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Why do strong acids dissociate (almost) fully on a molecular level?
In the Brønsted-Lowry theory, strong acids $\ce{HA}$ dissociate using water and react into hydronium ions and the conjugate bases $\ce{A-}$. However I never understood in the strong acids case what stops the products of the reaction(ions) into reforming chemical bonds (why the reaction is one-way) and what enables them when the reagents aren't strong acids. How is this explained macrocosmically and microscopically?

Deprotonation is always a reversible reaction. Hence why, even strong acids, have $\mathrm{p}K_\mathrm{a}$ values (which rely on the presence of equilibria).


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Mass of sucrose solutions
What is the mass of 150 ml of 40% sucrose, and what is the mass of 150 ml of 20% sucrose? I don't know exactly how to figure it out without using a device to tell me the mass. All I know is the mass of 10 ml of 40% sucrose, which is about 15.7 grams.

If the water volume was additive you could do it by adding more water w/ known density...  But it's not, is it...

If 10ml of 40% sucrose is 15.7 grams, then the mass of 150 ml of 40% sucrose is
$$ 15 \times 15.7 grams$$
You can not determine the mass of 150 ml of 20% sucrose from the information given.
You could calculate from this sucrose solution density table:
http://homepages.gac.edu/~cellab/chpts/chpt3/table3-2.html
Note that the table shows the premise "I know is the mass of 10 ml of 40% sucrose, which is about 15.7 grams" to be incorrect.

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What is lowest energy state, how does an atom gain it and what does it have to do with hybridization?
I understand that hybridisation occurs due to the atom wanting to exist in the most stable state possible, and the most stable state has the lowest energy. What I don't understand, however, is say, in CH4, the bond angle is 109°18', so how does this particular value mean that the atoms and bonds in methane are farthest from each other? And if they are indeed farthest from each other, how does that mean the CH4 molecule's in the lowest energy state?

VSEPR stands for Valence Shell Electron Pair Repulsion.  It's a complicated acronym, but it means something that's not difficult to understand. Basically, the idea is that covalent bonds and lone pair electrons like to stay as far apart from each other as possible under all conditions (to minimize electrostatic repulsions of different pairs of electrons) .  This is because covalent bonds consist of electrons, and electrons don't like to hang around next to each other much because they have the same charge. This, of course, allows the molecule to be in its lowest energy state.
This VSEPR thing explains why molecules have their shapes.  If carbon has four atoms stuck to it (as in methane), these four atoms want to get as far away from each other as they can.  This isn't because the atoms necessarily hate each other, it's because the electrons in the bonds hate each other.  That's the idea behind VSEPR.
Now, one problem with the whole VSEPR thing is that if you have four things stuck to carbon, for example, there are no orbitals that want to get 109.5 degrees apart from each other (109.5 degrees corresponds to the geometric maximum distance the atoms can get apart).  After all, s-orbitals go in a complete sphere (360 degrees) and p-orbitals are 90 degrees apart.
What happens instead of using s- or p- orbitals is that when covalent bonds are formed, the s- and p- orbitals mix to form something called hybrid orbitals.  "Hybrid" just means "mixture of two different things", and that's exactly what a hybrid orbital is.  When three p-orbitals with 90 degree angles combine with one s-orbital with 360 degrees, they average to form four sp3 orbitals with 109.5 degree bond angles.  Depending on the numbers of s- and p-orbitals that mix, you can get a bunch of different bond angles.

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Properties of f-orbitals
I am not a Chemist, but I took enough undergraduate Chemistry classes to understand the basic properties of s, p and d orbitals and how the behaviour of electrons contributes to different kinds of chemical bonds and elemental/molecular properties.
I stopped before we got to actual quantum mechanics - my maths wouldn't have been good enough to stand it. It still isn't.
However, I was always fascinated as to why none of the lecturers would ever tell us about f-orbitals. When asked, they muttered dark hints that they were significantly different from lower electron shells, and that explaining why was not only a waste of time but might spoil our existing understanding.
I left university 20 years ago, but I was suddenly blindsided by remembering this question out of nowhere. And now I want to know the answer. Google searches reveal a lot of maths that I can't follow. So can someone explain:

What are some of the basic properties of elements with electrons in f-orbitals, and how do they differ from lower orbital positions?
I assume this is related to the fact that most (all?) elements with f-orbitals manifest as metals, so how and why? And why do some elements without f-orbitals display metallic properties?
Why were my tutors so unwilling to try and explain this to non-Chemists who were supposed to stop at their appreciation of lower orbitals and basic covalent bonds?

Many times we stop at d-orbitals with undergrads because 1.) there are way more orbitals to visualize and they are more complex looking, 2.) they apply to elements that begin to act very 'strange' with respect to main group elements on the first couple rows (even transition metals are overbearing at times, breaking trends and offering a million exceptions to the general rule), and  3.) because these types of elements are just not that common in chemistry.  Makes for great discussion but not very practical. In 9 years I haven't used Actinides or Lanthanides.  There is bigger fish to fry.

Take a look at this somewhat related previous question.

I think there are multiple reasons contributing.
First of all, the atomic orbitals of the f-block and their properties and interactions are complicated. You have a lot of electrons, and they are far away from the nucleus, on many differently oriented orbital lobes. (well, that is the visualisation at least, in reality atomic orbitals are described by quantum mechanical wavefunctions, and those get more complicated as well for the f orbitals)
Since you have a lot of electrons repelling each other, and on top of that there are relativistic effects coming into play, it is rather difficult to do quantum chemical calculations with these elements to determine their energy levels, etc.
A lot of "rules" (and approximations) valid for low atomic number elements stop working.
Overall, heavy elements have special properties, that cannot be explained by the simple rules and approximations used for light elements. As you have said there is a lot of difficult math in the background. The rules and visualisations are just there to be an  easy to grasp abstraction of the underlying physics. The side effect of this is that you need more and more exceptions as the system gets more complicated.
Second, they are usually rare and expensive elements.(actinides are also radioactive) They have no known biological role, and usually only have few significant uses. So usually one does not encounter them unless one is working in some niche field.


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Why did HF disolve this lightbulb at the surface rather than more submerged points?
I watched this video where a lightbulb is suspended in a solution of HF.  The result is that the HF dissolves the glass, but it seems like the part of glass on the surface reacted faster than fully submerged parts.  Here is an image of the lightbulb after the experiment:

The glass breaks at approximately the 2:50 mark, and it looks like at that point the bottom portion was cut from a clean line.  Perhaps the larger portion dissolved afterwards or was damaged during cleanup, I'm not sure.
Is there a reason why the glass closer to the surface would react faster than more submerged parts?  Why would the very tip not dissolve first?
My best guess is that the specific line happened to be the thinnest glass.  I find it unlikely that the specific line is the thinnest part.  Wouldn't it be much more likely for there to be random holes in the submerged portion from the thinnest part of the glass?

Apologies if this is tagged poorly.  Chemistry is not a very strong field for me.

Near surface of the liquid the glass is a bit hotter. All glass of the bulb gains comparable amount of heat energy from tungsten wire, but parts above the liquid are cooled only by air, so they are much hotter. Part of this heat is transferred into slightly submerged part, so it is the hottest part that is under liquid, and all reactions are quickened by increase of temperature.
However, this exact experiment is a bit ambiguous, as glass here is under extreme thermal stress, and shattering near surface of the liquid may be addressed to that.


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How to list elements according to their radius? For example Ag, Cd, Au Can you check the accuracy of these sentences I've created?
I've asked a similar question before, but now I want to find out the real rule. Pleas tell me if my interpretation is correct: Cd, Ag , Au (increasing ray order). This happens because although we know that period and group effect the radius length, most important one is the group. So firstly we should list the elements according to their group and then list them in the same group... For example Zn, Cd (same group), then Pd, Pt, (same group), then Ph, ir (same group). From our lesson we know that from Cd to Hg the ray doesn't change but I think when we have to list them we should consider that it grows a little. Are these statements right? If not, please help me clear the concept. I'm not a native speaker of english, so I may not express correctly in some occasions

This is the radius general trend on the periodic table:

The main reason that explains how it changes is the nuclear effective charge experimented by the electrons when atracted by the nucleos. Briefly speaking, the more electrons you add as you go down a group (in different energy levels), the lower the nuclear effective charge that the electrons will experiment, so the electron denisity will be less "compresed", resulting in a larger radius.
Going from the right to the left, you add one by one, a new electron that experiments almost the same nuclear effective charge, so it will be atracted with a similar force, and also, you are adding another proton to the nucleus, so the overall atraction will be higher, the electron density will decrease and the atomic radius will decrease.

The OP's understanding is not correct. The Periodic Table does a very poor job of predicting radii, nor was such a use ever intended for it. There is no obvious reason that the radius, r, of the elements should show periodic or systematic patterns.
As it turns out, however, the Table does capture certain systematic trends, as the answer by Altered State indicates. Refer to the covalent (measured) radii from https://en.wikipedia.org/wiki/Atomic_radius and you will see that there are numerous exceptions to the general trends. For instance:

P, S and Cl all have roughly the same radii
Ru < Rh < Pd < Ag is opposite to what you'd predict from the general trend, even though they're in the same row and adjacent, and finally
Pt, Au, & Hg are all smaller than the elements directly above them (i.e. same groups) in the table.

