text:
Why are partial molar Gibbs energy and Gibbs energy per mol equal?
I'm not a chemist, but I need some insight into chemical thermodynamics.
In the book that I'm reading (Nevers, 2012), equation (4.28) relates the chemical potential of a pure species to it's Gibbs energy per mol:
$\qquad\mu_a^{(1)} = \bigg( \dfrac{\partial G}{\partial n_a} \bigg)_{T,P,n_b...} = g_a^{(1)}$
where the subscript stands for the species $a$, the superscript stands for it's phase, $\mu$ is the chemical potential and $g$ is the Gibbs energy per mol.
I understand the first equal sign. Though, I don't understand the second equal sign. What am I missing?

Partial derivative of the Gibbs free energy with respect to the amount of the species, all other species' concentrations in the mixture remaining constant, and at constant temperature. When pressure is constant, chemical potential is the partial molar Gibbs free energy. At chemical equilibrium or in phase equilibrium the total sum of chemical potentials is zero, as the free energy is at a minimum. Which is also defined in book in Example 4.1. It's physical properties of substances we can only observe them and understand rules, but why it is so, is question for particle physics.

@Sigrlami I disagree with your “why it is so, is question for particle physics” — sebastian is asking why, given the basic principles and axioms of thermodynamics, the second equality holds. There is a good reason for that, as explained in Max's answer… while one has to beware of why questions in science, they're not all bad!

@F'x I agree. I find Sigrlami's comment rather confusing. Max' answer below is what I was looking for.

The key is that the Gibbs free energy is an extensive quantity. This means that for a single-component system it can be written as
$$ G\left(T,P,N\right) = N\cdot g\left(T,P\right) $$
where N is the number of particles and g is the Gibbs free energy per particle. In other words, if you double the number of particles and double the volume (at fixed temperature and pressure), then G will also double.
From the above expression you can see that the chemical potential is just the Gibb's free energy per particle:
$$\mu\left(T,P\right) = g\left(T,P\right)$$

Excellent and clear answer! You write it for pure phase, but the same reasoning extends to mixtures and thus answers sebastian's question.


text:
What does this scale mean?

$$\log_{10}~\mathrm{pg}/\mathrm{ml}$$
I do not understand the scale.  How do I convert this to a standard concentration (without the logs)?

It's just http://en.wikipedia.org/wiki/Logarithmic_scale you can easily switch to numbers by calculating log10 of number, pg/ml stay as is.

To expand upon the comments, the y axis in this graph is simply the log of the concentration in pg/ml.  To convert you use the equation $$C = 10^{y}$$ where y is the y value and C is the concentration.  Note that in older literature, log was oftentimes assumed to be base e whereas nowadays we assume base 10.  The figure (thankfully) clarifies this potential discrepancy.
A final note: log values are unitless so the y axis has no units.  The units are reported here so that you know what the appropriate units of the converted value should be.

source: Stack Exchange

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text:
What happens in the reaction of ammonium dichromate and mercury(II) thiocyanate?
I found this reaction of $\ce{NH4Cr2O7 + Hg(SCN)2 ->[\Delta]\ ?}$ on youtube, and I'm curious as to how it works. I tried searching online, but didn't find much on it.
Thanks.

This is a combination of two reactions.  The first is the ammonium dichromate vocano and the second is called Pharaoh's snakes.
The volcano
The reaction occurring here is: $$\ce{(NH4)2Cr2O7(s) -> Cr2O3(s) + N2(g) + 4 H2O(g)}$$  The reference above provides several pictures and a description of the chemistry.
Pharaoh's snakes
Again, the reference above describes what is happening, so only a summary of the chemical reactions is given here:
$$\ce{2 Hg(SCN)2->2 HgS + CS2 + C3N4}$$ $$\ce{CS2 + 3O2 -> CO2 + 2SO2}$$ $$\ce{2C3N4->3(CN)2 + N2}$$ $$\ce{HgS + O2->Hg + SO2}$$
Pharaoh's snakes is a rather hazardous demonstration and should not be performed without the proper ventilation, protective gear and waste disposal equipment.

Which $C_3N_4$ isomer forms preferentially? This list of carbon nitrides provides such options as beta carbon nitride, graphitic carbon nitride, dicyanodiazomethane, nitrogen tricyanide and carbon bis(cyanamide).

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text:
Counting valency of atoms, in a molecule with python
The documentation for the indigo module can be found here
http://ggasoftware.com/opensource/indigo/api#inputoutput
So for instance if I have a molecule object for some SMILES string, e.g. "[C](=[O])", I wish to calculate the valency of each atom, for instance here the desired output would be [atom=C, unbound_electrons=2],[atom=O, valency=0]
If I consider the atom "[C]"
Can anyone explain why is this code printing [atom=C, unbound_electrons=0] not [atom=C, unbound_electrons=4]
from indigo import *
indigo = Indigo()

mol=indigo.loadMolecule("[C]")

print(mol.grossFormula(),"\n")

for atom in mol.iterateAtoms():
        print([atom.symbol(),atom.radicalElectrons()])

EDIT: I could work it out if I could generate a list of the types of bonds on the atom in conjuction with atom.atomicNumber(). E.g. if I could say [C] has a double bond I could take it's atomic number - 2 (second shell) - 2 (double bond)
EDIT#2: This might be useful for visualising what i'm talking about
from indigo_renderer import *
renderer = IndigoRenderer(indigo)
renderer.renderToFile(mol,"mol.png")

EDIT#3: I am not a chemist, so might have got some concepts wrong

I was looking into this, and the results I was getting seemed very anomalous.  Consider this iteration from a single carbon atom at -4 through +4, this is the output I receive (python code at the bottom of the post):
[C+4]
C valence= 4 radicalE= 0 charge= 4 implicitH= 0
[C+3] |^1:0|
C valence= 4 radicalE= 1 charge= 3 implicitH= 0
[C+2] |^3:0|
C valence= 4 radicalE= 2 charge= 2 implicitH= 0
[C+]
C valence= 0 radicalE= 0 charge= 1 implicitH= 0
[C]
C valence= 0 radicalE= 0 charge= 0 implicitH= 0
[C-]
C valence= 0 radicalE= 0 charge= -1 implicitH= 0
[C-2] |^3:0|
C valence= 4 radicalE= 2 charge= -2 implicitH= 0
[C-3] |^1:0|
C valence= 4 radicalE= 1 charge= -3 implicitH= 0
[C-4]
C valence= 4 radicalE= 0 charge= -4 implicitH= 0

I printed them using the canonicalSmiles format.  After I realized something strange was going on (neutral, and +/- 1 charges were 0, but everything else was fine), I dug into SMILES documentation to find out that currently it appears only up to divalent radicals are supported.  Something like a lone carbon atom has technically a tetra-valent radical, so it defaults to nothing, because it cannot be interpreted by SMILES.
I pulled this from ChemAxon which explains what the notation after the atom means. I'm going to assume that |^1:0| is the notation for a mono-valent system.  The 0 indicates the index of the atom, which is always zero for this example.

Radical numbers
  Atom indexes with:
  - divalent radical center are written after "^2:"
  - divalent singlet radical center are written after "^3:"
  - divalent triplet radical center are written after "^4:"
  - trivalent radical center are written after "^5:"

And from the Indigo website itself it states under the supported extensions

Radical numbers: monovalent, divalent singlet, and divalent triplet

So, I don't think you've done anything wrong, you've just simply hit a limitation of the SMILES system and the interpreter of the indigo software.  Here is the code I used below:
from indigo import *
indigo = Indigo()

from indigo_renderer import *

renderer = IndigoRenderer(indigo)
mols = []
mols.append(indigo.loadMolecule("[C+4]"))
mols.append(indigo.loadMolecule("[C+3]"))
mols.append(indigo.loadMolecule("[C+2]"))
mols.append(indigo.loadMolecule("[C+1]"))
mols.append(indigo.loadMolecule("[C]"))
mols.append(indigo.loadMolecule("[C-]"))
mols.append(indigo.loadMolecule("[C-2]"))
mols.append(indigo.loadMolecule("[C-3]"))
mols.append(indigo.loadMolecule("[C-4]"))

i = 0
for mol in mols:
    print mol.canonicalSmiles()
    for atom in mol.iterateAtoms():
      print atom.symbol(),"valence=",atom.valence(),"radicalE=",atom.radicalElectrons(),"charge=",atom.charge(),"implicitH=",atom.countImplicitHydrogens()
    filename = 'mol'+str(i)+'.png'
    renderer.renderToFile(mol,filename)
    i += 1

Thanks so much for the level of detail here! This is extremely helpful. This confirms my worries, I've developed a work around that examines the bond network and counts bonds per atom. It's a bit messy and roundabout, but it seems to work!

I have never used the indigo framework, but there's more than one way to process SMILES strings in Python. Open Babel and Pybel might be an alternative:
import pybel

# SMILES for flurazepam, taken from
# http://chemspider.com/chemical-structure.3276
flurazepam = 'CCN(CC)CCN1C(=O)CN=C(C2=C1C=CC(=C2)Cl)C3=CC=CC=C3F'

# SMILES for trimethylamine, taken from
# http://chemspider.com/chemical-structure.1114
trimethylamine = 'CN(C)C'

# SMILES for nitromethane, taken from
# http://chemspider.com/chemical-structure.6135
nitromethane = 'C[N+](=O)[O-]'

mol = pybel.readstring('smi', flurazepam)

for atom in mol.atoms:
    print '{:<5} {:3} {:3} {:3}'.format(atom.type, atom.formalcharge, \
        atom.implicitvalence, atom.valence)


text:
Safe, easily detectable gas or aerosol for testing fume hoods
I've recently started undertaking a project to construct a small fume hood. So far I'm considering its use for soldering, 3d printing with ABS and polystyrene, and PCB etching using a muriatic acid/hydrogen peroxide solution, however I'd also like to leave my options open. I want to test the fume hood once completed - I've seen some videos that use dry ice for this purpose, and I figure I might also be able to test the hood by smell using bathroom deodorant, but I'm wondering what sort of concentration these would have to reach in order for me to detect them, and whether there are any other tests I might consider.

Smoke generation is preferred over smell, as the visual test is a bit more objective.  See this reference for more info, although $\ce{TiCl4}$ is not available for home use.  An anemometer might be of help as well.

@bobthechemist That's darn close to being a full answer, so feel free to add it if you'd like.

