that could be applied to solve this problem, so for the sake of other future visitors to this site can you at least point me in the right direction because i'm completely lost

Suggestion 1: Add units. (That should be a given for all equations.) Then you should realise what you have just calculated, what that is equivalent to physically and maybe then you can see which step is remaining.

@Jan That calculates the energy needed to raise the temperate, I know the change in energy is would be E final - E initial so if I did that formula for both masses it would give me 12,540 for water and 2508 for Carbon, but subtracting them wouldn't give me 150.4 kj/mol which is the answer?

How did you just arrive at $12540$, what is its unit and how did carbon get in there?

What I assumed is Q must be the energies so I applied that equation for both water which would be (300 grams of water) * (4.18) * (10) which would give me 12540. I did the same for Carbon, I assume it's carbon because it has a "C" for it's symbol, and when I did the same for carbon it gave me 2508. So when I got the two energies I subtracted them but it wasn't even close to 150.5 jk/mol.

One last time: Please use units, it gets you so much further.

I see what you mean but now I'm even more confused now, you were initially confused on why I had factored in Carbon into this, so am I not suppose to use carbon? Because I wouldn't know how else to find moles of carbon. but even assuming that was true 12540 joules/ 5 moles of carbon wouldn't be 150.5 kj/mol?

Ah, we’re almost there. Why are you using $5~\mathrm{mol}$ of carbon, because in the question there clearly are not $5~\mathrm{mol}$ of carbon. Check out which type of carbon you are using.

You have 60 g of $\ce{C60}$ (which has a molecular weight of 720.6), so that means you'll need to multiply your answer by 720.6/60 which (cleverly) turns out to be 12.01. Another way of saying it is that you have calculated the heat of combustion for 1/12.01 moles of $\ce{C60}$, so you'll need to multiply your answer by the inverse in order to get the desired (per mole) quantity.
Using your notation, numbers, and noting that a change in one degree Celsius is equivalent to a change in one kelvin:
$$\Delta E_{\rm molar} = -12.01\;\mathrm{mol} \cdot 300\;{\mathrm g} \cdot 4.18\;{\mathrm J} \cdot {\mathrm g}^{-1}\, \cdot {\mathrm K}^{-1}\, \cdot 10\;{\mathrm K} = -150.6054\;\mathrm{kJ}\cdot\mathrm{mol}^{-1}$$

thank you for your answer, I've been trying to figure this out for hours


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What symmetry elements mustn't a molecule possess to be considered achiral?
Undergraduate textbooks say that a molecule mustn't posses a reflection plane and an inversion center to be considered achiral. As mentioned by Ron in his answer to this question, molecules that don't possess a plane nor center of symmetry, but do possess an $S_4$ improper axis are considered to be achiral. Are there any other symmetry elements that need to be taken into consideration when determining whether a molecule is chiral or not?

Short answer: Anything that can be expressed in terms of $\mathrm{S}_n$ axes makes a molecule achiral (Schoenflies definition). Or alternatively, anything that can be expressed as $\bar n$ makes a molecule achiral (Hermann-Mauguin definition).

Actually, any reflection in a plane is just an $S_1$ operation, and an inversion through a point is just an $S_2$ operation. So, @Jan's comment is indeed the most concise yet complete definition: any $S_n$ axis present will make the molecule achiral.

Okay. I would like to see some exotic example :).

@Marko that's hard to find. Your best bet is looking for a molecule in the $S_4$ point group, which has the operations $E$, $S_4$, $C_2$, and $S_4^3$ - i.e. no mirror plane and no centre of inversion, but is still achiral because of the improper rotation axis. You can find some examples and pictures here http://csi.chemie.tu-darmstadt.de/ak/immel/tutorials/symmetry/index7.html

Nice examples, especially the coronane.

Someone may post an answer (with the examples).


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PTFE coating - how? Alternatively, lowering surface energy
Another part of our experimental setup requires me to lower the surface energy of a small sealed hollow glass sphere (30mm dia) in order to minimize adsorption.
Any idea how I can do this? There seem to be some PTFE sprays, but how well do they adhere to (fragile) glass? Any alternatives ideas?

More details on the experiment? It seems to me that you have the experiment wrong. I suspect you're trying to change the surface tension.

Measuring density changes in gas using the buoyancy of a glass sphere and a microbalance. Unless the glass is heated between measurements adsorption can skew the results if density change is low.


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Why are the melting points of Sb and Bi lower than that of As?
My book says it has something to do with the inert pair effect. And unlike As, Sb and Bi tend to form 3 covalent bonds rather than 5. I don't understand what's really going on here.

see this:http://chemistry.stackexchange.com/questions/33815/trend-in-the-melting-point-down-the-group