We can conclude that covalent radii cannot be confidently predicted using the Periodic Table. It is beyond the scope of this answer to explain the details behind various measurements of radii, but three common ways radii are tabulated is using covalent, ionic, or van der Waals measurements, as well as calculated values (see Wikipedia link, above). The results depend on the way the values are measured. Because of the atomic nature of the elements, and because atoms are held together principally by electrostatic forces (ignoring how the nucleus is held together), a change in the electric field due to valence and chemical bonding (either ionic or covalent) has a dramatic effect on an atom's radius.
Since the valence of an element varies, it is also incorrect to believe that an element has a fixed (or nearly fixed) radius. Iron will have a substantially different radius in FeCl3 compared to that in Fe2O3. It may be of some interest to compare the radii of the elements in their 0 valence state as solids, but even then the crystal structures (hence the electronic interactions between atoms) vary. For instance in comparing the radius of Fe in bcc iron with N, do we cool the N2 to form the hexagonal crystal (down to -210 °C) and search for values for Iron at that temperature? Worse, do we compare the N-N molecular triple bond distance or the van der Waal distances? And then there's He, we need to cool He down to 1 K and also subject it to 2.5 MPa pressure to get a solid. I'm sure it would be quite difficult to obtain radii of the other elements under these conditions. ("Difficult" in the sense of finding reliable measurements in the literature.) In estimating the comparative size of two atoms, it is generally only useful to do so when they are in the same environment. This usually means with the same bonding, physical state, etc. In these situations, periodic trends are not very useful; detailed analysis of the quantum mechanics, especially the orbital (shell) occupancies, is required.
We all learned in physics that like electrical charges repel and opposites attract. How does this help us understand the trend in atom size with increasing atomic number? An increase in atomic number of 1 increases the number of protons and electrons by 1 (and adds variable number of neutrons). Radius is frequently assumed to be the distance from the (point-like) nucleus to the distance at which the valence electron(s) are 75%, 85%, or 90% contained. (This is arbitrary.) This distance depends on the repulsion from all of the other electrons that the valence electron(s) experience as will as the attraction between in and the positive nucleus. Since all of the electrons spend the majority of their time in positions other than directly between a valence electron and the nucleus, the shielding by those other electrons felt by the valence electrons is less than 100%, meaning it feels more attraction than repulsion (if this were not true, the electron would wander (or be pushed) away.)
What this means for the radius is that while yes the radius increases with increasing number of electrons (and protons), but the increase is not as great as you might expect since the outer-most electrons are attracted a little bit more than you'd expect. There are effects from half-full shells, as well as the differences in shielding of the various orbitals (s, p, d, f). This makes it quite difficult to confidently predict sizes once you get below row 2 in the Periodic Table.


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How can BN be paramagnetic?
I read something recently that said "BN has been observed in the gas phase, is paramagnetic, and has a vibrational frequency lower than N2. These properties can be explained by the molecular orbital diagram of BN".
However, when I drew such a diagram, I failed to see how it described any of these properties, particularly paramagnetism, because there were no lone pairs of electrons. Any ideas?

This is the general MO diagram you need to fill with the valence electrons of BN

Boron has 3 valence electrons, and nitrogen has 5 valence electrons, this makes 8 electrons.
You have to start filling the orbitals from those with lowest energy to those with higher energy.
So, 2 electrons on $\sigma_\mathrm{2s}$, two electrons on $\sigma_\mathrm{2s}^*$, two electrons on $\sigma_\mathrm{2p}$. You have now 2 electrons left, and two orbitals of the same energy.
In this case, you need to follow Hund's rule, which states that  if two or more orbitals of equal energy are available, electrons will occupy them singly before filling them in pairs. So you will have to put one electron on the $\pi_x$ orbital and another one on the $\pi_y$ orbital.
So you end up with 2 umpaired electrons, and paramagnetism of the molecule is explained.

Does interference lead to sigma/pi 2p  order switch when some atoms are n <= 7?


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What does the Clausius Inequality mean in thermodynamics? (entropy)

$$\delta S\ge\frac{\delta q}T$$

I don't understand how a quantity (the change in entropy in this case) can be greater than the very thing that defines it ($\mathrm{\frac{\delta q}{T}}$).
Also what is the significance of this inequality?

Your definition of entropy is incorrect. The significance of the Clausius inequality is that it shows that the definition of entropy, i.e. $\mathrm{\delta S=\cfrac{\delta q_{rev}}{T}}$ (note that entropy change is defined for a reversible process) is consistent with observed reality: the entropy of an isolated system does not decrease spontaneously.
We know that $$\mathrm{\delta w \geq \delta w_{rev} \Rightarrow \delta w-\delta w_{rev}\geq 0}$$
and according to the First Law of Thermodynamics
$$\mathrm{\delta U = \delta q + \delta w=\delta q_{rev}+\delta w_{rev}}$$
Therefore $$\mathrm{\delta q_{rev}-\delta q=\delta w-\delta w_{rev} \geq 0}$$
Now using the definition of entropy yields
$$\mathrm{\delta S=\cfrac{\delta q_{rev}}{T} \geq \cfrac{\delta q}{T}}$$
Taking $\mathrm{\delta q=0}$ (since we have an isolated system no heat can flow into it), produces
$$\mathrm{\cfrac{\delta q_{rev}} {T} \geq 0 \Rightarrow \delta S \geq 0}$$

this just derives it, not answers the OP's question

why is dw(rev) less than dw?

Also how does this tell you the direction of spontaneous change?

I don't understand how a quantity (the change in entropy in this case) can be greater than the very thing that defines it.

Actually it is not greater, but greater than or equal to:
$$\delta S\ge\frac{\delta q}T$$
If heat is provide irreversibly, the term on LHS is greater otherwise in case of reversible process it is equal.

Also what is the significance of this inequality?

This inequality when rearranged becomes:
$$TdS-dq\ge0$$
For constant volume or pressure it becomes(recall $q_v=U,q_p=H$):
$$TdS-U\ge0\\TdS-H\ge0$$
both of which say the common equations,(for a spontaneous change):
$$dA=U-TdS\le0\\dG=H-TdS\le0$$

The reason that there is so much confusion over this is that the usual form of the Clausius inequality appearing in most references is not completely and properly specified.  In an irreversible process, the temperature within the system is not going to be spatially uniform, so there is considerable ambiguity as to what value of the temperature T to use in calculating the integral of dq/T.  The correct value to use is the temperature at the interface between the system and surroundings (at which the heat transfer is occurring), $T_I$.  So the correct form of the Clausius inequality should more properly read:
$$\Delta S \ge \int {\frac{dq}{T_I}}$$
So, for example, if the surroundings consists of a constant temperature bath, the T in the Clausius inequality is the temperature of the bath (i.e., $T_I=T_{bath}$).
If the path between the initial and final equilibrium states of the system is reversible, then the system temperature is uniform throughout and equal to the interface temperature, such that $T_I=T$, and the equality in the Clausius relationship applies.  The integral then gives the change in entropy.  For any irreversible path between the same two thermodynamic equilibrium states, the value of the integral is less than the integral for a reversible path, $\Delta S$.

Are all these inequality defined for irreversible process... How to cslculate if temperature is also changing

@harambe Are you asking how to calculate the entropy change for an irreversible process, or are you asking how to evaluate the integral on the right hand side of the equation for an irreversible process?

The latter... How to evaluate for irreversible process where there is no thermodynamic equilibrium.. And the system is not in equilibrium with itself so it's temperature is changing

@harambe In that case, for most situations, you would have to solve the transient mass-, momentum-, and energy balance partial differential equations for the system, which include heat conduction and viscous momentum transport, to establish the variation in local heat flux and temperature at the boundary; and you would have to integrate this over the area of the interface and over time.


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Show which group the element with Z=31 belongs

and I understand it belongs to the 13 th group, but my teacher wants to express it as IIIA, so how should I convert it in my mind?

This is the old periodic table in which are still used these groups. You should avoid its use, but if your teacher needs you to learn them anyway, here they are:

See wikipedia for more information. The key publication for renumbering should be free access. Also see rule IR-3.5 in the IUPAC red book.


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UV-Vis Spectra: Multiple peaks for one component
I am working on a project in my lab, and we are analyzing capsaicin using HPLC. I am trying to find an appropriate wavelength for the HPLC using a UV-Vis spectrometer. When analyzing the pure capsaicin standard dissolved in methanol, I of course use methanol as a blank. However, I got two peaks, one around 230nm and one at 280nm. Since the system automatically subtracts the methanol, I can't understand why I am getting two peaks.
Anyway, 230 yielded the maximum absorbance, so I used that in my HPLC. I didn't get great results and decided to read some literature on the project. I saw that most other people had a UV-Vis spectrum like mine, but they decided to use 280nm. Why is that?

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site.

Different functional groups may show different absorptions and even a single functional group may exhibit two signals in your UV-VIS spectrum, e.g. corresponding to $\pi,\pi^*$ or $n,\pi^*$ transitions.
While HPLC grade methanol is typically much cheaper than acetonitrile, it might not be the best choice here. You might want to have a look at the cutoff wavelengths of different solvents.
I'd definitely give measurements at $\lambda = $ 280 nm a try.

Many organic compounds give more than one maximum peak when its UV-Vis spectra is analyzed. Each peak correspond to a electron transition from a ground state to an excited state, and more than one different transitions (with different energy, and therefore, different wavelenght) are allowed.
To determine the concentration of a compound, in principle, any fairly strong peak would be valid, but depending on the experimental conditions, sometimes one gives a stronger signal, and other times that one gives a worst signal.
The peak employed on your determinations can also be selected based on your instrument conditions, i.e., if one of the peaks is close to your measurement limits, you may prefer to use another, or if one of them has a better shape, etc.


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What is the difference between Dielectric and Capacitive Polarizable Continuum Model?
I have good background knowledge on Quantum Chemistry calculations but my knowledge on Solvation Models is not that amazing.
Many times I read a paper that uses TD-DFT, they also assign some solvent to their calculations. Some papers use Dielectric-PCM and some others use Capacitive-PCM. I know that D-PCM assumes the solvent as Dielectric media and C-PCM assumes as Capacitive Field and I have some very basic knowledge on them but what I can not understand is when we use which?
For example what is the criteria for selecting DPCM or CPCM for Methanol solvent?