@16 If you have any doubt as to whether this is a safe setup, please consult a professional or at minimum someone with experience in this area.

Smoke generation is preferred over smell, as the visual test is a bit more objective. See this reference for more info, although $\ce{TiCl4}$is not available for home use. There are hobbyist alternatives such as smoke matches which will help you test the fume hood performance. (The above link is just an example, I did see better prices at less-known websites.)  An anemometer might be of help as well.

Why not use dry ice in water, as a cheap alternative to a fog machine?  Or rent a fog machine from the local party supply store?

@AbelFriedman certainly possible, depending on what one has available in the lab.


text:
How do I determine the crystal field splitting for an arbitrary point group?
How can I figure out the degeneracy of the d orbitals for a site that has a given point group? Specifically I'm interested in $D_{3d}$ and $D_{3h}$, but it would be good to know how to do it in the general case.

If you already know the symmetry of your site then it is quite easy. In a lot of books (e.g. this one) and on this web site you can find the character tables of the point groups supplemented with two additional columns which show the transformation properties of the basis vectors (e.g. $\ce{p}$ orbitals), their rotations and their quadratic combinations (e.g. $\ce{d}$ orbitals). In the following picture I highlighted the important parts in the character table of the $\ce{D_{3\mathrm{h}}}$ group.
So in the $\ce{D_{3\mathrm{h}}}$ group the $\ce{d_{z^2}}$ orbital transforms as the irreducible representation $A_{1}^{'}$ and the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals transform as the irreducible representation $\ce{E^{'}}$.

After you have identified the irreducible representations of the $\ce{d}$ orbitals you can tell their degeneracy by identifying which $\ce{d}$ orbitals belong to the same irreduzible representation. If two or three $\ce{d}$ orbitals belong to the same irreducible representation they are degenerate. In the case of the $\ce{D_{3\mathrm{h}}}$ group shown below this means that $\ce{d_{z^2}}$ forms an energy level of its own, while $\ce{d_{x^2-y^2}}$ and $\ce{d_{xy}}$ (both belonging to $\ce{E^{'}}$) are degenerate and $\ce{d_{xz}}$ and $\ce{d_{yz}}$ (both belonging to $\ce{E^{''}}$) are degenerate. So, your $\ce{d}$ orbitals split into three energy levels, two of which are doubly degenerate, when your site has $\ce{D_{3\mathrm{h}}}$ symmetry.

It wasn't asked for in the OP, but including some comment about how to determine the relative ordering as well might be helpful to future visitors.

@bobthechemist As far as I know it is not possible to determine an energetic ordering of the levels by symmetry alone. What is possible is to reason that, the more directly the orbital lobes of a certain $\ce{d}$ orbital point towards the ligands the higher is the respective $\ce{d}$ orbital's energy (though this method is not very exact). Do you mean that?

Yes.  I didn't expect that symmetry alone would be able to determine the energetic ordering.  Unless there's a more clever approach out there without resorting to calculations, visualizing the orbital and ligand positions seems to be the best approach.


text:
At what temperature will proteins and fats boil in a vacuum?
A question was asked on another stackexchange site:

This is not nice perspective, but eventually it will happen. An astronaut falls out of spaceship because of damage caused by collision with other object, or because of suit decontamination. The fluids from the body would evaporate, and if any bacteria would survive, than only as spores. Does it mean the perfect mummification of the body? Or there will be some decay, caused by enzymes from damaged cells, for example?

Given that the water in the body will outgass taking with it much of the rest of the body, are there any cited sources that specify the pressure at which proteins, fats and bones will break down and become vapour?

You're talking about the human body being exposed to outer space and asking about what would happen to it's internal structure. Bones, fats and proteins are within the enclosed system that is the human body, they are not subjected to the low pressures of space. The answer to the pressure at which proteins, fats and bones will break down and become vapour is not related to the answer to what would happen to them in the human body if the body is exposed to the vacuum of space. Nevertheless, I'll try to provide both answers.
The answer to the second question (short-term) can be found on NASA's website at "ask an astrophysicist". It's actually quite surprising for someone used to sci-fi movies:

You do not explode. Your blood does not boil. You do not freeze. You do not instantly lose consciousness.

The question you linked talks about "decay", or the effects of long time exposure. That's hard to predict, but at some point the body would be attracted by a planet and collide and you wouldn't have anything to boil anymore. Now to the chemistry part of it:
There isn't much (if any) data on the boiling point or vapor pressure of the compounds you're asking - it's hard to reproduce and it's more a curiosity than anything else. So let's talk about what would happen if you have fat, bones and proteins directly exposed to outer space pressures and temperatures, which seems to be your point anyway. Temperatures in space go from -270°C (in outer space, unexposed to sunlight) to +120°C (near the earth, exposed to sunlight). Since you're asking about boiling point let's talk about what will happen at the highest temperature.
Let's start with the easier part - the bones. It is safe to assume that the inorganic components will be tougher than the organic ones. The inorganic part of bones is mostly a specific type of apatite, which is a rock. It can be found in moon rocks. It will not boil. So at least the inorganic part of the bone would remain somewhat intact.
Fat is made of triglycerides. It'll probably break down to glycerol and the correspondent acid. Then it would decompose and boil.
Proteins denature at harsh conditions, you'll be left with amino acids at some point. It's hard to tell what would happen since there isn't much data on it, but given the harsh conditions, which include exposure to all sorts of radiations, the amino acids certainly wouldn't last long and at some point you'd be left with more basic molecules that don't resemble a protein anymore. So what can we conclude? Exposed as a whole, the internal structure of the human body would remain somewhat intact for a while (physically speaking). It's hard to predict what happens then, but given that the outer space is such a hostile environment, it's unlikely that the body would last long. Exposed as individual components they would last much, much less. From bones, fat and proteins, it's likely that only the bones stand a chance.


text:
What are the oxidation states of sulfur in the tetrathionate ion?
What are the oxidation states of the sulfur in the tetrathionate ion $\ce{S4O6^{2-}}$?

Consider the structure of tetrathionate. The two central sulfurs each have two lone pairs and are assigned half of the electrons from the two bonds they make, since the electrons of bonds between atoms of the same element must be distributed evenly (due to there being, by definition, an electronegativity difference of zero between two atoms of the same element). A neutral sulfur atom has six valence electrons, so the oxidation state of the central sulfurs can be calculated as follows:
$$6 - 4 - \frac{1}{2}(4) = 0$$
That is, six electrons in neutral sulfur, minus four from the lone pairs, minus half of the four sulfur-sulfur bonding electrons, gives zero.
The terminal sulfurs, on the other hand, have no lone pairs, and all the electrons from the sulfur-oxygen bonds are assigned to oxygen, since it's the more electronegative element. The only electrons assigned to those sulfurs are half of those from the sulfur-sulfur single bonds. Hence, their oxidation state is:
$$6 - \frac{1}{2}(2) = +5$$
That is, six electrons in neutral sulfur, minus half of the two sulfur-sulfur bonding electrons, giving an oxidation state of +5.

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text:
Cathodic protection of active metals
Cathodic protection prevents the removal of positive form the metal surface by applying a negative charge to the metal.
Is is possible to prevent the reaction of an active metal (i.e. sodium) with water by aggressively forcing current into the metal using a DC source? We allow the "water" to be a dilute solution (~1%) of the metal hydroxide. There is also an anode that completes the circuit but its OK if the anode disintegrates.
Edit: I will let the answer stand but it begs one question: If the power supply forces the sodium metal to be at a strongly positive potential, it may end up coated in layer(s) of hydroxide (or even oxide?), which may prevent the water from reaching it. But if the metal is liquid (NaK) then it may cause the metal to have "negative surface tension" making it impossible to maintain a high potential as the capacitance runs away to "infinity". So it's not completely settled.

I guess the mighty power source may boost the fire after the hydrogen explosion. Just for curiosity, Czech chemists discovered recently alkali metals ignite hydrogen by microsparks caused by charge disbalance during the reaction with water.

It is unlikely that a potential applied to a sodium electrode in water would prevent the spontaneous oxidation of sodium by water.  Consider the following reduction potentials:
$$\ce{2Na+ + 2e- -> 2Na(s)}\ E^0=-2.71\ \mathrm V\tag{1}$$
$$\ce{2 H2O + 2e- ->H2 + 2 OH-}\ E^0=-0.83\ \mathrm V\tag{2}$$
The reduction of water will always occur before the reduction of sodium ions.  If we reverse equation (1) so we can combine the two and get the full redox equation for your system, we get:
$$\ce{2 Na(s) + 2 H2O <=> 2Na+(aq) + H2 + 2 OH-}\ E^0= 1.88\ \mathrm V\tag{3} $$
A positive potential indicates a spontaneous reaction under the given conditions so thermodynamically, it will be near impossible to prevent the reaction from being spontaneous.
How impossible?  We can do a thought experiment and ask the question at what pH would this reaction be non spontaneous? In other words, when would the Nernst Equation for this system evaluate to zero:
$$E=E^0-\frac{RT}{nF}\ln\frac{\text{Products}}{\text{Reactants}}\tag{4}$$
Entering our values, converting Ln to Log and substituting $\mathrm{pH} = -\log[\ce{H}]$ (and assuming all other concentrations are fixed at 1 M or 1 atm):
$$E=1.88+0.059(14-\mathrm{pH})\tag{5}$$
The reaction would not be spontaneous until the pH reached 45 units.  Something that can't be done in water.

Thanks. Metals above 0.83 V can't be protected at neutral pH, and anything above ~1.2-1.3V can't be protected even by both altering the pH of the water and cathodic protection.

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text:
Would dissolving a salt and letting it sit for 48 hours have an effect on the solution?
If one were to dissolve magnesium citrate (about $100~\mathrm{mg}$) in water (about $70~\mathrm{ml}$) and let the solution sit for 48 hours, would the solution be exactly the same as one dissolved and consumed right away?
How does one explain the difference in taste from a 48-hour solution and a 5-minute solution?