Short answer: It matters most that you are using a modern solvation model than which one
Slightly longer answer: If you have the time, try a few solvation models, compare with experiment and see which gives you the best results. In my experience, modern models do fairly well for TDDFT calculations. Be careful about the atomic radii used and whether the computational program generates a reasonable solvation surface.
Also keep in mind that not every program supports both D-PCM and C-PCM. Many packages just implement one solvation model.
There are a number of excellent reviews on polarizable continuum solvation models. As you're probably aware, they are a really efficient way to handle molecular solvation in calculations, since the details of the solvent are largely ignored. Instead, the solvent is treated as a dielectric environment, and you handle polarization between the molecule and the solvent self-consistently. That is, the solute molecule is treated at normal atomistic detail, creates a solvation shell (usually treated by a set of overlapping spheres) and the electrostatics of the molecule influences the solvent, and vice-versa.
Now, I've never seen "C-PCM" as a "capacitive" PCM. Usually people refer to the "C" as a conductor-style model. That is, in a C-PCM model, the dielectric constant is set to $\infty$ but then you re-scale the charges on the molecule later.
There are more than just these techniques. The main problem is that it's very hard to bench-mark solvation energies.
Some great recent reviews:

"Thirty years of continuum solvation chemistry: a review, and prospects for the near future." Theoretical Chemistry Accounts (2004) v. 112(4) pp. 184-203
"Quantum Mechanical Continuum Solvation Models." Chem. Rev., 2005, 105(8), pp 2999–3094
"Perspective on Foundations of Solvation Modeling: The Electrostatic Contribution to the Free Energy of Solvation" J. Chem. Theory Comput., 2008, 4 (6), pp 877–887
"A Universal Approach to Solvation Modeling" Acc. Chem. Res., 2008, 41 (6), pp 760–768
"The COSMO and COSMO-RS solvation models" WIREs Comput Mol Sci, (2011) 1: 699–709
"Polarizable continuum models." WIREs Comput Mol Sci (2012) 2: 386–404.


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How are hot and cold defined?
My teacher asked me this question a few days ago. "How to define hot and cold?" And I said, "The higher the temperature, the hotter you are. The lower the temperature the colder you are." But my teacher said, "That's not 100 percent correct."
So, how do you really define hot and cold?

This question (and its answer) might be helpful as well: http://chemistry.stackexchange.com/questions/10227

Um, I guess you are asking the wrong people about the wrong question. The right question would be how your teacher defines that, and the right person to ask is your teacher, he/she gets payed for that. Nevertheless, you are right to ask for a second opinion here. ;-)

The kinetic energy of the substance's molecules also determines how hot or cold the substance is.

For life, boiling is clearly too hot and freezing is too cold.  Actually, you get protein denaturing much above $37^oC$ and metabolic processes slow down much below that.  Is it a chemistry teacher, a physics teacher or a biology teacher, or a culinary arts teacher?

The concept of temperature arises from the zeroth law of thermodynamics:
If two systems are in thermodynamic equilibrium with a third system, the two original systems are in thermal equilibrium with each other. Basically, if system A is in thermal equilibrium with system C and system B is also in thermal equilibrium with system C, system A and system B are in thermal equilibrium with each other.
From here, many hypothetical scales for temperatures are stablished.
It's said than one body has higher temperature than another when there is a heat exchange from the former to the latter.
But hot and cold are just sensations or feelings defined by people. Something is called hot when it's at higher temperature than our body (~37 °C), so a positive heat exchange will take place on contact. The opposite for cold (heat transference from our body to the cold object).
So, you are right, the higher the temperature of an object, the hotter it will be, but the concept hot is just to define the thermal perception.

As an interesting little addition: the fact that your body temperature is the reference for hot/cold also means that if the outside air is above 37$^o$C, blowing air over your skin with a fan actually makes you feel MORE warm instead of less (as people are accustomed to, because the outside temperature is almost always cold with respect to our body temperature)

@Michiel it's more complicated than that. If the outside temperature is above 37, you are typically all sweaty, and blowing air over your skin makes the sweat evaporate more quickly, resulting in a cool feeling. The effect you mentioned does occur, but only when the air temperature is much higher than 37 Celsius and the evaporation effect cannot compensate for the heating by the new air. It is well observable in dry saunas, for example.

@rumtscho Good point, I indeed ignored the evaporative cooling effects of sweat in my comment.

Hot and cold are subjective terms that can vary from person to person. This means that what I think is warm, you might think is cold, meaning there is no true value for these definitions.
If you were a reference point however, we could that boiling water is hotter than you, and ice is colder. But what about something more subtle like the difference between hot water for green tea (maybe 180 $^\circ$ F ) and coffee (about 198-201 $^\circ$F)? They are both hot to you, but to the coffee, green tea would be cold. Get what I'm saying?

I finally find out the real answer. It is actually the amount of heat absorb or release which determine how hot or cold you are.
I find this answer by a simple experiment. The experiment just simply putting one in cold water and the other one in hot water. Then, put both hands in warm water. You will find the the hand with the hot water will feel cold and the other will feel hot.
It is because
1) The temperature of the cold water is lower than the warm one. So, when we put the hand in the warm from the cold water, our hand will absorb heat, causing the hand to feel hot.
2) The temperature of the hot water is higher than the warm one. So, when we put the hand in the warm from the hot water, our hand will release heat; causing  the hand to feel cold
Thus, we can conclude that the more heat it absorbs, the more hotter it is. Whereas the more heat it releases, the colder it is.

You don't recognize the difference of "heat (energy) absorbed" and "temperature change". Think you got a piece of hot potato in your mouth. Case one: you rest it on the tip of your tongue. The heat exchange is minimal, but the temperature change will burn your tongue. Case two: you flush the hot bit with lots of saliva. The heat exchange will be much greater than in case one, but the temperature of the potato will decrease rapidly and the temperature change inflicted on your tongue will be much less.

Hot and cold are relative terms, nothing is absolute hot or cold. When you say something is hot then you are defining it as having a higher temperature with a reference body.


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Which electrons will be taken when oxidations take place and how does the resulting orbital diagram look like?
In detail, what I really mean is which electron in which orbital is being taken when a oxidation happen? Let me give a example: chromate ion(II)
Two electrons are taken but from which orbital:

The highest one

There is no trivial way to predict the resulting configuration with 100% accuracy. Here's a related past answer of mine with a useful link, and a related question.

Why dissenter??????

Which of the electrons in or orbital 3d will be taken?

For transition metals (specially for those on the second and third rows), like chromium, or iron, which electrons are taken away in oxidation is not straightforwardly deduced. The ones that will be lost are those which are higher in energy and lead to a more stable situation.
For iron, which is $\mathrm{[Ar]3d^6 4s^2}$, the higher energy electrons are $\mathrm{4s^2}$, and those that will be removed on oxidation, leading to a $\mathrm{d^6}$ configuration for iron (II).

Can you answered to the example? The case of iron is too easy. Of course electrons in s orbitals will be removed.

I took "not straightforwardly deduced" as having to look up...

For the first transition row, the ion configuration can be predicted. Sorry, I forgot to answer to your example. Cr is $\mathrm{[Ar]3d^54s^1}$, the first eletron to be taken away will be the higher in energy one, ie, the only s electron. Then you have to eliminate another one, and you only have d electrons left, so the $\ce{Cr^2+}$ configuration will be $\mathrm{[Ar]3d^4}$. And yes, many transition metals of the 2nd and 3rd rows, will require you to look them up.

One of the d orbital electrons will be taken out, in terms of orbital diagram, does it matter to take out the electron from the first or the last box?

For the isolated ion, which is the case, it does not matter, all the five d orbitals have the same energy.

What explains that there is no +1 oxidation for chromium? Losing just one makes it even more stable.


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crystalline structure of Si wafers used for MEMS
When they say that we have a {1 0 0} single crystal wafer, I know that the surface normal to the wafer is oriented in the <1 0 0> family direction, but if I draw a line on the wafer, parallel to my primary flat, does it indicate a crystallographic direction family too? Like <1 0 0> or <1 0 1>. I mean it's obvious that it cannot be <1 1 1> due to the FCC lattice of Si(I am making the assumption that the thin plane of the primary flat is at exact right angles to the wafer surface, if that angle is allowed to be 54 degrees, it can also be a <1 1 1> ), but it can surely be directed at none of the crystallographic directions at all, so the question is, if I draw a line on the wafer surface parallel to the primary flat, will it be directed towards the <1 0 0> or the <1 0 1> family, or does the manufacturing process just slice an arbitrary flat out of the Si ingot before slicing it into wafers?

The flats on Si wafers are not arbitrary.  The largest, or primary flat, is perpendicular to the <110> orientation in the plane of the wafer. If it exists, the secondary flat helps identify the wafer orientation and doping type (and can easily vary based on wafer maker and the company ordering the wafers). All modern wafers (anything 200mm or bigger, almost all 150mm) are (100) and boron doped, so the secondary flat isn't used.
The primary flat has a specific orientation since the chips need to be aligned along the <110> crystallographic orientations so the wafer can be diced up.  This usually uses a diamond scribe and careful breaking rather than sawing, because the saw is wider than a scribe line (so fewer chips/wafer) and dirtier .
So, you wafer has <001> pointing up, and <110> in plane towards the flat.


text:
Can the 1st generation Grubbs Catalyst be used for ring closing metathesis on non-terminal alkenes?
I'm wondering if the 1st generation Grubbs catalyst can be used to do ring closing metathesis on a non-terminal alkene.