The only reactions that I would expect to occur to any significant extent with pure water are acid-base reactions between the citrate and water, where water, being amphiprotic, could act as either an acid or a base:
$$\ce{C6H6O7^2- + H2O <=> C6H7O7- + OH-}$$
$$\ce{C6H6O7^2- + H2O <=> C6H5O7^3- + H3O+}$$
Citrate, being the conjugate base of a weak acid, is itself a weak base (according to the citric acid Wikipedia page, the $\mathrm pK_\mathrm a$ for the dissociation of the second proton is 4.75, and for the third between 5.41 and 6.40). Citrate is tripotic, and in the magnesium salt citrate possesses one undissociated acidic proton, which makes it an amphiprotic molecule as well. Given that the $\mathrm pK_\mathrm a$ of water is approximately 15.7, and the $\mathrm pK_\mathrm a$ of hydronium ($\ce{H3O+}$) is −1.74, both reactions above favor the reactants over the products by many orders of magnitude, though the second reaction is favored above the first (i.e., the dibasic magnesium citrate salt should still be very slightly acidic, overall).
Acid-base reactions typically reach equilibrium quite quickly, so there should be very little change in pH and no pH-attributable differences in taste over time. If any change does occur, I would expect it to be the result of a change in concentration as the water gradually evaporates. Another possibility is that, if the water is impure (which tap water certainly is), additional side-reactions might occur. What those are would depend on the exact composition of the water as well as environmental conditions. It's possible for the magnesium to form a precipitate, for example, in combination with other anions dissolved in the water (unlikely given the quantity in your example, but possible). If the magnesium citrate is part of a commercial product containing other ingredients, then it may well be that other agents are responsible for any effects you observe.

The difference in taste is due to a difference in molarity of the magnesium citrate.
Its solubility in water is 20 g/100ml (http://en.wikipedia.org/wiki/Magnesium_citrate)
Over a 48-hour period a significant amount of water will evaporate from the container meaning the concentration will increase. Based on the numbers provided it is unlikely that the concentration will get the the point of precipitation; the solution will visually be the same.
Evaporation is dependent on surface area so the larger the surface area of exposed water the more profound the difference in taste will be.

Would keeping the container closed (to avoid evaporation) still have an effect on taste?

Following the logic presented in my answer yes. Preventing evaporation would prevent a change in the molarity of the solution. It there is some type of chemical  decomposition of an actual molecule covering the solution would have no effect.


text:
(Organic Chemistry) Common Atoms in Organic Molecules
What are 6 atoms commonly found in organic molecules?

Could you elaborate more on your question? If you provide more context, you will get better answers.

Carbon, Hydrogen, Nitrogen, Oxygen, Phosphorus, and Sulfur. Best way to remember this is CHNOPS.
http://wiki.answers.com/Q/The_six_most_common_elements_in_organic_molecules_are


text:
Exothermic or Endothermic reactions
Can we determine whether a reaction is endothermic or exothermic? For example if we are given the following reactions:
$$\ce{2SO2 + O2 -> 2SO3}$$
$$\ce{H2 + I2 ->2HI}$$
$$\ce{N2 + 3H2 -> 2NH3}$$
Can we determine weather these reactions are endothermic or exothermic or do we have to just memorize this? Are there any general trends that could make memorizing this information more manageable?

In general just by looking at these equations there is no surefire way to know which is endo or exothermic. However you can look at the bond strength or the ∆H of formation for each molecule. In short you have to know something.

The energy stored in the molecules(specifically in the bonds) is lower before the reaction in an endothermic reaction and higher in an exothermic reaction. This means if you know relatively how strong the bonds or how much energy it takes to form each of the molecules then you can know whether the reaction is endo or exothermic.
If you have a few ∆H of formation values you can use Hess's Law

As anglinb pointed out, the only completely reliable method of determining whether the reactions are endo- or exothermic is by calculation. That said, certain generalizations may allow reasonable predictions about changes in thermodynamic parameters to be made:

Reactions that produce a net increase in $\sigma$-bonds between particular pairs of atoms that were $\pi$-bonded in the reactants tend to be exothermic, since $\sigma$-bonds tend to be more stable than $\pi$-bonds. Conversely, reactions in which the generation of $\pi$-bonds and reduction in number of equivalent $\sigma$-bonds occurs tend to be endothermic.
Reactions that involve a net reduction in the total number of molecules of products as compared to reactants, and/or involve the generation of products in more highly ordered states (for example, the generation of solids from liquids or gases), tend to occur with a reduction in entropy. Reactions that involve the generation of heteroatomic compounds from homoatomic substances tend to increase entropy. The spontaneity of a reaction is governed by the equation $\Delta G = \Delta H - T\Delta S$, where $\Delta G < 0$ for a spontaneous process. Therefore, if a reaction occurs spontaneously while the entropy change ($\Delta S$) is negative, the reaction must be exothermic by default; an endothermic process with negative entropy change cannot be spontaneous.

Remember that these are very rough guidelines, and ultimately only performing calculations with concrete values for the relevant thermodynamic parameters can be totally definitive. The first and third reactions from your question can be predicted to have negative $\Delta S$ by the above guidelines, and therefore must be exothermic (if they're supposed to be spontaneous at all). Crunching the numbers using thermodynamic data does in fact confirm that.

In the first guideline, you have mentioned that if there is a net increase in the sigma bonds between particular pairs of atoms that were pi bonded in the reactants then reactions tend to be exothermic. But to break pi bond, we will need to provide energy so, shouldn't the reaction be endothermic then?

Yes, some energy is required to initially break the bonds (which is factored into the energy of activation), but the formation of bonds always releases energy; the only question is whether that energy released upon completion is greater than the initial input of energy required to set the reaction in  motion. When a sigma bond between some generic pair of atoms forms, it's typical that more energy is released than would be required to break a pi bond between the same two atoms, so the reaction is exothermic overall.


text:
Bond of interest
What does it mean for our bond of interest if we have low bond length and high frequency? Trying to make this connection but am not understanding. If anyone could explain it intuitively, that would be much appreciated. Say we have the bond of interest of heptane which is C-C. What is the variable that is changing (bond order, atom mass, etc.)? Also if the IR frequency is 1400, what is the effect of variable? Also what does this say about the physical bonding in the molecular bonds of interest?

Are you referring to vibrational (IR) spectroscopy?  Your question (and its title) are a bit unclear.  Please revise and include an example of the phenomenon you are trying to understand.

Can you narrow down or more exactly specify the parameters of your question? I presume by the reference to frequency you're referring to the vibrational frequency of bonds in spectroscopy?

Is this much more clear?

In IR, the vibrational frequencies of a simple diatomic molecule can be described by the following equation:
$$
\bar\nu = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}
$$
Here, $k$ is the force constant, $\mu$ is the reduced mass of the molecule, and $c$ is the speed of light (which results in appropriate units for reporting the wavenumber, rather than ordinary units of frequency, e.g., Hz.). This equation is derived from treating the molecule as a simple harmonic oscillator, using Hooke's law, Newton's laws of motion, and the classic technique of reducing a two-body problem to a one-body problem.*
The calculation of the force constant is complex, but suffice to say that it can be viewed as being roughly proportional to the strength of the bond. Since the wavenumber is proportional to the square root of the force constant it can be said that, ceteris paribus, stronger bonds tend to yield higher wavenumbers. Since higher bond multiplicity and shorter bond length correlate positively to bond strength, those factors can also be correlated to higher wavenumbers. (There are also positive correlations between metrics such as, e.g., bond energies and wavenumber, etc.) These factors can be used to rationalize the trend that wavenumbers of triple bonds tend to be greater than those of double bonds, which in turn tend to be greater than those of single bonds. Similarly, the diminished wavenumber of conjugated bonds vs. analogous unconjugated bonds can be explained by the bonds of conjugated systems having lengths intermediate between single and double bonds. One source indicates the following approximate proportionality relationship between bond length, $r$, and the force constant, $k$:
$$
k \propto \frac{1}{r^6}
$$
Clearly, even minute differences in bond length produce large variance in the force constant.
Conversely, wavenumber is inversely proportional to the square root of the reduced mass. Hence, as reduced mass grows, wavenumbers decrease. This largely explains, for example, why the highest wavenumbers are those of bonds between hydrogen and some other element, since hydrogen is the lightest element.
*I think the math involved is probably well beyond the necessary scope of this answer, so I won't elaborate on it here. I dealt with these topics in introductory physics and differential equations classes. McQuarrie's physical chemistry textbook specifically deals with the harmonic oscillator model vis-a-vis spectroscopy, and I can recommend it for anybody interested in further details.

To add on to Greg's well-written answer (since this is too long for a comment) we can write Hooke's Law in a slightly more practical form for purposes of predicting IR stretching frequencies:$$\bar\nu (cm^{-1}) = 4.12\sqrt{\frac{k}{\mu}}$$ with $k = 5, 10 \text{, or } 15$ $\times10^5$ dynes/cm for a single, double or triple bond.  For example, a $\ce{C=C}$ bond: k = $10 \times 10^5$, $\mu=\frac{(12)(12)}{12+12}=6$ giving $\bar\nu=1682 cm^{-1}$ where the experimental value is $1650$.  We can leave it to the OP to see that a $\ce{C-H}$ bond has a calculated frequency of $3032 cm^{-1}$ (experimental is $3000$) and a $\ce{C-D}$ bond has a calculated frequency of $2228 cm^{-1}$ (experimental $2206$).

+1, a very nice addition. I'm sure I've seen that formulation before, but can you explain briefly the significance of the 4.12 factor?

@GregE. $\frac{\sqrt{N}}{2\pi c}$ with speed of light in cm/s and N = Avogadros number so that the values used in calculating the reduced mass can be in AMUs.

Thanks. I assumed 1/(2pi*c) would be factored in but it hadn't occurred to me to think of conversion to amus.


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How does toluene react at higher temperatures and why?

Reaction of toluene at high temperatures gives us (o-m)toluene; whereas at normal conditions of electrophilic attack, it gives us (o-p) directing.

Why does this happen at high temp although as methyl group is activating and o-p directing?

I could imagine that at high temperatures the higher activation energy for reacting at the meta position could be overcome by the higher number of meta positions (2) available for the reaction compared to the para position (only 1). But that is just a guess...

What specific reaction are you talking about?'

Alkylation at high temp of toluene

Assuming you're talking about Friedel-Crafts alkylation of toluene, this can be explained by the reaction being reversible and thermodynamic control taking precedence at sufficiently high temperatures. The directing ability of substituents in electrophilic aromatic substitution reactions is a function of the reaction being under kinetic control at lower temperatures. According to Fleming's book on MO theory, a single electron-donating substituent on benzene conjugated to the ring's $\pi$ system disrupts the normal degeneracy of the $\psi_2$ and $\psi_3$ molecular orbitals of unsubstituted benzene. Comparing the intermediates of an EDG-substituted benzene undergoing electrophilic attack, the HOMO of the $\pi$ system, $\psi_3$, is higher in energy when attack occurs at the meta- position than when it occurs at the ortho- or para- positions. Using the concept of resonance, this can be illustrated in an abstract sense by considering the resonance structures of the intermediates resulting from attack at the various positions; an additional resonance structure exists for ortho- and para- attack if the substituent is conjugated to the $\pi$ electron system of the ring.
With alkyl groups, the electrons are donated by hyperconjugation, which is a weaker effect (and not apparent from resonance structures), but nonetheless has a similar impact on regioselectivity and the $\pi$ molecular orbitals. Additionally, the coefficients on the atoms of the HOMO of the $\pi$ system are lowest at the meta positions when an electron donating group is present (including alkyl groups), indicating electron density is higher at the ortho- and para- positions.
(I should clarify, of course, that resonance is not an actual physical process/phenomenon, merely an abstraction which provides a qualitative picture of the extent of electron delocalization and which often correlates well with the actual distribution of electron density in the molecular orbitals.)
Ultimately, at low temperatures, when the reaction is kinetically controlled, the favored pathways are the ones in which the transition states and intermediates are stabilized, leading to ortho-/para- substitution.
At higher temperatures, the preferred product is the thermodynamic one, and the advantage of meta- substitution is that there is less crowding and reduced steric hindrance in, e.g., a 1,3,5-trisubstituted benzene by comparison to a 1,2,4-trisubstituted benzene. Additionally, Friedel-Crafts is actually reversible at high temperatures. Even if the kinetic product forms initially, then it can decompose by dealkylation in the presence of Lewis-Acid catalysts and acid at high temperatures, at which point equilibrium can be established with the thermodynamic product.
According to March, experimental evidence suggests that methyl groups will mainly migrate intramolecularly in the isomerization reaction to give the thermodynamically preferred meta-substituted benzenes, while both intramolecular rearrangements and complete dealkylation with subsequent intermolecular reactions are possible with other types of alkyl groups.

Why is meta substituted benzene thermodynamically preferred product in reactions like friedel crafts and sulphonation nitration?


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What is the heat energy required to perform electrolysis equivalent?
The question may be off-topic here. If my concept is flawed, please poke me in the rib (+:
Sources on the internet indicate a minimum 1.2Volts is adequate for electrolysis of water; nominally 6V. Applying a DC source to water provides adequate energy to break the hydrogen bond. Apparently some heat is also released in the process..
Say, one were to collect 100ml (an arbitrary figure) oxygen by electrolysis applying 6V DC to 1 Litre of water.
Can the energy to break the hydrogen bond be provided by applying heat to water? All other things being identical, how many calories would be required to collect the same amount of oxygen from the same volume of water?

I won't do any rib poking, but there is some flawed thinking here.  Certainly not off topic though.
Nomenclature: First, when you pass a current through water, you are not breaking the hydrogen bonds, which are intermolecular forces, but (a) breaking the bond between hydrogen and oxygen and (b) forming molecular oxygen and molecular hydrogen, $\ce{2H2O -> 2H2 + O2}$, and this reaction can be broken down into two half reactions: $$\ce{4H2O + 4 e- -> 2H2 + 4OH-}$$ $$\ce{2H2O -> O2 + 4H+ + 4e-}$$
Potentials: The 1.2 V you refer to likely arises from the standard reduction potential of oxygen, which is the reverse of the second reaction above.  I won't go in to the thermodynamics here, but suffice it to say that a brief look at a standard reduction potential table indicates that the reduction of water (first equation) has a potential of -0.83 V and the oxidation of water has a potential of -1.23 V for a grand total of ~ -2 V. (Note, these values are pH dependent and standard reduction potential tables report values at a pH of 0.)
Electrochemistry When you apply an overpotential to a system (the potential beyond what is needed to get the oxidation/reduction reactions to go forward) then two things will happen (a) the reaction will go faster and (b) heat will be generated.  This heat would be analogous to heat due to friction in a physical system and is considered wasted energy.
Thermochemistry Finally, we get to the answer to your question, which is "probably not, but the way you asked it, yes"  (hunh?)  Recall the nomenclature issue above.  Adding heat to water will break the hydrogen bonds in water eventually leading to the water changing from a liquid to a gaseous state.  However, adding heat to water would not result in the oxidation/reduction reactions outlined above because an important ingredient is missing: electrons.


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To model early earth chemical reactions which types of reactions should be included?
I'm currently working on building a simulator for the early earth to examine the spontaneous occurrence of food generated autocatalytic sets. No one in my group is a chemist, so the bottle neck for the project is implementing some sort of realistic chemistry in our model.
We are considering using the software outlined in this paper:
http://arxiv.org/abs/1304.1356
This will allow us to create a reaction network from an initial set, based on a set of general rules. The paper demonstrates well how the formose reactions are executed, which we intend to include. What other reaction groups would it be important to include? Our initial set of molecules would be comprised of the food molecules for each set of reaction rules included, we don't intend to start from H,O,N,C for example, so fundamental chemical rules should not be too crucial, as they should be generalised into the major reaction groups we will use.

The idea is questionable at best. Too much work to do.

For the reactions look up Stanley Miller's experiment. The experiment used water (H2O), methane (CH4), ammonia (NH3), and hydrogen (H2). I think you should start with that.

The formose reaction is one of the proposed pathways that carbohydrates could form. Two equivalents of formaldehyde ($\ce{CH2O}$) react to form glycoaldehyde ($\ce{HOCH2CHO}$), which reacts with further equivalents of formaledehyde or gycloaldehyde in aldol condensations to produce the 3-, 4-, 5-, and 6-carbon sugars.
$$\ce{CH2O + CH2O -> HOCH2CHO}$$
$$\ce{CH2O + HOCH2CHO -> HOCH2CH(OH)CHO}$$
$$\ce{HOCH2CHO + HOCH2CHO -> HOCH2CH(OH)CH(OH)CHO}$$
Formaldehyde can be produced from hydrogen and carbon monoxide or dioxide, themselves the products of steam reforming of methane and the water-gas shift reaction.
Steam-reformation:
$$\ce{CH4 + H2O <=> CO + 3H2}$$
Water-gas shift:
$$\ce{CO + H2O <=> CO2 + H2}$$
Fomaldehyde synthesis:
$$\ce{H2 + CO <=> H2CO}$$
Once you have the sugars, the carbonyl group and the alcohols give you the option to bring all of organic chemistry to bear and convert the sugars into anything you want through substitutions, additions, eliminations, and redox. Consider the Chichibabin pyridine synthesis, which can operate with as simple molecules as acetaldehyde ($\ce{CH3CHO}$), formaldehyde, and ammonia:
$$\ce{H2CO + CH3CHO <=> H2C=CHCHO + H2O}$$
$$\ce{H2C=CHCHO + CH3CHO + NH3 + 1/2 O2 <=> C5H5N + 3H2O}$$


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Less maintenance-dependent continuous pH testing
I am creating an automated chemical component mixing system (DIY project). One of the features required is continuous monitoring of pH value in my main solution. I do not need it to be very precise, precision of 0.2 would be enough. Also, I want it to be as low maintenance as possible (cleaning the probe once a month would be great). pH of the solution will vary from $pH = 4$ to $pH = 7$.
So, the question is, which of this testing ways will require less maintenance of pH probe:

Constant submersion in the solution.
Once per hour dipping in the solution.

The most common approach is to measure periodically.
pH glass electrodes (the most common ones) need to be "wet" in order to be used, so if you are going to remove it from the solution every hour you'll need to keep it submerged in the appropriate solution. Not only that, you need to wash it (preferably with the most pure water you can get) between measurements. Read section 2.2 of this monograph to understand why it needs to be "wet" and what solution to store it in. There's information on a lot of types of electrodes. This explains how to wash it. Another problem you might face with continuous measuring is not the electrode, but the meter. From that monograph:

The condition of the pH measuring assembly is subject to fluctuations over the course of time. Depending on the accuracy demands, a pH calibration should therefore be performed periodically, e.g. once a day.

You'd have to test and see what accuracy you get with your equipment.


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Does an onion completely remove "fresh" paint VOCs and other chemicals emitted?
I have read that paint contains VOCs(Volatile Organic Compounds), and solvents generally, and I could wait for them to dissipate, but i've read there are ways to remove the smell of paint such as half an onion in a bucket of water, such as slices of lemon on a plate, such as small bowls of distilled white vinegar.
Do these techniques actually remove all the nasty elements from the paint fumes, or just some of the nasty elements from them?  Do they remove all the paint fumes?
I wouldn't want to remove the smell but leave the VOC or other bad elements there.  If that's the case i'd rather just leave a window open in the room, closing it periodically, seeing if there's finally a time when the paint smell doesn't build up when the window is closed, then i'll know the paint isn't emitting any more fumes.

Would a lighted candle work?

@fp  it'd be unattended so probably not such a good option but if it's much more effective than e.g. lemons, then i'd consider it.

I've never heard of onions, lemons or vinegar removing a paint smell, but I would guess that they'd just cover up the unpleasant odor with something nicer.
As to your later comment:

I wouldn't want to remove the smell but leave the VOC or other bad elements there.

VOC stands for "volatile organic compound." (See the Wikipedia page on VOCs) Volatile compounds are ones that evaporate easily. Technically, the smell of that onion (or lemon, or vinegar) is a VOC, just not a particularly harmful one. (Not knowing the particular chemicals that make up a onion's scent, I can't say for sure that it's a volatile organic compound, but it's a pretty good bet.)
That fresh paint smell comes from VOCs (and potentially from other volatile compounds), so by the time the smell is gone, any VOCs you couldn't smell are likely gone too.