Yes, it appears that that ring closing metathesis of non-terminal alkenes has been performed using the 1st generation Grubbs catalyst several times:

Asymmetric syntheses of enantiopure C(5)-substituted transpentacins
via diastereoselective Ireland–Claisen rearrangements
Stereoselective Synthesis of Deuterium-Labeled (2S)-Cyclohexenyl
Alanines, Biosynthetic Intermediates of Cinnabaramide
Ru-catalyzed metathesis of octadienylether xyloside Synthesis
of (−)-Agelastatin A by [3.3] Sigmatropic Rearrangement of Allyl
Cyanate
A chiral auxiliary cleavable by ring-closing alkene
metathesis — Efficient synthesis of chiral nonracemic
cycloalkenes
Conversion of chiral unsaturated cyanohydrins into
chiral carba- and heterocycles via ring-closing metathesis
Total Synthesis of Pancratistatin Relying on the [3,3]-Sigmatropic
Rearrangement
Stereoselective Chelate-Controlled Addition of Grignard Reagents to Unsaturated Medium-Ring Heterocycles


text:
Is a metal hydroxide fully or partly oxidised in insufficient base?
Firstly, my question only applies to transition metal compounds with more than one hydroxide group, otherwise it would be more obvious.
The reason I'm asking is in an effort to oxidise some nickel dihydroxide into nickel oxyhydroxide for practical use in an electrode. I've only come across somewhat laborious methods (any help in this would be appreciated) so I've considered the oxidation like in a NiMH cell (the second equation).
With nickel in a basic solution:
My question could stem from the following situation; If I stir a 2:1 ratio of NaOH:Ni(OH)2 together I would assume the following reaction would be favoured:
$$\ce {4 Ni(OH)2 + 8OH- -> 1O2 + 2 Ni2O3 + 8H2O}$$
But, if I stir a 1:1 ratio of NaOH:Ni(OH)2 and ground the solution:
$$\ce {Ni(OH)2 + OH- -> NiOOH + H2O + e-}$$
*I may or may not be misssing some electrons...I'm just starting out in electro-chemistry.
Is this assumption true? Why or why not? Am I missing the bigger picture? Cue Gibbs?  I did try it, but with the setup I had, I couldn't check for gaseous oxygen, and unfortunately both Ni2O3 and NiOOH are black and powdery.
Similarly with iron in an acidic solution:
In insufficient acid, which one is what would happen and why?:
$$\ce {Fe(OH)3 + H+ -> [Fe(OH)2]+ + H2O}$$ vs.
$$\ce {Fe(OH)3 + H+ -> Fe^{3+} + 3 H2O + excess \,\,\, Fe(OH)3}$$

I guess the question boils down to; How do reduction / oxidation potentials change with multiple individual reduction / oxidations?
I hope everything makes sense. If not I will gladly try to explain my question further.
Thank you for your help!

"How do reduction / oxidation potentials change with multiple individual reduction / oxidations? "; i can't understand what you mean by "multiple individual reduction / oxidations?"

@eaxdpiotnyeantial Say for Nickel dihydroxide, is Ni(OH)2 or NiOOH harder to oxidise with the base? Why?

What you are worried seems to be in your comment, which I'm explaining[maybe not]
Oxidations:

$$\ce{NiOOH + H2O + e- -> Ni(OH)2 + OH-} \quad E^{\circ}=+0.49\\
\ce{Ni2O3(s) + 6H+ + 2e- -> 2Ni^2+ + 3H2O} \quad E^{\circ}=+1.753\\$$

I suppose if you know about electrode potential you might be able to understand now,If you wish I might explain it, if you can't on your own.


text:
Cyanide from seeds breaking down in a pond
I have a large loquat tree that overhangs a pond. The loquat seeds apparently contain cyanide of some sort, so I'm trying to find the effect the seeds have when they fall in the pond.
Over time I imagine the seeds would break down.
What happens to the cyanide? Does it escape into the air, or remain in the water?

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Loquat seeds contain Amygdaline, which is also found in apricot kernels. Amygdaline belongs to the group of cyanogenic glycosides.
In these compounds, a potentially cyanide releasing aglycon, such as mandelic nitrile, is bound to a carbohydrate (= a "sugar"). These compounds are quite abundant in nature; other known examples are Prunasin from the japanese bush cherry or Sambunigrin from elderberry seeds.
In some cases, cyanide is released by enzymatic cleavage of the glycoside, that is when the plant material is digested by animals. Consequently, some of the plants (or at least some parts of them) are known to be toxic. In other cases, the glycosides do degrade in water.
If hydrogen cyanide is released to the pond, I'd assume it stays there - $\ce{HCN}$ is highly soluble in water.
However, I wouldn't be very concerned. Seeds aren't like paper bags full of toxic material.

It will likely take a while until the seeds break up to release Amygdaline.
To my knowledge, Amygdaline is rather stable in water, hydrogen cyanide isn't released until a sugar-splitting enzyme or acid acts on the molecule.

... 3. HCN is rather reactive and therefore unstable in the environment, it will never build up


text:
Why is PbCl₄ covalent?
My answer: The inert pair effect in $\ce{Pb}$ causes it to pull back the electrons, resulting in polarisation.
My teacher's answer: An ionisation state of $+4$ is too difficult to achieve, and it is not gained even when 4 chlorine atoms get reduced, as the later ionisation enthalpies of lead (or any metal) are high.
My point is, if we don't think of the inert pair effect, we are generalising this, and moreover, high ionisation energies of lead should be explained by the inert pair effect.
Which explanation do you think is better?
As a side note, I would like to add that my teacher pointed out that the inert pair effect is an outdated concept. Is it so? If yes, why is it rejected?

The high covalent characted is just explained by the fact of $\ce{Pb^{4+}}$ being very polarizing and $\ce{Cl-}$ fairly polarizable.
The inert pair effect would give an explanation about why it is difficult to reach that oxidation state, and why it has high tendency to get reduced, but there is no need to require to this concept to explain the covalent character of the compound.
Regarding your last question, I do not see why the inner pair concept should be rejected. Is just another model that can explain in a simple way properties or behaviors of certain elements. There are better models and orbital calcualtions that can also do the job? Of course, but that is no reason for avoiding the use of the model. Models that are still useful to explain stuff easily, can be employed even if more accurate ones do exist. One good example is ligand field theory, which is still useful even when molecular orbitals theory is more accurate and can still explain the same (and many more) things than the former.
Even though, a theorem or model HAS to be rejected if a new one is found that explains properties better and is incompatible with the old one.

+1, Agreed, I see absolutely no reason why the inert pair effect should be abandoned/ should be outdated, it is a wonderful example of relativistic effects. The crystal field theory on the other hand should be avoided, since the superior ligand field theory is not much more complex.

Yes, I intended to write LFT, I will edit that. Thanks for the correction.

Halogenides with metals in high oxidation state are rarely truly ionic on their own, as a strong positive charge on the cation must be stabilized. To do so, many metal-halogen contacts are required, and the metal cation is usually too small to coordinate many halogen ions. So, usually higher halogenides are either molecules with highly polar bonds or (di/oligo/poly)mers with highly polar bonds where some halogen atoms participate in two metal-halogen bonds.
In case of $\ce{PbCl4}$ NBO analysis reveals -0.4 NBO charges on chlorine atoms, so it is a highly polar bond, even though it is a molecular compound. So, I would say that due to unfavorable ionic radiuses, $\ce{Pb^{4+}}$ is incapable to form enough contacts to stabilize effectively, forming instead a tetrahedral polar-covalent molecule.

In $\ce{PbCl4}$ the lead is in $\ce{Pb^{IV}}$ state and $\ce{Cl}$ is having a considerable, not very large - but enough to be polarised easily, and the high charge density on lead polarizes the anion much more efficiently, so that the electron density is shifted and partially shared, in effect the compund is much covalent.You would like to see Fajan's rule, if you didn't knew them before.


text:
Can I replace potassium nitrate with sodium nitrate in a smoke bomb?
A very famous experiment goes like this:
$\ce{KNO3}$, molten wax and sugar are mixed and later burnt; the resulting mixture produces quite a lot of smoke.

After hearing about the experiment, I threw in the idea, "What would happen if $\ce{NaNO3}$, or other alkaline compounds were used instead?". However, the teacher wants us to be sure that the building doesn't explode. So here I am. Has this experiment been attempted before? What would happen if it was used?
Here is a link with the smoke bomb experiment.

In pyrotechnics, $\ce{NaNO3}$ is normally avoided because it is very hygroscopic.
With dried $\ce{NaNO3}$, the effect should be identical.
In any case, the usual safety rules apply!

@MikhailTal Have fun, but play safe and protect your eyes!

We didn't do it, we'll do it later, but still, I'll accept it.

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text:
Why is CO a major species only in fuel-rich combustion?
In fuel-rich combustion, my understanding is that enough fuel reacts and consumes all of the oxygen present. The combustion products $\ce{CO2}$ and $\ce{H2O}$ undergo dissociation until equilibrium is established and one of these is $$\ce{CO2 -> CO + 1/2O2}.$$
From the graph below, the equilibrium mole fraction of combustion products are plotted vs. equivalence ratio. Why is there a sudden rise in $\ce{[CO]}$ as we move into fuel rich combustion? Is this due to the drop in $\ce{[O2]}$ such that $\ce{[CO]}$ must increase to compensate for equilibrium to be established?

I think it is quite simple: the higher the proportion of the compound that will burn in the mixture, the more oxygen will be required to a complete combustion. If you have a constant amount of oxygen (in the reactor), and more is requires while increasing the amount of fuel, the complete combustion will gradually decrease, and the ratio $CO/CO_2$ produced will increase as a result of that.
As it can be seen on the graphic, from the point that the equivalent ratio of fuel and oxygen is larger than one, the $CO$ starts being formed in a considerable proportion, and over a ratio of 1:3, starts being the major product.

In fuel rich combustion, fuel and existing O2 reacts, and CO2 produced dissociates into CO and O2, and the O2 is consumed again since there's fuel left (and produces more CO2 -> more CO etc.), which results in CO accumulation and not very much O2. Equilibrium is established since the large [CO] compensates for the low [O2] in the reverse reaction to form CO2. Is this logic correct?

Yeah, I believe that is correct.

A followup question, is equilibrium essentially the point where O2 is depleted enough such that the rate in which it reacts with remaining fuel (and forms CO2) is less than the rate in which O2 is reformed from CO2 dissociation? This would imply that as t -> infinity the excess fuel will eventually all be used up?


text:
How do I achieve same crystallinity when cooling polymers?
I am trying to make "films" of polyethylene glycol by melting them, pouring them into a mold and then cooling the mold down. The plan is to do this with different molecular weights. The problem is that depending on the molecular weight, the crystallinity will be different. Is there a way of ensuring that the crystallinity of the polymer is the same for all of them?
Thanks

Is there a way of ensuring that the crystallinity of the polymer is
  the same for all of them?

First you need a method to assess the degree of crystallinity.  Samples of the same material that differ in degree of crystallinity should show different light transmission intensities due mainly to increased light scattering in the more crystalline samples.  If you have access to a spectrophotometer you could insert the films and measure the light transmission.  If you don't have access to a spectrophotometer, a "poor-man's" approach would be to use visual comparison of the transmitted light (use a black piece of paper and a pencil beam flashlight in a dark room).  It is important that the film thicknesses all be the same.  If there is variability in your film thickness, then cast thicker films so that the percentage variation will be smaller.
By cooling your sample more quickly, you will tend to preserve more of the amorphous nature found in the solution.  Conversely, cooling more slowly or annealing your sample will tend to increase the degree of crystallinity.
Adjust your rate of cooling on each of the films until they all produce the same transmission intensity, then the degree of crystallinity should be similar in your films.