VOCs, as BLHaas said, are "volatile organic compounds"; that is, compounds that have a high vapor pressure and so readily evaporate into the surrounding air, and are mainly hydrogen-oxygen-carbon. The ones we smell and identify as being VOCs are most often "aromatics", which have one or more six-carbon rings (phenyl/benzyl, napthyl, etc). These aren't all bad; the smell of cooking meat, of flowers, herbs, fresh-cut grass, lemons, and yes even onions, are all VOCs. The VOCs in paint are what you smell, and also what are bad for you (at least some of them), so theoretically, getting rid of the smell means you've gotten rid of the problem.
The VOCs typically associated with latex and oil paints are mineral spirits, typically an amalgam of 7- to 12-carbon hydrocarbons produced by petroleum refinement. Oil paints require a high percentage of these in order to properly dissolve the pigment and achieve the desired consistency. Latex paints are based on an acrylic resin, which for many decades now has required relatively low VOC content in favor of a simple water base. However, acrylic acid, the basis of the acrylic resins holding the pigments and binders of the paint, can easily degrade into another example of a stinky organic chemical class; aldehydes. Aldehydes are caused by hydrolysis and subsequent hydrogenation of carboxylates, of which acrylics are a member. The simplest one is formaldehyde, and it's not great news; it's carcinogenic and acutely toxic in sufficient concentrations (higher than those in the average freshly painted room). Aldehydes also have a relatively low odor threshold; humans can detect aldehydes in concentrations measured in the low parts per billion. So it doesn't take much to stink up a room.
To actually answer your question, the most common trick I could find was to cut an onion in half and place each half, cut-side-up, in opposite corners of the room. The chemistry behind that is apparently inherent in the active ingredient of onions that irritates your eyes when you cut them. This ingredient is produced by enzymatic alteration of amino acid sulfoxides, and is called Syn-propanethial-S-oxide. Notice the "thial" portion of the name. Thials are to aldehydes as thiols are to alcohols; the switch from an oxygen to a sulfur, as with thiols, typically increases a thial's reactivity and lowers its boiling point compared to an aldehyde. So, you have a highly reactive sulfur-based chemical that yearns to evaporate into the air and irritate your mucous membranes.
Another thing it will apparently do, interestingly enough, is undergo a dipolar cycloaddition reaction with aldehydes[1], like the ones produced by evaporation of impurities in acrylic paint. This rearranges the thial and the aldehyde into a combined substance with a five-membered, oxygenated ring, which ostensibly does not offend the senses as much as either precursor. Because aldehydes are offensive in such low concentrations, it similarly doesn't take much Syn-propanethial-S-oxide to remove them and reduce the level below the odor threshold.

so onions will make the room safe and remove the odour by addressing the cause(aldehydes) rather than by masking anything.  Is it only Aldehydes that are toxic in the paint fumes?  If not, what are the other things that are toxic or smell and will onions remove them too?

Acetone is in paint and so are benzene and toluene. All of these are toxic but the acetone is less so since it naturally occurs in our blood.

This answer might go some way towards providing information that might help in answering the question, but I don't think it answers it.

For me, this answers the question of “does onion neutralize formaldehyde”. If formaldehyde is an aldehyde, then it’ll react with (= be neutralized by) onion’s evaporated thial, and if the problematic concentration of aldehydes is lower than one of thials then it might be a viable tradeoff. However, generic tVOC monitor will probably not notice a difference and the question then would be is onion’s outgassing also harmful.

Anecdotally, a few cut onions seemingly reduced the stark chemical smell upon entering a flat with hardwood flooring much more than limited ventilation means available, and it doesn’t feel like it’s simply the matter of overpowering one smell with another (as would be the case with, for example, incense sticks—which would additionally introduce a whole lot of PM into the air). However, it’s an ongoing experiment.


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Material that is flexible when cold but becomes brittle when hot
I'm looking for a material that I can use to make a mesh netting.  It should be flexible and have reasonable tensile strength when cold but brittle when hot (>$\pu{400^\circ F}$).  I want to use it to contain charcoal when cold but disintegrate when charcoal is burning and some force is applied to it.
Any ideas?

Does Rope work? (If by brittle you don't mind burns to ash.)

The usual solution to the problem it sounds like you're trying to solve is thick and nested paper bags -- they'll store the charcoal while not on fire, and then burn with the charcoal once you set it on fire. Plus, if you want, you can soak the inner paper and the top in wax to make the whole bag a bit easier to light.

But will it maintain its shape until coals are ashed over?

@RodrigodeAzevedo I rejected your edit because this isn't the first time you haven't made a complete edit. Tag-only edits are fine as long as there's nothing else in the post to be changed. Please address all issues in a post.

It's indeed not common, as the brittle/ductile transition is usually the other way around. You can, however, find counter-examples (in a limited range of temperature) by searching for “inverse brittle-to-ductile transition”.

Clay or ceramic materials. Not flexible, but worth a try. It has to be completely dry before heating or it can explode on heating projecting pieces everywhere.


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"Drinking water" plastic - can I store beer in this?
This is an outdoors & chemistry question.
I have an old and functional water bladder, shown below. On it, there are markings that coffee, beer, wine, hot drinks etc. must NOT be stored inside, only water.
While I realize that wine is acidic, damaging the bladder and becoming poisonous, beer seems to me quite close to water. Furthermore, I indend to test it with weisbeer, which does not contain such a large amount of gases as to explode.
Would be safe for me to store non-fizzy beer for up to a day every day in a "do-not-store-beer-inside" water bottle and then safely drink it?

You probably will need to specify the type of plastic used (usually it's embossed on the body)

In order to be able to give useful and reliable advice, we really need to know the type of polymer that the container is made from. If it's manufactured in the US, then there should be a triangular recycling symbol molded into the plastic. The number in that symbol corresponds to the plastic used in the container. As an aside, if you're concerned about acidity, beer is also typically more acidic than water and contains a variety of phenols and carboxylic acids. It's entirely possible that those compounds could accelerate the degradation of the plastic and leech unhealthy substances.

Unfortunately, no indication of the used plastic is indicated on the case, and I have thrown away the packaging. Please give me general/worst case advice. Should I discover the used plastic, I shall post a new, more concrete question.

I doubt that this is a chemistry problem. Food safe plastics which can hold wine and coffee are so abundant and cheep, there is no reason that they went to the trouble of finding one which is not suitable for these, but suitable for water, and made the bladder out of it.
All beverages beside water are a good growing medium for one microorganism or another. And the bladder is hard to get clean, the sucking straw even more so. So, it looks to me that they are warning you to not put in anything beside water because you'd risk starting your own bacterial or fungal colony. Beer is as bad as other drinks - don't store it there.

From the description, the plastic will capture volatile compounds and you will never eliminate the smell of beer and the alcohol could disolve the plastic a bit and give the beer a plastic smell/taste.
Not a good idea.

Proposed experiment: I fill the thing with beer and keep it 48 hours. Then taste the beer and if it has changed taste, then it is probably poisonous. If it has not changed taste noticeably, it is not poisonous for sure. Correct?

@Vorac, not correct. You're operating under false assumptions: (a) the chemicals will be detectable by taste, when in fact they could be tasteless below some concentration; (b) if noticeable degradation doesn't occur within 48 hours it won't occur over longer time scales, which is inaccurate. If it's a polymer that degrades by acid-catalyzed hydrolysis, that can be quite a slow process. It strikes me as improbable that anything significantly toxic would leech out over short time scales, but, given that the product specifically warns against such use, why risk it?


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Reaction with carbon
Usually maximum time we see then after cooking by gas there is no black spot on the pot. But continually burning can affect the pot surface and makes black powder.
My question is:

What is the most acceptable reason that black powder(carbon) take places on the pot surface. Is it related to the burning temperature? or fuel reacts with the pot surface?

Please clarify your question with an example.  Is the pot empty?

Will the answer differ for empty and filled pot? how?

Your question is not clear - are you burning something that is in the pot, or are you concerned with the flame from the stove forming soot on the bottom of the pot?

I'm concerning with that, how black spot is on the pot. What reaction behind this.

See this question

For combustion to occur you need 3 things at the same place: fuel, oxygen and heat.
If the air intake is dirty or not big enough, you get a flame that has an orange color, and on a cooler surface deposits C before it has a chance to burn.
The reaction with the pot is only a physical one of heat transfer and maybe C deposition.

Can you write some carbon depositions reactions which incorporated with the pot?

Temperature is not a thing in the sense that it is a substance like fuel or oxygen; it is a measure of thermal energy.  It is more syntactically correct to say fuel, oxygen and heat.

One possibility for propane: $\ce{C3H8 + 4O2⟶ 4H2O + 2CO2 + C}$ but there are different mixes for the gas used for cooking. @bobthechemist: you're right.


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How does alum improve the settling of particles to produce 'clean' water?
How does the addition of alum — potassium aluminium sulfate or $\ce{KAl(SO4)2}$ — help settle the dust particles in water?
There was such a statement in my Year 9 textbook, and so I was wondering how exactly alum does it.
Edit: I have posted an answer below based on what I've found, but I'd love it if some else could verify and fill in the missing gaps in it.

OK, so after a bit of researching this is what I've found.
First of all,why does the dust not settle in the water in the first place? Well, dust forms a kind of a suspension.

The particles are all similar and ionized, so therefore they have a similar charge (positive, I think) causing them  repel each other and stay separate.
And because they're so small, the gravitational force acting on them isn't very strong either.
Also, the dust particles, previously dispersed in the air, have a lot of kinetic energy already.

All these 3 reasons coupled cause the particles to remain suspended for so long.

So what does adding alum do? Alum has a negative charge and tends to disperse in water very fast and very well (Why so? Because it's a colloid?).
This causes it to join up with all of the offending particles and neutralize them. Now that the particles don't have any repelling charges, they tend to clump together into 'flocs'.
The increased size as well as the lack of repelling charges cause the alum particles to settle down at the bottom or rise up and float in the water.

Why do the particles clump together and settle? Apparently even this isn't as simple as it seems.
After the particles are neutralized, they clump together because of something called the London Dispersion Force which are part of the Van der Waals forces. Basically, weak intermolecular force arising from quantum-induced instantaneous polarization multipoles in molecules causes even nonpolar particles to attract each other due to the correlated movements of the electrons in interacting molecules. More info can be found inside.
As to why they then settle, I think it can be sufficiently explained by gravity.

Oh well, that's all. All for explaining something even the ancients knew.


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How does water deionization process work?
What are the technical/scientific details behind the process of water deionization? How does deionization machines like those manufactured by Millipore work?

Check out this page: http://www.millipore.com/lab_water/clw4/type1&tabno=2

@Michiel - If you could summarize/explain the process, that would be a good answer.

As pointed out by Michiel, there is a lot of info on Merck's Millipore website. You can get a pdf with some technical specifications for each equipment.

@BenNorris - I don't have that much knowledge on the topic to expand on the information already given on that website and I don't think summarizing a different webpage is going to do any good in helping the OP. Typically a bit more extensive explanation is easier to follow.

Most cost/energy efficient method to make deionized water is reverse osmosis. The mechanism is to apply external pressure to water and force it to go through a layer or layers of semipermeable membrane.


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bond energy calories
I am attempting to optimize interactions for a series of random and test molecules to a protein using docking software. The software uses QM/MMGBSA to calculate free energy in joules (yes it is not a measurement but it is very corollary). Larger molecules automatically have more free energy because they have greater potential energy. I would like to divide the free energy by the amount of atoms in the molecule or the molecules total Dalton. The idea would be to normalize the energy and provide the amount of calories associated with the bond. I asked a professor at a university about measuring the efficiency of a bond in this manner and he said he was unsure what joule/mass would provide. It does appear that the amount of bond calories is able to properly classify with correct magnitudes of viability 2 positive controls and 1 negative but I need this idea to pass peer review. Is my rationalization of the joule/Dalton reasonable?
I am measuring the binding energy of azttp (AZT triphosphate), gtp (Guanine triphosphate) and AZTppppA (AZTP+AMP) on HIV Reverse Transcriptase wild type (8ICO). I am used 8 high quality conformers and independent docks with QM/MMGBSA for my negative sample (AZT+AMP), 5 conformers for my major positive sample (AZTTP) and 1 conformer for the minor positive sample (gtp). I have two or three high quality conformers through the same processes for each of my test samples.
HIV RT has a tunnel where nucleotide analogs pass through with a pocket where if a molecule binds effectively it is accepted but if it fails it is phosphorylated and excreted. I am attempting to create a metric which can be used to measure the effectiveness of a binding and allowing me to infer (only) probability of rejection. Since the window/box for the docking is small and does not change for each test subject I would like to consider HIV-RT a constant.

Could you be a bit more specific about what you are measuring and what you are trying to measure? Typically docking calculations yield the CHANGE in free energy between the bound and unbound state of a ligand-protein complex.
As you pointed out this can be different for larger/smaller molecules but is more highly influenced by the types of groups interacting, conformation of the ligand in the pocket etc.

@CTKlein i updated the question with more information as per your request

Is my rationalization of the joule/Dalton reasonable? || No. Standard scheme for measuring bond energies is to calculate free energies for complex and subtract energies of two separate molecules. In your case you probably have to calculate free energies for biological conditions and then derive constant of reaction. After that you have to solve classical problem of amount of bound/unbound molecules using some known (medically meaningful) starting concentrations

@permeakra thank you for your response but larger molecules have a greater probability of forming larger complexes as CTKlein points out

@caseyr547 first, masses are irrelevant in the case then =) for obvious reason. Second, chelate effect does exist. And third, I don't think 'mass' or 'size' has any meaning. AFAIK, in medicine new substance has to be 1) efficient ligand 2) specific ligand 3) not unreasonably costly per use case. 4)not extremely toxic. Joule/Dalton does not address any of this issues.

@permeakra I'm sure your great with chemistry but as CTKlein pointed out and i observe mass effects the efficiency to some measurable degree. While I'm sure your comments about the constant of reaction rate would work well in non static situations that is not the case here. Also calculating the free energy in a biological condition is impossible given our current technological advancement.

That's not really the point I was making. I was pointing out that molecules of different size will have to adjust accordingly in a binding pocket and/or the binding pocket may change its conformation. Inherently, this has nothing to do with the mass of the ligand.
Also I'm still not clear what exactly you are measuring when you say "free energy". As mentioned, a docking calculation will give you a the CHANGE in free energy. In your case that would be G(complex)-G(ligand)-G(protein). This value is a metric for how tightly the ligand binds which may or may not be what you're interested in.

@CTKlein The formula for deta G which I am using is dG = E_complex-minimized- (E_ligand_minimized + E_receptor_minimized). My hypothesis which I observe to be correct in 3 cases is: The probability of forming more complexes and generating a lower dG is higher when there are more atoms to form complexes in the ligand molecule.

@caseyr547 Once more. ligands with several points of binding binds more efficiently if the centers can assume fitting positions. It is not an artificial effect and is quite well-known in coordination chemistry. If you want metric measuring effectiveness of binding it is exactly free energy of binding. There is completely different matter, that you have to compare it with binding energy of water at same sites and to measure actual preference, but that's different matter. So you probably have either to live with what you have or reformulate the question

@permeakra summed it up nicely. If the pocket is big enough to accommodate the larger ligands you are testing then they are likely binding in more locations than the smaller ligands, hence lower free energy of binding. There is a multitude of interactions that need to be considered when measuring binding free energies (water in the pocket, types of chemical groups on the ligands/pocket etc.). While it would be wonderful if a metric as simple as mass were a good indicator for binding energy, it simply isn't the case.

@CTKlein ok the pocket is big enough to accommodate massive ligands and they are binding in more locations. Can I divide the energy by the amount of binding locations?

@CTKlein the goal would be to average the energy contribution per bond.

@permeakra yes the energy alone gives the metric of quantity but it does not give the metric of quality of the bond. 6 random bonds with 10 atoms total is less than 6 specific bonds with 10 atoms and is less than 3 specific bonds with 4 atoms total. The energy for the 6 random bonds would be greater than the energy of the 3 specific bonds but on average the quality would not be as good.

@caseyr547 I see. In this case I'd considered two options. Option one is to dock as many molecules of water simultaneously as I can and then check how many water molecules cover binding site of the ligand in question. Another option is to check the surface of contact between the binding ligand and the host protein. Both metrics are hard to measure, but they have some meaning at the very least. The last option is to divide by number of atomic contacts between the ligand and the protein.]

@permeakra yes ok dividing the energy by the number of contacts is possible but it has to be done by hand. i have a program which will count the number of contacts for individual ligands. if i did it right, gtp has 178 contacts and aztamp has 400. gtp has a complex dg of -36 and aztamp likewise is -79. From this, Can I safely conclude that aztamp contacts are individually no better than gtp?


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Why are fluorides more reactive in nucleophilic aromatic substitutions than bromides?
In nucleophilic aromatic substitution reactions, why do fluorides react faster than bromides?
Ordinarily bromide is a better leaving group than fluoride, e.g. in $\mathrm{S_N2}$ reactions, so why isn't this the case here? The only thing I can think of is that fluorine is more electron-withdrawing (via the inductive effect), which could stabilise the Meisenheimer complex formed as an intermediate.

The key point to understanding why fluorides are so reactive in the nucleophilic aromatic substitution (I will call it SNAr in the following) is knowing the rate determining step of the reaction mechanism. The mechanism is as shown in the following picture (Nu = Nucleophile, X = leaving group):

Now, the first step (= addition) is very slow as aromaticity is lost and thus the energy barrier is very high. The second step (= elimination of the leaving group) is quite fast as aromaticity is restored. So, since the elimination step is fast compared to the addition step, the actual quality of the leaving group is not very important, because even if you use a very good leaving group (e.g. iodine), which speeds up the elimination step, the overall reaction rate will not increase as the addition step is the bottleneck of the reaction.
Now, what about fluorine? Fluorine is not a good leaving group, but that doesn't matter as I said before. It is not the leaving group ability of $\ce{F-}$ which makes the reaction go faster than with, say, bromine or chlorine, but its very high negative inductive effect (due to its large electronegativity). This negative inductive effect helps to stabilize the negative charge in the Meisenheimer complex and thus lowers the activation barrier of the (slow) addition step.
Since this step is the bottleneck of the overall reaction, a speedup here, speeds up the whole reaction.
Leaving groups with a negative mesomeric effect (but little negative inductive effect) are not good at stabilizing the Meisenheimer complex, because the negative charge can't be delocalized to them.

+1, I posted my response about 10 seconds after yours, but I think yours settles the issue w/r/t the -I effect of fluorine, which I wasn't sure would be significant enough to make a major difference here.

@GregE. +1 for your answer, too. It contains a few nice points I don't have in my answer. May I ask about you opinion on something: Do you think that a group with $-M$ effect in meta position to the leaving group would lower the activation barrier of the addition step? I would say no, because in meta position the negative charge cannot be delocalized into the group with $-M$ effect and thus wouldn't be stabilized, right?

I think that's probably right. My understanding is that EWG need to be ortho/para for a -M stabilizing effect in the Meisenheimer complex. Still, the electronegativity of the group might have some stabilizing inductive effect, although at that distance it would be very small.

@GregE. Thanks. Should I add to my answer, that leaving groups with $-M$ effect (but little $-I$ effect) are not as good, because the negative charge can't be stabilized by them via delocalization?

Sure, I think that's a valid point to add. It does occur to me, now that I think more about it, there are some strong EWGs that function almost exclusively by -I effect (quaternary amines) or by a combination of induction and negative hyperconjugation (perhalogenated alkyl groups), and those might be strong enough to have some measureable impact. That's purely speculative, though.

@GregE. I think I have read somewhere that perhalogenated alkyl groups and quaternary amino groups can be used as leaving groups in a nucleophilic aromatic substitution but unfortunately I don't remember where.

Yes, I think that's possible. Even if they don't act as the L.G., I think it's likely that they might be strong enough to adequately stabilize the complex purely by induction in any position on the ring (though less effective certainly than ortho/para groups with strong -M effect), but as I said that's speculative. I may look for literature on the subject later.

Fluorine is the most electronegative element, and the fluoride anion is also much smaller and less polarizable than any of the other halogen anions, making its activity much more dependent on solvent effects. In protic solvents, fluoride tends to be very strongly solvated as a hydrogen bond acceptor and is thereby stabilized, resulting in high leaving-group tendency and very low nucleophilicity. The larger halogens can be better leaving groups in aprotic solvents, however, where their bulk and polarizability results in greater comparative stability, and the stabilizing effect of hydrogen bonds is not present. I guess it's conceivable that the presence of fluorine atoms would better stabilize the Meisenheimer complex than other halogens, though I don't know for sure. Still, since all halogens are electron-withdrawing by induction or negative hyperconjugation, the presence of more strongly electron-withdrawing groups (e.g., nitro) conjugated to the $\pi$ system of the ring would still be necessary.
What follows is not a direct answer to your question, but a slight segue into a related subject:
The rate-determining step in SnAr mechanisms is typically the initial nucleophilic attack resulting in formation of the Meisenheimer complex, or occasionally the final proton transfer if neutral nucleophiles are used. The loss of the leaving group is normally a fast step, given the high energy of the Meisenheimer complex intermediate. Once the complex is formed, loss of the leaving group should be rapid. Therefore, I suspect aprotic solvents would be preferable in many situations (particularly with neutral nucleophiles), irrespective of the effect on the leaving group, since they don't hamper the activity of the nucleophile in the way that protic solvents can.