So once the crystallinity is found for each sample (via your described method or any other) the only way would be to change the rate of cooling via trial and error?

That's how I see it, unless you're running these experiments in a research environment where other high dollar value equipment (x-ray equipment, etc.) is available.  It wouldn't surprise me if you see a trend between cooling rate, crystallinity and molecular weight, that might make it possible to estimate or project cooling rates for later samples.

This way you might get the same degree of crystallinity with one method, but very likely the morphology will be different, so other methods give different values (there is no method that gives a undoubtable absolute value). The whole thing sounds rather pointless to me.


text:
Alkalis and Bases
I just learned about acids and bases during class. I noticed that my textbook mentions the term "alkali/alkaline" and sometimes it mentions "base/basic". As far as I know, alkalis and bases are metal oxides hydroxides. But why does my textbook make use of two words that mean the same thing instead of just one of them? Do these terms (alkali and base) mean something different?

These words are in many cases interchangeably used: an alkaline solution/compound is the same as a basic solution/compound.
But you can specify "alkali" to refer to those bases that are soluble in water and that are salts of alkaline or alkaline earth metals too.
Alkali is like a subset of bases which fulfills these conditions.
E.g.:

Sodium hydroxide is a base, and also is an alkali metal salt, and is also soluble in water, so it is also an alkali.
Ammonia is a base, soluble in water, but not a salt, so it is not an alkali.
Magnesium hydroxide is a base, is a salt of an alkaline earth metal, but it is not soluble in water, so it is not an alkali.

I think an alkali is a substance that dissolves in water to produce $\ce{OH- (aq)}$ ions. Since the $\ce{OH- (aq)}$ ions can react with acids, that makes alkalis bases as well. Therefore alkalis are soluble bases.
Ammonia can also dissolve in water to form ammonium, $\ce{NH4+ (aq)}$ ions and $\ce{OH- (aq)}$, that is, it undergoes hydrolysis, in which case hydroxyl, $\ce{OH- (aq)}$ ions are produced. Therefore in that respect ammonia qualifies as an alkali, and we have to remember the very high solubility of ammonia in water as demonstrated in the fountain experiment.
Sodium carbonate is an alkali because it also dissolves in water to form hydroxyl ions.
Therefore, from this summary there are two very important criteria for  compounds to be considered an alkali:

It must dissolve in water
It must produce $\ce{OH- (aq)}$ ions in the aqueous medium.

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text:
Does dissociation require energy to occur?
I assume spontaneous separation to ions of salt solution (or acid) requires energy to occur, right?
Where does that energy come from? Is it from the heat of the solution?
Now, if someone devised a way to draw potential energy from those separated ions, could he make this process continuous (assuming solution would draw heat from its surroundings), or would it eventually stop to work?

I assume spontaneous separation to ions of salt solution (or acid)
  requires energy to occur, right?

Yes, it requires energy to occur.  This energy is usually referred to as the activation energy.  It is a measure of the energy barrier that must be surmounted for the process to occur.  In the case of salts ionizing in solution, the barrier is usually (but not always) quite small.  Salts that dissolve and give off heat (exothermic) will have a lower barrier than salts that require energy input (endothermic) to dissolve.

image source: http://web.campbell.edu/faculty/nemecz/323_lect/enzyme_mech/mech_chapter.html

Where does that energy come from? Is it from the heat of the solution?

The energy comes from the surrounding environment.  The surrounding molecules have a thermal energy that can be transferred through collisions with the reactants.  The reactants will convert to products once enough energy has been imparted to them to cross the barrier.

Now, if someone devised a way to draw potential energy from those
  separated ions, could he make this process continuous

I think you're asking if the energy withdrawal can be made continuous.  If so, then the answer is "yes" - at least until all of the starting material has been consumed and converted to product.  This is done today, examples would include batteries, gasoline, power generators in general, etc.

When the ions of an ionic compound disolve, do not stay isolated. They form new bonds with the solvents. This makes the overall process spontaneus when a salt is soluble.
Dissociation of course requires energy. But bond formation between ions and solvent molecules release energy.
This can result on a possitive (endothermic disolution, heat is absorbed in the process, like for example, $\ce{KI}$ in water. Bonds formed release less energy than bonds broken) or negative disolution enthalpy (exothermic disolution, like the one for $\ce{H_2SO_4}$ and water. Bonds formed -hydrogen bonds, stronger that ion-dipole interactions-, release more energy that is consumed by bonds broken).
Remember that if you want to know if a process is or not spontaneous, you do not have to take into acount just enthalpy, but Gibbs free energy, which is $\Delta G= \Delta H + T\Delta S$, which also takes into account entropic factors and the influence of temperature.

authors: ᴡʜᴀᴄᴋᴀᴍᴀᴅᴏᴏᴅʟᴇ3000


text:
What is 'equality' of chemical bonds?
Why are all the $\ce{C-O}$ bonds in $\ce{CO_3^2-}$ equal but not those in $\ce{H2CO3}$?

For $\ce{CO_3^2-}$ you have the following possible resonance structures:

The structure bellow, is the one which represents better the real ionic molecule, in which all the bonds are equal. You can delocalize the negative charge through all the ion.
For $\ce{H2CO3}$ the structure is:

As you can see, in this case, bonds are fixed, and there is no possible resonance delocalization, so you have two single bonds (less strong) C-O and one double bond C=O (stronger). For $\ce{CO_3^2-}$ you have that all bond lenths the same showing an intermediate situation between single and double bonds.


text:
Lowest CAS Registry Number on record
I've been looking for a while and I still can't get a straight answer. I just need to know the lowest CAS Registry Number on record and the corresponding chemical.

Why? What are you going to do with it?

It's for an organic chemistry homework assignment. I already know how to use CAS and SciFinder, but I don't think it would be possible for me to search every compound to find which one is the lowest.

Currently the CAS Registry identifies more than 81 million organic and inorganic substances and 64 million protein and DNA sequences, plus additional information about each substance. The Registry is updated with an approximate 15,000 additional new substances daily.Well are you afraid of doing it by hit and trial :-)

Look in the index of a print Aldrich catalog. They have a sequential listing of available compounds. The first one is probably the lowest CAS assigned...

I knew most of that since those were the other questions on the homework. And that's why I think it's practically impossible.

I also tried that, and the lowest I found was much higher than my current lowest. Right now I have Formaldehyde at 50-00-0

I do not have a citable resource, but to my knowledge, 50-00-0 is the first and lowest CAS RN assigned in 1965.

I find this a very strange homework assignment. What exactly are you supposed to learn or master by doing this ?

We have to familiarize ourself with CAS, SciFinder and ChemDraw to make us more well rounded for formal report research. The question I asked was, by my guess, supposed to link the check digit formula and some common group areas together, as well as learn a "fun fact."

sounds more like advertizing than educating

I work at CAS and recently answered a similar question on Twitter (@CASChemistry). In fact, there are 8 pre-50-00-0 CAS Registry Numbers in our database:

35-66-5
35-67-6
36-51-1
36-88-4
37-71-8
37-82-1
37-87-6
38-26-6

So, the lowest CAS Registry Number on record is 35-66-5 for "Benzacridine, 7,9-dimethyl- (9CI)".

"Carotenes" has a CAS Registry Number of 36-88-4.
If that chemical is "too vague", Naphthol AS-BI $\beta$-D-glucuronide has a CAS Registry Number of 37-87-6
According to semi-independent source Organic Chemist's Desk Reference 2nd ed. (2010) pg 6 sec 1.1.1.5, CAS 35-66-5 DOES seem to be the lowest, but how do we know "Benzacridine, 7,9-dimethyl- (9CI)" exist? Cl- and methyl- are occupying the same position at the aromatic ring.
Also "Carotenes" (CAS 36-88-4) by the name itself does not seem to exist as a single compound but a set of compounds.

Isn't 50-00-0 smaller than that?

The compound is 7,9-dimethyl-acridine.  I'm guessing that (9CI) refers to the CAS 9th collective index.


text:
What are the specific disadvantages of using a bio-catalyst (called Lipolase) in the synthesis of the drug pregabalin?
I am studying the synthesis of a drug pregabalin, for which both chemical and bio-catalysts are used. Using a biocatalyst is way better than chemical route, as explained here. In fact, The invention of the bio-catalyst route got 2006 IchemE award in Green Chemistry and Engineering.
I am interested in knowing is this: are there any disadvantages of using bio-catalysis in this reaction? (Some general disadvantages can be seen here) Suppose you are in a debate and you have to favor the chemical side, what would be your arguments?

Most enzymes work in aqueous solutions, so if your starting material or product is poorly soluble in water, it might not get into or out of the enzyme. Enzymes are also temperature sensitive, so if you have to heat the reaction for any reason you could destroy enzyme activity. Additionally, enzymes can be difficult and expensive to produce in large quantities. But these are general issues with enzymes, I don't know about any specific to pregabolin synthesis. But if Pfizer adapted the enzymatic approach they must have control over these issues.


text:
How ‘lively’ would this ‘health’ potion be?
I’ve stumbled across a rather peculiar recipe for a allegedly ‘health’ potion in a computer game, composed of an unspecified amount of each of the following:

rust (iron oxide)
saltpetre (potassium nitrate)
oil of vitriol (sulphuric acid)
aqua vitæ (ethanol)

To make things clear: I’m not suicidal, I do in no way intent to actually mix that stuff (the initial tagging was maybe a bit misleading), but I’m interested in what would happen if someone actually would.  Drinking such a potion vile concoction is of course out of the question because of the sulphuric acid alone.
From what I know from highschool chemistry (and a generous helping of Wikipedia), saltpetre and sulphuric acid would be used to make explosives by way of nitrosulphuric acid.  Oversimplifying, by adding cellulose or tuluol one would end up with nitrocellulose or TNT, respectively.
But I lack the terminology to go further than that.  What would be the result of adding alcohol to that, and then iron oxide, or the other way round?  Is there some known mixture that would fit, however roughly, along those lines?  From the context of where that formula came from that wouldn’t surprise me.  It may of course as well be that the formula was simply chosen because of it’s obvious unhealthyness.
I hope this clarifies the intent of my question.  The idea may have come from a pseudo-fantasy/alchemy setting, but I’m interested in the ‘real’ chemistry behind it.

Is such a question even on-topic?  I’m not really sure—if not, please don’t hesitate to tell me, then I’ll take it down myself.

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I think the question is ok even though it might be considered a little borderline. We generally don't give any medical advise but in this case you are not really asking for that. Discussing the potential dangers (or safety) of such a mixture should be on-topic here.

@Philipp Absolutely agreed. I added the [tag:safety] tag to emphasize this. || On the account of the potion, I would not expect sulfuric acid to have any nice effect on the body at all.

@Philipp:  You are right, I should have explained how far I have gotten by myself.  I’ve tried clarifying my question.

Explosives: It is true that sulfuric acid and sodium/potassium nitrate are used to make various explosives (by creating nitric acid), but the key is something to nitrate. The worst this mixture could generate is nitrated ethanol, and I was unable to find any source suggesting that could be done with nitric and sulfuric acid alone. Nitrating a molecule that small with that simple of a structure seems unlikely without extreme conditions or reagents. If it were possible, the most likely product does seem to be something unpleasant.
Poisons: What definitely would be produced is nitric acid, and the production of nitric acid via sulfuric acid and a nitrate salt produces nitrogen dioxide, a poisonous and caustic gas. This is some seriously bad stuff you do not want to come in contact with.
Based on that, I'm going to conclude this mixture is probably not explosive, but definitely could become so if an easily-nitrated compound was a contaminant. Those are more common than you might imagine. It definitely would create a nasty gas.

A concoction quite like this (with Fe-EDTA complex instead of iron sulphate) is sold as aquarium fertilizer.  Dosed in small amounts it would at least promote plant growth...

A glass of that would just kill you on the right proportions, not much of a health potion...
(Honestly, I was worried about you when I first saw the title, the list and the home-experiment tag)

A potion to end all pains, lol.


text:
Strong and weak acid
I know sulfuric acid is a strong acid and acetic acid is a weak acid. However when I write them in their chemical symbol.
Sulfuric acid: $\ce{H2SO4}$
Acetic acid: $\ce{CH3COOH}$
I realise that acetic acid has 4 atoms of hydrogen whereas sulfuric acid has only 2 atoms of hydrogen. So, shouldn't acetic acid be stronger than sulfuric acid?

Looking at the structures of those two acids might help. In both cases, the acidic hydrogens are part of hydroxyl groups. Because of polarity and the inductive effect, the O-H bond is weaker than C-H bonds which are very strong. As a result, it's nearly impossible for Carbon to lose a hydrogen without some kind of elimination reaction, whereas the O-H bond can autoprotolyze with other O-H bonds (even water). The weakness of the O-H bond is the basis for the acidity of those hydrogens. C-H bonds are essentially non-acidic, having pKa values around 45-50.

Sulfuric acid is a strong acid; when placed in water, it completely dissociates into $H^+$ and $HSO_4^-$. While acetic acid dissociates into $H^+$ and $CH_3COO^-$; but being a weak acid, some of the molecules remain as $CH_3COOH$.
The difference between a strong and a weak acid is not necessarily their level of acidity; it is the degree to which they dissociate. Also as the previous answerer said, due to the nature of the bond of the 3 hydrogen molecules in acetate, we wouldn't consider it an acid because they don't break into $H^+$ ions as easily.

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The pKa of an acid tells you how strong the acid is.
$pKa=-log([H^+][A^-]/[HA])=-log(Ka)$ where $A$ is the conjugate base of the acid (the species left when the proton dissociates). A low value for $pKa$ gives the stronger acid since a high value for $Ka$ indicates that the equilibrium $\ce{HA<=>H+ + A-}$ lies well to the right.
There are a number of factors that make an acid's equilibrium position lie far to the right. By far the most important is the stability of the conjugate base ($A^-$). In your example, the reason $\ce{H2SO4}$ so much more acidic than $\ce{CH3COOH}$ because is can form more resonance structures than ethanoic acid so it distributes the negative charge better. Also the inductive effect of the alkyl group destabilizes the negative conjugate base. There are more reasons but this is the most important.


text:
Reactivity series of metal
I have learned about metal and their reactivity series. However, why metals have different reactivity? What are the factors of different reactivity rate for metal? All I can think of is the number of electrons which causing it.

I suppose that with reactivity you mean a metal's tendency to become oxidized.
There are several reasons and it depends on the metals you are analyzing.
If you are referring to alkaline or alkaline earth metals, reactivity increases as you go down the group. They all have a large oxidation potential since it's relatively easy to remove an electron (or two) of its orbital. As you move down the group, atoms are larger and the valence electrons will be less atracted by the nucleus. Therefore, they are easier to remove and the metal will be more reactive.
For d-block transition metals (most of them are considered "transition" metals) it's a whole different story. There are many factors that contribute to its reactivity and it's harder to specify general trends. Many of them (especially the ones on the second and third transition rows) can be oxidized to different oxidation states depending on the conditions. You need to look at their electron configuration and reduction potentials to start rationalizing its properties. Depending upon the electron configuration, they will be able to lose (or even gain) less or more electrons and with different ease.
I don't know your background with chemistry, so I won't write a more detailed explanation but if you clarify that you have a more concrete interest in any explanation, just ask.

For the transition metal, is it the lower the oxidation number of a metal, the higher the reactivity of the metal?

No, not necessarily at all. For example, oxidation potential of $Cu$ to $Cu^+$ is about -0.5 V (an stronger oxidant than $H^+$ is required), and the one for $Mn$ to $\ce{Mn^2+}$ is about +1.2 V (easily reduced by $H^+$). It's way harder to make predictions for transition metals. There are also some important exceptions with very low reactive metals, (noble metals) like $Au, Ir, Pt$. Like I told you,you should check their electron config and a list of reduction potentials for metals. The lower the reduction potential, the higher is the oxidation potential (ie higher reactivity).

There are different factors that affect the re-activity of any element. These are as follows:

The atomic radius, or the distance of the outermost shell from the nucleus.
The nuclear charge (atomic number), because the greater the number of protons, the greater the pull on the outermost shell.
The electron shielding due to the repulsion of electrons from the inner shells, pushing the outermost shells away from the nucleus.

Some factors can 'overpower' others depending on the position in the Periodic Table.

The atomic radius or distance of the out most shell to the nucleus
The atomic number. The greater the number of protons, the greater the pull on the out most shell.
The electron shielding due to the repulsing of electrons from the inner shells, pushing the outemost shells away from the radius.

Would you please elaborate on each factor too?


text:
The pH of a neutralized solution
If pH is defined as the concentration of hydrogen ions in solution, then how can a ‘neutralized’ solution (defined as having an equal amount of hydrogen and hydroxide ions) have a pH other than 7?
Wikipedia writes: “In a reaction in water, neutralization results in there being no excess of hydrogen or hydroxide ions present in solution.” In my mind, an equivalent amount of hydrogen ions and hydroxide ions means a pH of 7 (in water), but Wikipedia carries on to write: “The pH of the neutralized solution depends on the acid strength of the reactants.”
This question doesn’t seem to answer my question.

Are you sure is it completely neutralized? Because it's impossible

When a solution is neutralized, the pH is equal to the point of equivalence, which is not 7 if either the acid or base is stronger (as with titration experiments).

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site.

The reason why I have never accepted an answer to this question is because, when I first read the answers, they were too advanced for me to understand, and subsequently my direction of study went strongly towards mathematics and economics.

TL;DR
The neutral point is not the same as the equivalence point.
A neutral aqueous solution at room temperature, $25~^\circ\mathrm{C}$ and standard pressure $1~\mathrm{atm}$, has always $\ce{pH}=7$.

If you look for credible sources of definitions of neutral solutions on the internet, most likely you will find something along the lines of this:

Neutral solution
  (Science) Has a $\ce{pH}$ level of $7$: a solution in which the concentration of hydrogen ions and hydroxide ions are equal (biology-online.org)

Or even worse:

Neutral Solution Definition:
  An aqueous solution with a $\ce{pH}$ of $7.0$ $(\ce{[H^{+}]} = 1.0 \times 10^{-7}~\mathrm{M})$. (chemistry.about.com)

The second one especially is negligent of a lot of features.
Unfortunately the IUPAC does not provide an actual definition of a neutral solution. There are two possible definitions, which come to the same conclusion (compare the above), given the same external conditions.

The lazy definition:
The concentrations of hydronium and hydroxide ions are identical. $$c(\ce{OH-})= c(\ce{H3O+})$$
In terms of $\ce{pH}, \ce{pOH}, \mathrm{p}K_w$ one arrives at a more complete side.
Pure water is a neutral solution. It's autoprotolysis provides
\begin{align}
\ce{2H2O &~<=>~ OH- + H3O+}.\tag{1}\\
\end{align}
Now we can formulate the equilibrium constant and furthermore assume that autoprotolysis is small compared to the overall concentration of water.
\begin{align}
  K_c &= \frac{c(\ce{OH-})\cdot c(\ce{H3O+})}{c^2(\ce{H2O})}\\
  K_w &= c(\ce{OH-})\cdot c(\ce{H3O+})\tag{2}\\
\end{align}
Equation $(1)$ also provides the lazy definition. $$c(\ce{OH-})= c(\ce{H3O+})\tag3$$
We plug that into $(2)$ and we arrive at
$$c(\ce{H3O+})=\sqrt{K_w}\tag{4}$$
Consider the actual definition of
$$\ce{pH} = −\lg a(\ce{H+}) = −\lg\left( m(\ce{H+}) \gamma_m(\ce{H+}) / m^\circ\right)\tag{5}$$ rewritten using concentrations
$$\ce{pH} = −\lg a(\ce{H+}) = −\lg\left( c(\ce{H+}) \gamma_c(\ce{H+}) / c^\circ\right)\tag{5a}$$
The identities used here are $a(\ce{H+})$ for the activity of a proton in aqueous solution, $\ce{H+ (aq)}$ and we will consider $\ce{H+ (aq) = H3O+}$ to be the same. Now we are going to assume, that the activity coefficient is $\gamma_c(\ce{H+})\approx 1$ for very diluted systems. Using the standard concentration $c^\circ$ we do not have to care about units.
$$\ce{pH} = −\lg\left( \frac{c(\ce{H+})}{c^\circ}\right)\tag{5b}$$
Now we can substitute in $(4)$ and we will have a nice definition of the $\ce{pH}$ of a neutral solution.
$$\ce{pH} = −\lg\left( \frac{\sqrt{K_w}}{c^\circ}\right)\tag{6}$$