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Electrochemical characterization of benzenesulfonamide
I'm trying to elucidate current peaks for benzenesulfonamide, using an acetic acid-sodium acetate 0.1M buffer at pH 5.5 as electrolyte. When doing cyclic voltammetry, I get a voltammogram similar to the platinum (Electroanalytical Chemistry and lnterfacial Electrochemistry, 43 (1973) 9-36), because I use platinum as a working electrode. So, I can not see any current peak related to benzenesulfonamide, and I have not found any article related to this. I've been thinking about using a glassy carbon electrode instead of platinum.

I'd change the solvent to acetonitrile (don't forget the quarternary ammonium salt) and give a glassy carbon electrode a try.
It should be possible to observe the (irreversible) reduction of the sulfonamide.
And once you have optimized the conditions, please perform a final run with ferrocene as an internal reference.

Edit:
Let me explain the last suggestion. The literature is full of electrochemical data of organic compounds measured in non-aqueous solvents, mostly acetonitrile.
It was once common (bad) practise to report the potentials vs SCE (saturated calomel electrode), although the latter, an aqueous(!) reference electrode, was atually never used in the measurement. Often rather exotic conditions plus a shaky conversion of data turned good experimental work into a pile of data of limited value.
Using an internal standard is the obvious solution to obtain solvent-independent electrochemical data for publication.
Note that the use of ferrocene as such a standard isn't my idea! It was originally proposed by GRITZER and KUTA in Pure Appl. Chem. and is a IUPAC recommendation nowadays.


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Equivalent weight of metal
A metal oxide contains $60$% metal.The equivalent weight of metal is
(1) $12$
(2) $60$
(3) $40$
(4) $24$
I know that the equivalent weight of metal oxide = equivalent weight of metal + 8.But how can i arrive at the answer with this.
Any help appreciated.
Thanks.

Please see our homework policy. We expect homework and homework-like problems to have some effort put into them, and deal with conceptual issues. If you edit your question to explain (1) What you have tried, (2) the concept you have trouble with, and (3) your level of understanding, I'll be happy to reopen this. (Flag this message for ♦ attention with a custom message, or reply to me in the comments with @Manishearth to notify me)

Equivalent mass of a substance(element or compound) is defined as the number of parts by mass of the substance which combine or displace directly or indirectly $1.008$ parts by mass of hydrogen or $8$ parts by mass of oxygen or $35.5$ parts by mass of chlorine or $108$ parts by mass of silver.
The equivalent mass is a pure number.When a known mass if element is changed into oxide directly or indirectly , the mass of oxide is noted.
MASS OF OXYGEN= Mass of oxide-mass of element
Thus,
$ Eq. Mass = \frac { Mass of element * 8} {Mass of oxide} $
Here in this question you need ti assume the mass of oxide to be $100g$.
So, the mass of metal is $60$ and mass of oxygen is $40$.
Substituting these values in the above formula you get $\frac {60*8} {40}$$ = 12$.
Hence the ans is $12$.


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How chemists find out the electronic properties of elements (atomic numbers)
How did chemists find how many electrons, neutrons, and protons each element had?
I mean I can understand the Mendeleev's table but without really knowing how you make one...

One could do it in the following way.

The atomic number $Z$ can be determined by examining the x-ray spectrum of an atom.
The frequency $f$ of the so-called K-alpha line is related to the atomic number $Z$ by Moseley's law
$$
    \sqrt f = k_1 \cdot \left(Z - k_2\right) \, .
$$
The atom is electrically neutral, so the number of electrons is obviously calculated from the atomic number.
And the number of neutrons is also simply to choose: the only requirement at this stage would be the correct atomic weight for the element in question.

Thanks based on your answer I was able to found the history http://en.wikipedia.org/wiki/Atomic_number#History . Also it's important to understand http://en.wikipedia.org/wiki/Atomic_weight


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Azeotropes and separation by distillation?
I have just recently learnt the theory of fractional and normal distillations and the basics of maximum and minimum boiling azeotropes.
Most books say that the components of the mixture forming an azeotrope cannot be separated by distillation because on heating, the residue (in case of negative or maximum boiling azeotropes) and the distillate (in case of positive or minimum boiling azeotropes) is closer in composition to the azeotropic composition and always approaches it.
This being understood, we at least separate one component from the mixture as the residue (for positive azeotropes) and the distillate(for negative azeotropes) is richer in one component than the other since the remaining part approaches the azeotropic concentration thereby increasing the concentration of one of the components depending on which side of the azeotropic "point" are we on the graph of boiling temperature and composition.
If this is true, for any mixture, we can at least theoretically obtain one purified component (provided we have enough quantities of the mixture)??
Reference azeotrope basic

You seem to have the right idea.
To elaborate, take the specific example of the positive azeotrope of (roughly) 96% ethanol and 4% water. If you begin with a mixture containing less than 96% ethanol, distillation will result in a distillate more abundant in ethanol and nearer to the azeotropic composition. That implies the residue will necessarily be proportionally lower in ethanol. If you continuously repeat the distillation procedure using the remaining pot residue, it will approach 100% water. On the other hand, if you begin with a concentration above 96% ethanol, the resulting distillate will actually be lower in ethanol (and therefore nearer to the azeotropic point), while the residue will actually become more concentrated in ethanol. Hence, in theory, with many repeated distillations you can approach purity in the residue when dealing with positive azeotropes, and the ultimate composition of the residue will depend on which side of the curve you're on to start with.
When dealing with negative azeotropes, the situation is analogous but the particulars are basically reversed. Take the example of hydrofluoric acid and water, which form an azeotrope at approximately 37% HF. If you initially begin distilling a mixture comprising less than 37% HF, you'll obtain a distillate more dilute in HF, while the residue will contain more HF and be nearer to the azeotrope. If you were to collect the distillate and repeat the distillation, you would again end up with an even more dilute distillate, with the concentration of the distillate approaching 100% pure water. If you begin distilling a mixture containing more than 37% HF, then the residue will end up being more dilute as it approaches the azeotrope, while the distillate will actually be richer in HF. If you repeat this process many times over, you can eventually obtain a distillate approaching 100% pure HF (in theory).
So, in summary, you're correct to suggest that one of the components can be isolated with something approaching complete purity from an azeotropic mixture, assuming a sufficient volume to enable an adequate number of simple distillation cycles or theoretical plates in fractional distillation. Of course, the problem is that you really have no choice in determining which of the two components you can isolate, as that depends entirely on the starting composition of the mixture. In practice, you can often separate azeotropeic mixtures by modifying the distillation process in various ways.

That's fine. I just wanted the reassurance that mixtures forming azeotropes are not hopeless regarding their purification. On a side note, would all my (or 'our') arguments be valid if we used 'pressure' distillation (using reduced pressure to obtain the vapour)?

The specific proportions of some azeotropic compositions can be altered significantly by pressure changes, while others will remain relatively constant. In principle, the above arguments should still hold under any given constant pressure. For some mixtures, it may be possible to alter the azeotropic composition by varying pressure to improve the efficiency of the distillation process.

Alter the azeotrope by changing the ambient pressure.  Add a third component to remove one component a(e.g., ethanol, water, benzene), leaving the other.  Use another mechanism for separation (membrane permeation).
DOI: 10.1021/ba-1973-0116.fw001
Azeotropic Data—III


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What is the best way to detect the presence of tear gas?
I realise that tear gas can be several different things (Pepper spray / CS Gas / CR gas, PAVA, etc)
Is there a good way to detect some or any of these in the air?
Ideally using an electronic sensor that can be reused?

Owlstone make an electronic multi-agent detector - I don't know how much it costs but I would guess somewhere in the range \$10K - \$100K. There might be cheaper single-agent electronic sensors or chemical tests.


text:
Why doesn't diffusion take place between immiscible liquids?
If we take a test-tube of bromine and invert an upside-down test tube of air on top of it, it'll diffuse with the air, until the bromine is thoroughly 'mixed' with the air in both test tubes. But why doesn't the same thing happen with two immiscible liquids? Say oil and water.
When you put a layer of oil on top of water, the two don't mix, but they should. By diffusion.

In gaseous phases, we generally assume that there's very little or no bonding between individual molecules (or atoms in, say, a noble gas). This isn't strictly the case, but it's a pretty good approximation, and many good approximate techniques rely on it. A gas that behaves entirely in this way is known as an 'ideal gas'. (See also: ideal gas law.)
In contrast, liquids have bonding between individual molecules, and their behaviour is strongly affected by molecular properties.
So, in the examples you gave:
1. Bromine and air
In this example, imagine that you have a bunch of ping pong balls bouncing around at a reasonably high speed. (Ping pong balls are often used as examples for ideal gases.) Since the balls are already travelling at high speed (given a sensible temperature), and they only get in each other's way by bouncing off each other, it's relatively easy to get the bromine and the air 'balls' to mix, and there's really nothing apart from gravity to make them separate again.
2. Water and oil
Water molecules form pretty good bonds to other water molecules. (See: hydrogen bond.) Bonding in oils is generally weaker, but also occurs via a very different mechanism to the bonds in water. (See: Van der Waals force.) This means that oil does not form good bonds to water, and any water to diffuse into the bulk oil would have to break the bonds to all the neighbouring water molecules, and any oil to diffuse into the bulk water would have to separate a few water molecules to make a space to do so. Not just that, but since the water and oil molecules can reform their bonds if they manage to connect to other water and oil molecules, if you mix them up, the droplets will tend to recombine into the separate phases.
As a comparison to this example, let's check out two liquids that do mix.
3. Water and ethanol
For the water, we have basically the same situation as before -- water molecules forming good bonds to each other. The ethanol, though, has an -OH group that can form bonds to the water in the same way that the water does (though not as well). This means that ethanol that mixes with water (and vice versa) will tend to stay mixed, and given that the liquids are being mixed around just by random motions, means that you'll get one mixing with the other just as a matter of statistics.

So, when water and ethanol mix, is ethanol bonding to the water? And, in general, any substance that mixes with a solvent bonds to the solvent?