The second formulation includes a very crucial point. The autoprotolysis $(1)$ is temperature, $T$, dependent, which is obvious following the definition of the standard equilibrium constant
$$ K^\circ = \exp\left\{\frac{-\Delta G^\circ}{\mathcal{R}T}\right\}\approx K_c.$$
Therefore also the ion product of water is temperature dependent, $K_w(T)$.
Wikipedia actually has a couple of values for the $\mathrm{p}K_w$. Reaction $(1)$ is endothermic. A higher temperature means providing energy, means a higher $\mathrm{p}K_w$. For example, in the human body, $37~^\circ\mathrm{C}$, this value is slightly higher than for the one we usually assume as the "neutral point" $\mathrm{p}K_w(25~^\circ\mathrm{C})=14$.
Since $(6)$ is more correct in the following form,
$$\ce{pH}(T) = −\lg\left( \frac{\sqrt{K_w(T)}}{c^\circ}\right)\tag{6a},$$
the neutral $\ce{pH}(T)$ therefore changes with temperature.
Another note is that it is also pressure dependent, but that might take it a little too far.
The equivalence point is usually defined for an arbitrary chemical reaction between an acid and a base, and it refers to the situation where there are stoichiometric quantities of acid and base are present.
$$\ce{AH + B <=>[\ce{H2O}] A^- + HB+}$$
This means, that at the equivalence point the following statement holds:
$$c_\text{initial}(\ce{HA}) = c_\text{final}(\ce{HB+}).$$
Therefore the $\ce{pH}$ at the equivalence point will only be governed by the reactions of $\ce{HB+}$ and might not be neutral.

The issue of solvent dependency is addressed by Fred Senese quite concise on antoine.frostburg.edu:

pH is often used to compare solution acidities. For example, a solution of pH 1 is said to be 10 times as acidic as a solution of pH 2, because the hydrogen ion concentration at pH 1 is ten times the hydrogen ion concentration at pH 2. This is correct as long as the solutions being compared both use the same solvent. You can't use pH to compare the acidities in different solvents because the neutral pH is different for each solvent. For example, the concentration of hydrogen ions in pure ethanol is about 1.58 × 10-10 M, so ethanol is neutral at pH 9.8. A solution with a pH of 8 would be considered acidic in ethanol, but basic in water!

For me, the key to understanding how the pH at a point of equivalence is not necessarily 7 is as follows: a neutralization reaction produces a salt and water, but the salt is not necessarily neutral.

But I'm still not sure about whether a salt can have an excess of $\text{H}^+$ or $\text{OH}^-$ or not.

@ahorn A salt of any weak acid or base will be in equilibrium with the dissociation of water, thus producing protons and/or hydroxide ions. Therefore the pH will differ from the neutral point.

Just as a point of reference (though I think it should be clear to people by context), $\lg(x) \not= \log_2(x)$ here. This is a common notation in the US.

@Tyberius I personally prefer having the charge where it should be, that's why I made the effort of writing $\ce{^{-}OH}$. I agree with the recent additions to mhchem, $\ce{H3+O}$ looks a bit silly.

In addition to all other good answers, I'd like to actually show you why the "neutralized solution" is not equal to a neutral solution.
Whenever we want to describe a neutralisation reaction (that is nothing else than a titration) we first need to find out where protons come from and where they go. Let's examine a simple titration of a monoprotic acid with a monoprotic base:
$$c(\ce{H3O+})=c(\ce{OH-})+c(\ce{A-})-c(\ce{BH+})$$
Protons come from the autoprotolysis of water and the acid residue anions and they will react with the titration base. Extending the parts of this equation by their $\mathrm{pH}$, $c_0$ and $k_\mathrm a$ containing counterparts, we get:
$$c(\ce{H3O+})=\frac{k_\mathrm w}{c(\ce{H3O+})}+\frac{c_0(\ce{HA})~k_\mathrm a}{k_\mathrm a+c(\ce{H3O+})}-\frac{c_0(\ce{B})~k_\mathrm b~c(\ce{H3O+})}{k_\mathrm w+k_\mathrm b~c(\ce{H3O+})}$$
At the point of inflection, the equivalence point, the amount of both moles need to be equal and the total volume is the addition of previous acidic volume plus the added base volume $V_\mathrm{tot}=V_{\ce{A}}+V_{\ce{B}}$. The ratio of both concentrations $c(\ce{HA})$ and $c(\ce{B})$ is therefor equal to $1$.
$$\begin{align}
n_{\ce{A}}&=n_{\ce{B}}\\
\Leftrightarrow c_\mathrm A~V_\mathrm a&=c_\mathrm B~V_\mathrm b\\
\xrightarrow{V_\mathrm{tot}=V_\mathrm A+V_\mathrm B} \frac{c_B}{c_A}&=1=\tau~\text{or}~\gamma\end{align}$$
Knowing this, rearranging the second equation to $c_\mathrm A/c_\mathrm B$ gives us:
$$\tau=\frac{c_\mathrm B}{c_\mathrm A}=\left(1+\frac{k_\mathrm w}{k_\mathrm b~c(\ce{H3O+})}\right)\left(\frac{k_\mathrm w}{c(\ce{H3O+})}-c(\ce{H3O+})+\frac{k_\mathrm a}{c(\ce{H3O+})+k_\mathrm a}\right)$$
which we now somehow have to solve for the proton concentration after setting $\tau=1$. As this is rather annoying I did that for the titration of any more or less relevant acids for $\mathrm pK_\mathrm a$ values between $-2$ and $16$, with $\ce{NaOH}$ giving the following graph.
$\hskip1.2in$
You can see that strong acids will be neutralized a pH 7 and extremely weak acids at pH 14. In between there is a linear range in which one can calculate the pH at the equivalence point by the simple equation $$\mathrm pH_\mathrm{eq}=7+0.5~\mathrm pK_\mathrm a$$
This can be expanded for the neutralization of every monoprotic acid with every monoprotic base and again there is a simple equation for calculating the pH at the equivalence point.
$\hskip1.2in$
$$\mathrm pH_\mathrm {eq}=7+0.5~\mathrm pK_\mathrm a-0.5~\mathrm pK_\mathrm b$$

Simple contradiction is salt hydrolysis:
The neutralization Reaction:

$$\ce{CH3COOH +NaOH->CH3COONa + H2O}$$

Now the following reaction happen due to low acidity of the acid, thus the reversible reaction a little favored:

$$\ce{CH3COO- + H2O->CH3COOH + OH-}$$

Though the reaction reaches equivalence/neutralization point, the solution will be little basic, as even though none, hydronium or hydroxyl ions, is in excess but one may become in excess with the advent of salt hydrolysis(kinda in situ).


text:
Oxygen dissolve in water
I know that $\ce{CO_2}$ and $\ce{O_2}$ are slightly soluble in water ($\ce{H_2O}$)
In my understanding, when carbon dioxide dissolves in water, it will react to form carbonic acid just like the equation below.
$\ce{H2O + CO2 -> H2CO3}$
So, my problem now is, I can't find the equation for oxygen.
 Will they form hydroxide ions or something else?

Carbon dioxide is not stable in water in its original form, so it undergoes an acid-base reaction to give the carbon dioxide.
On the other hand, oxygen is stable and soluble in water in its $\ce{O_2}$ form, so it will stay like that, no reaction other than solubilization. Besides, it could react with organic matter or other compounds in water, oxidizing them, and getting reduced to $\ce{OH^-}$

This is not correct - 99 % of the carbon dioxide dissolved in water appears in the form of CO2 and only 1 % as carbonic acid.  Moreover, the hydration reaction is rather slow, rate is on the order of seconds.


text:
What is the difference between carboxylate ions and groups?
I would like to know the difference between carboxylate ions carboxyl groups? An example is is $\ce{COOH}$ and $\ce{COO}$.

Carboxyl
The carboxyl group $\ce{-COOH}$ is an organic functional group consisting of a carbon atom double bonded to an oxygen atom and single bonded to a hydroxyl group.

Compounds with a carboxyl group are called carboxylic acids (e.g. acetic acid, $\ce{CH2COOH}$).
Carboxylate
A carboxylate $\ce{-COO^-}$ is a salt or ester of a carboxylic acid. For example, acetic acid reacts with sodium bicarbonate to form sodium acetate.
Compounds with this group have the carboxy- prefix or -ate suffix (e.g. sodium acetate $\ce{CH3COONa}$). When these salts are ionised (e.g. dissolved in water), the anion with the carboxylate group is called a carboxylate ion.
The carboxylate ion is the negative ion (anion) formed as the conjugate base when a carboxylic acid loses a proton.
$$\ce{R-COOH -> R-COO^- + H+}$$
Being its conjugate base, sodium acetate mixed with acetic acid makes a useful buffer solution for many biochemical reactions where a 'mildly acidic' pH of 4-6 is desired.

I suppose it must be hard not to see the big $\ce{H}$ difference.

You mean answer: a proton


text:
How to predict the outcome of a crossed Cannizzaro reaction?
My question is, how do we predict the products of a cross-Cannizzaro disproportionation in presence of a concentrated base? i.e, How to judge which of the aldehydes (both don't have any alpha hydrogen) is reduced and which one is oxidized?
If we are asked
$$\ce{CH3-O-C6H4-CHO   + HCHO ->[OH^- (conc)] ?}$$
Case: if we see acc to RDS, which is hydride transfer, we must see which compound is better hydride donor. Then it will be $\ce{CH3-O-C6H4-CHO}$. This will get oxidized, and $\ce{HCHO}$ reduced.
Case: if we see which is a better nucleophilic substrate, $\ce{HCHO}$ will get oxidised, and $\ce{CH3-O-C6H4-CHO}$, reduced.
Now my problem is, which of the above cases is correct? i.e, do we see which one of the substrates is better $\ce{H-}$ donor, or which one of them is a better nucleophilic substrate?
Please advise which case is correct, and how to approach such problems.

Please have a look at this tutorial shows you how math and chemical formulae can be nicely formatted on this site. In their current form the chemical equations in your question are rather hard to read.

Philipp, I know a bit of latex, typed this in a hurry, sorry about the horrible equations...

No problem, thanks for taking the time and improving the formatting.

Oh, its the least I could do...

To answer this, first let us look at the mechanism of the Cannizzaro reaction in general. It is depicted in scheme 1 as a cyclic transition state. The cycle itself isn’t as important, the realisation that one aldehyde is present as a (deprotonated) hydrate after the action of $\ce{OH-}$ is.

Scheme 1: General mechanism of a Cannizzaro reaction.
In your reaction, you are reacting formaldehyde with a (substituted) benzaldehyde. The question boils down to ‘can we predict, which aldehyde will form a hydrate mor easily?’ And the answer is: we can!
$$\begin{array}{cc}\hline
\text{carbonyl} & \text{hydrate ratio} \\ \hline
\ce{H2CO} & 99~\% \\
\ce{MeCHO} & 60~\% \\
\ce{Me2CO} & < 1~\% \\ \hline\end{array}$$
We have a clear tendency for formaldehyde to almost exclusively form hydrates under standard conditions. Thus, we can conclude that formaldehyde will alwys be the hydrate partner, end up oxidised and reduce its reaction partner as shown in scheme 2.

Scheme 2: Cannizzaro reduction of an aldehyde with formaldehyde as sacrificial aldehyde.

The values are fine, but can you elaborate on how the RDS step being easier for the methoxybenzaldehyde doesn't matter?

@HarryHolmes If what you assumed is the rate determining step (RDS; I had to look it up so I’m writing it out here) isn’t actually the rate determining step then I don’t really see what any elaboration would do.

It seems to say here that the hydride transfer step is indeed the rate determining step. Am I interpreting it wrong?

It's in the fifth paragraph, by the way.

@HarryHolmes The issue here is that methoxybenzaldehyde won’t come anywhere close to transferring any hydrides anywhere (except very slowly as part of a non-crossed Cannizzaro reaction giving the benzoic acid side product) so any discussion about what a hydrate of methoxybenzaldehyde would do is beside the point. In the case at hand, the hydride transfer is still rate determining, sure; but the selectivity is due to something entirely different. (Also, OP’s thoughts about nucleophilicity are equally irrelevant.)

Canizzaro Reaction

$$\ce{A +B->[\text{conc. OH}^-]A' +B'}$$

If A has more partial positive charge, equivalently better hydride donor or more reactive, it will reduce and B will oxidise. Exception is formaldehyde where it will always get oxidised to formate as it's hydrate is more stable in aqueous medium.

So you mean to say that the one which is a better nucleophillic substrate will be reduced, and not oxidized?

Moreover, if A has more positive charge, how is it a better $$\ce{H^-}$$ donor? Shouldn't we look for a higher electron density for a better donor?

It has been found that a crossed Cannizzaro reaction occurs on mixtures of aromatic
aldehydes and formaldehyde, which leads to the formation of formic acid and the
corresponding aromatic alcohol (Ref.1) as precise prediction of Jan's answer:
$$\ce{R-CHO + HCHO + H2O -> R-CH2OH + HCOOH}$$
This method offers a particularly convenient method of preparing certain aromatic alcohols of interest of which the corresponding aldehydes are readily available. In this work, the mole ratios of aromatic aldehyde to formalin to $\ce{NaOH}$ was $1:1.3:3$ and yield of corresponding aromatic alcohol was reported to be 85-90%. The reference also reported that about 2-5% of corresponding aromatic carboxylic acid was also discovered.
In addition, it is also possible that crossed Cannizzaro Reaction with with two aromatic aldehydes (Ref.2). For example, when a mixture of 4-chlorobenzaldehyde and 4-methoxybenzaldehyde (p-anisaldehyde) was allowed to react with alkali, a definite disproportionation has taken place (chloro acid to methoxy acid: $2.3:1$ and chloro alcohol to methoxy alcohol : $1:2.5$).
Interestingly, total of seven reactions have been studied in this manner where one member of the aldehyde pair was halogen substituted: 4-chlorobenzaldehyde and 4-methoxybenzaldehyde; 4-chlorobenzaldehyde and benzaldehyde; 4-chlorobenzaldehyde and 3,4-methylenedioxybenzaldehyde; 4-bromobenzaldehyde and 4-methoxybenzaldehyde; 4-bromobenzaldehyde and benzaldehyde; 3-bromobenzaldehyde and 4-methoxybenzaldehyde; and 3-bromobenzaldehyde and benzaldehyde. In all seven systems, a crossed Cannizzaro reaction has taken place between corresponding aromatic aldehydes and in every case, the halogen substituted molecule showed a tendency to be oxidized to the acid at the expense of the other aldehyde. Also note worthy that when the second aldehyde is benzaldehyde, proportionating ratio has been reduced (halo acid to benzoic acid: ~$1.4:1$ and halo alcohol to benzyl alcohol : ~$1:1.4$). Yet, 3-bromobenzaldehyde and benzaldehyde mixture gave halo acid to benzoic acid ratio of ~$2.4:1$.
References:

David Davidson and Marston Taylor Bogert, "The Preparation of Aromatic Alcohols by the Crossed Cannizzaro Reaction with Formaldehyde," J. Am. Chem. Soc. 1935, 57(5), 905-905 (DOI: https://doi.org/10.1021/ja01308a036).
John C. Bailar, Jr., Allan J. Barney, and R. F. Miller, "The Action of Alkalies on Mixtures of Aromatic Aldehydes," J. Am. Chem. Soc. 1936, 58(11), 2110–2111 (DOI: https://doi.org/10.1021/ja01302a009).

The Cannizzaro reaction, is a chemical reaction that involves the base-induced disproportionation of an aldehyde lacking a hydrogen atom in the alpha position. The oxidation product is a salt of a carboxylic acid and the reduction product is an alcohol.
I believe that your fist compound is an isomer of methoxybenzaldehyde, but the other compound, "hcho" is... Formaldehyde?
Methoxybenzaldehyde would disproportionate in their corresponding carboxylic acid and alcohol if a base were present, which is not, so it will not undergo Canizzaro disproportion.
I'm not sure if formaldehyde can undergo Canizzaro reaction to yield methanol plus formic acid, but a base would be surely needed.
If you talk about a crossed Cannizzaro reaction, in which an aldehyde is used in combination with a more valuable chemical. In this variation, the reductant is formaldehyde, which is oxidized to sodium formate, and the other aldehyde is transformed into the correspondent alcohol. But you also need base.

I am sorry, I should have mentioned that I needed the answer for cross cannizaro only.

And yes, that is formaldehyde

So there you have it :) Formaldehyde is always acting as reductant, and it's oxidized to sodium (or whichever counterion your base would have) formate.

Please see the question once again, maybe you will be able to understand my basic doubt, which is, How to predict the products of any cross cannizaro dispropotionation? Thank you for answering the particular example, but please let me know which of the cases(reasoning) is correct?

useless; he's not asking what canizzaro reaction is?

He modified the question after Ive already answered, I did not understand what he was refering too. You already answered, so thats it.

oh i see, sorry


text:
What is the temperature dependency of p/T in the ideal gas?
This is about the ideal gas equation $$pV=nRT$$ to plot ${p}\over{T}$ against $T$, where ${nR}\over{V}$ is constant. How would this graph look like?

So... what is the temperature dependence of (p/T), where (p/T) = (nR/V)?

I'm sorry, I don't know what 'temperature dependence' means, and it's not used at this level.

In other words, if you double T, what will be value of (p/T)?

A straight line parallel to temperature axis.
It will be of the form $y = c$,  where $c$ is a constant, mathematically.
Also if you take any value of $\frac{p}{T}$ it will always give same value, because $\frac{nR}{V}$ is a constant.

Firstly, rearrange the equation to get $\frac{p}{T}$ in the left hand side:
$$\frac{p}{T} = \frac{nR}{V}$$
As the question says, $\frac{nR}{V}$ is always a constant. So, $\frac{p}{T}$ is also a constant. In other words, I double the temperature, I double the pressure as well, still giving me $\frac{p}{T}$, to make sure $\frac{nR}{V}$ is always constant.
So if one were to draw graph, because is always one value, even if we change the value of $T$, this has to mean that a graph of $\left(\frac{p}{T}\right)$ against $(T)$, will give a straight line graph with gradient $0$, and intersecting the $y$ axis at magnitude $\left(\frac{nR}{V}\right)$, since $\frac{p}{T}$ remains constant.

See the following:
$$PV = nRT \implies \frac PT = \frac{nR}{V} = \text{const.}$$
So whatever you do with $T$, $P/T$ won't change, so it would be a constant horizontal line.


text:
Cation exchange capacity via IR spectroscopy? Is it as good as barium exchange? What is the intutition on how it works?
Cation exchange capacity is often regarded as the single best summary statistic for soil fertility.  Years ago, I learned how to measure it myself, in a soil lab.  I used compulsive barium exchange, whereby I knocked off all of the cations using barium, and then I somehow (I forget) got all the barium off, before measuring how much barium is in solution.  It makes sense as a means of measuring CEC.
I'm now in a situation where I need hundreds of soil samples analyzed, primarily for CEC.  The lab with the best price does the CEC measurements by IR spectroscopy.
My question:  how can you measure the potential of a soil to exchange cations using IR spectroscopy?  It's based on correlations, right?  Under what conditions does this method break down?  I assume the prediction error is non-constant over the multivariate space comprising soil properties.  In what regions of this space should I expect greater bias and/or variance in the prediction, and what would the sign of such bias be?
Or, in simpler terms, when do compulsive barium exchange and IR spectroscopy agree and when do they disagree?