@user1305192: To the first question: yes. The bonds aren't standard two-electron covalent bonds, though -- they're much weaker than that, and weaker than the similar bonds water molecules form to each other in a liquid. To the second question: generally, any two molecular liquids that are miscible will have some sort of bonding interaction between each liquid's molecules, whether it's by an electrostatic or van der Waal's interaction, but the important factor is whether each of the liquids bonds much more strongly to itself than it does to the other liquid.


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Why is O2 the supporter of combustion?
It's there in every high-school textbook: $\ce{O2}$ is the supporter of combustion. Without $\ce{O2}$ combustion cannot take place. Why? And why only $\ce{O2}$? Why not some other element?
And, what happens when a combustible gas burns in air? Say $\ce{H2}$?

Oxygen is not strictly needed, it just happens to be very, very good at what it does!
Specifically, combustion requires three things:
1) Fuel: The thing that burns. This is is often a hydrocarbon, or other organic molecule. The simplest possible fuel is pure hydrogen gas.
2) Energy: What gets the reaction started, which is true of most chemical reactions.  Note that this so-called activation energy is usually much less than the energy ultimately released from combustion.  This is like rolling a boulder some distance in order to roll it down a hill, which releases much potential energy.
3) Oxidizer: The molecule that accepts electrons.  It turns out that combustion requires the fuel to be oxidized, that is, it donates electrons.  So we need something to accept the electrons, and that's the oxidizer, which is then reduced.
Oxygen is a great oxidizer because it is so electronegative, which means it really wants to accept electrons.  Only fluorine is more electronegative, and fluorine is a superb oxidizer: blow fluorine gas at nearly any substance and it bursts into flames.  For example, see this video: Fluorine.
Oxygen is in many ways a perfect oxidizer for supporting life: it doesn't set things on fire like fluorine, but otherwise can oxidize very many things.  When you eat, oxygen is oxidizing the food (fuel) you ingested, to generate energy you need to live.  So there is "combustion" going on inside you!

Does the fact that it's a diradical not play a role?

The main benefit of oxygen is that its the most commonly available oxidizer.
The main component of air that support combustion is oxygen.
However titanium burns in nitrogen and so does lithium, and also some other active metals such as finely divided aluminium. These form nitrides. Similarly most of these active metals also burn in hydrogen forming hydrides such as lithium.


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Why does sodium react with water?
The reaction of sodium with water is quite infamous.
$\ce{2Na + 2H_2O \rightarrow 2NaOH + H_2}$
But exactly why does $\ce{Na}$ displace $\ce H$ in $\ce{H_2O}$?

Alkali metals in general are extremely reactive metals. They are highly reducing in nature. Hence, they react with water forming their corresponding hydroxides evolving dihydrogen gas.
In case of sodium - as cited in your question - its reduction potential is $\pu{-2.7109 V}$ hence it easily reduces hydrogen.
Also, as the density of sodium is quite low it floats on the water. It immediately reacts with water to form a white trail of $\ce{NaOH}$ which further gets dissolved to give a colourless solution.
This reaction is highly exothermic evolving a lot of heat which is sufficient enough to melt sodium due to its low melting point.
Sodium moves in water as the dihydrogen gas formed below the waterline pushes sodium. If sodium is trapped in the container the increasing temperature could induce dihydrogen to catch fire.

Sodium is more "electropositive" than hydrogen as it is down on the periodic table. This makes it more reactive to say elements like fluorine or oxygen. In this same way it is reactive to ions like hydroxide. Since water can turn into hydroxide and hydrogen ion the sodium ion is able to displace the hydrogen ion. This causes sodium to react to hydroxide to form sodium hydroxide and hydrogen. The two hydrogen atoms merge to form a hydrogen molecule.


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How would I write an input Lammps command for continuous change in temperature?
I'm going to study the thermal expansion of graphite. For this goal, I should change the temperature continuously. But I do not know how to code it by Lammps?

Have you tried something or are you just starting out with LAMMPS?  There is a section in the documentation about thermostating.

(Disclosure: I'm not a LAMMPS user, but I've used other MD packages.)
You'll probably want to use one of the thermostat settings such as the Nose-Hoover type described here in the LAMMPS documentation, but there's a more general HowTo guide on temperature control and calculation in the docs as well that looks pretty useful.
I would strongly recommend reading the sections of the manual concerning any given thermostat type you're going to use fully before running serious calculations with it, though, since there are some things to look out for, such as the tendency for setting the thermostat interval too low to cause large oscillations in effective temperature.
The Nose-Hoover type (and some of the others, but I only took a quick look) also lets you step up or down the temperature with a given interval. Depending on exactly what type of thermal expansion you're looking at, I'd just like to note: you really need to set the stepping interval appropriately to let the structure equilibrate at each temperature you want to look at, otherwise your results won't be meaningful. (Or, at least, won't have the meaning that you might want.) Relatedly, I'd also suggest that if you weren't planning to already, you run each simulation both ways: one stepping the temperature up, and one bringing the temperature down. If the behaviour matches at the temperature steps in each, you can be relatively confident that you've given the structure enough time at each temperature.

you can use some of this commands
fix     1 all nvt temp 200.0 300.0 0.01
or
fix    1 all npt temp 300.0 300.0 100.0 iso 0.0 0.0 1000.0
check the lammps manual


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Determine concentration Using Beer's law
How do you determine concentration of a sample using wavelength and absorbency? Say our wavelength for a sample of Red 40 is 508.50 nm and absorbance is 0.283 when we use 6ml water and 4 ml Red 40. What would be our concentration and the slope ε? This is really frustrating me so any help would be appreciated.

Seems to me you should have measured more than one sample.  What happens when you plot those points?

Do you know what the extinction coefficient is for Red 40 at 508.5nm?

@TomD No that is what i am confused about. I don't know how to determine the extinction coefficient and the concentration from the given information I measured in the lab.

It seems like you are missing some information, given that Beer's Law is A = e b c and you want to know c, you need to know A, e and b.  If you were supposed to perform only one measurement, then e must have been given or you are to find it from some resource.

I don't know anything about Red 40, but you will need the extinction coefficient.  Perhaps you determined this from a plot of concentration vs absorbance in the lab?
I am going to attempt to answer you question with reference to NADH, which has an accepted extinction coefficient of 6220 M-1.cm-1 at 340 nm.
Say I have a solution of NADH in a test-tube, I take 4 ml of this and dilute with water to 10 ml.  I then take this diluted solution and measure its absorbance at 340 nm in a spectrophotometer using a cuvette with a path-length of 1 cm.  The value I obtain is 0.4.
(Edit.  In the above measurement, I take care to 'blank' the spectrophotometer against water).
What is the concentration of NADH in the original solution?
From the Beer-Lambert law (where c is concentration)
0.4 = 6220 x c x 1
Therefore
c = 0.64 x 10-6 M  = 0.64 µ M
But the solution on which the measurement was made was
diluted 4 in 10 or 1 in 2.5
Therefore the concentration of NADH in the original solution is
c = 2.5 x 0.64 x 10-6 M = 160.77 x 10-6 M = 160.77 µ M
So, the concentration of NADH in the original solution is about 161 µ M (micro-Molar).
I am making a number of assumptions here, the main one being that the spectrophotometer obeys the Beer-Lambert law in the absorbance region of interest.  You should determine that this is so by measuring  the absorbance of series of NADH solutions and make sure that the resulting graph of absorbance vs concentration is linear and through the origin.
At the very least, doubling the concentration of NADH should exactly double the absorbance (0.8) and halving it should give exactly half the absorbance (0.2).  In addition, measurement of water (or whatever you are 'blanking against) should give zero - that is the spectrophotometer should be properly 'blanked'.  If not you will need to subtract this value.
So what happens if we do not dilute our original sample, but just measure the absorbance at 340nm directly?
From Beer-Lambert
A = 6220 x 160.77 x 10-6 = 1
The original solution is predicted to have an absorbance of 1, but IMO even with modern spectrophotometers you are 'pushing your luck' to assume that Beer-Lambert is obeyed at such a high absorbance.  A 'rule-of-thumb' is to keep the absorbance below 0.8.
One final comment.  It is very difficult to accurately determine an extinction coefficient from absorbance versus concentration plots, unless you accurately know the concentration of the original material.  If this figure relies on a 'dry-weight' measurement, you will need to be sure that the compound is pure, for a start.
Reading your question again, I think we need a bit more experimental detail, and if you could supply it, I'll have another go, maybe.
NADH is widely used in the biochemical sciences where (among many other things) the relatively high extinction coefficient of 6220 M-1.cm-1 is exploited in continuous enzyme assays.  If an enzyme produces or consumes NADH (or NADPH) then it is (usually) very easy to design an assay.
EDIT
It would appear that the accepted extintion coefficient (Molar Absorption Coefficient) for Allura Red AC (Red 40) is about 25 000 M-1.cm-1 (or 25 000 mol-1 L cm-1) at about 505 nm. The best reference I can get it here. Azo dyes tend to have very high extinction coeffficients, so this 'looks about right'
The diluted solution has an absorbance of 0.283
Taking the above as the extinction coefficient, and in addition assuming the measurement was made in a cuvette with a path length of 1 cm (l =1), one can calculate from Beer-Lambert that the concentration of Allura Red in the diluted solution is about 11.32 x 10-6 M.
The dilution factor is 4 in 10, or 1 in 2.5.  Therefore, the concentration of the original solution is about 28.3 x 10-6 M.


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How to estimate HCN content in a volatile mixture?
How can you accurately qualitatively and quantitatively determine the HCN in a volatile product from water?

please, describe composition of the product in question at least in generic terms.

This is a volatile gas produced from cassava leaf ditillate in water and shows colour test for cyanide but quantitative and confirmatory tests not obtained

Did you test it with iron sulfate (forms prussian blue, intense blue pigment) ?

the amount is too little to be determined by this method. I ma looking forward to some GC-MS or similar method

From a quick search of the literature, it seems that as you correctly thought, gas chromatography mass spectrography is a method that can be used to quantitate small amounts of cyanide in vapour.
While I am certainly no expert on this field or the precise method that you would use, this paper, which describes quantitation of cyanide and other chemicals in tobacco smoke may be of interest: A simple GC–MS technique for the analysis of vapor phase mainstream cigarette smoke

A specific $CN^-$ electrode will detect and measure from about 50 ppb of cyanide in water.