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What criteria have to be met to term a carboxylic acid a 'fatty acid'?
So we were doing the chapter Biomolecules in class the other day, and my teacher told us that every carboxylic acid that consists of an aliphatic chain and a total carbon atom count that's four or more, is a fatty acid.
Now owing to my teacher's (rather annoying) habit of grossly over-simplifying things, I decided to look up 'fatty acids' in my trusty Oxford Science Dictionary (printed in 2003). According to it,

Chain length ranges from one hydrogen atom (in methanoic acid) to nearly 30 carbon atoms.

But a quick Wikipedia search, gives me this:

Most naturally occurring fatty acids have an unbranched chain of an even number of carbon atoms, from 4 to 28.

So now I'm completely lost. Fine, chuck what my teacher said, but the other two sources seem to contradict each other. So I'd really appreciate it, if someone can help me get the following straight (I'd prefer a reliable source, if you're citing anything):
Q1: What is the minimum number of carbons in a carboxylic acid as a whole (includes the carboxyl group)? So is methanoic acid or butanoic acid that's the smallest fatty acid?
Q2: Is there an 'upper limit' on the total carbon atom count for fatty acids? As in; can I synthesize, say, an 80 carbon aliphatic carboxylic acid and still correctly call it a fatty acid?
Q3: Is there any other criterion that's necessary to classify something as a fatty acid that hasn't been mentioned by my teacher or the other sources? If so, what is it?

I think trying to come up with a specific definition of what's a fatty acid is a slippery affair. See Is activated carbon classified as organic or inorganic?

The old adage about tomatoes comes to mind: "Knowledge is knowing a tomato is a fruit; wisdom is not putting it in a fruit salad." Likewise, wisdom would be not referring to methanoic/ethanoic acid as "fatty acids", regardless of if they technically meet the definition.

I suggest to consult the IUPAC Compendium of Chemical Terminology, aka the Gold Book for a definition of fatty acids!
In a narrow sense, fatty acids are defined as

aliphatic monocarboxylic acids derived from or contained in esterified form in an animal or vegetable fat, oil or wax.

The definition continues to mention that

natural fatty acids commonly have a chain of 4 to 28 carbons […]

The paragraph closes with a wider definition, namely:

By extension, the term is sometimes used to embrace all acyclic aliphatic carboxylic acids.

Personally, I have used the term fatty acids only for natural fatty acids.

So in other words, the Oxford dictionary is incorrect?

@AaronAbraham I think that the definition in the dictionary is somewhat half arsed and arbitrary. I'd go with the Golden Book. Note that they mention that natural fatty acids commonly have up to 28 carbons. This is very reasonable and doesn't exclude compounds like lacceroic acid, which has been isolated from plants, from the family of natural fatty acids.

And I have even heard colleagues use the term for hydroxylated linear acyclic aliphatic carboxylic acids. On that note, I would have added the adjective linear, too; but I realise the term is not defined well.


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Making solid fuel out of sugar
I recently tried to make a cheap solid fuel out of sugar. First I attempted burning the sugar cube directly, which did not work well even using a blowtorch. This was expected. Then I attempted to use paper ash, and it could be lit with a lighter. However the flame went out quickly and that would not be usable for anything.
How can I make sugar burn completely? My aim is to have it burn as well as hexamine, although I realize that is probably quite impossible to do at a reasonable cost or else it would already have been done.
One idea I have had is to mix a small amount of oxidizer into the sugar - not enough to make it into a rocket-candy type fuel, but just enough to accelerate combustion. Another idea I have thought of is to mix some charcoal, combustible oil, or wood into the sugar as a secondary fuel. I won't be able to do these experiments for quite some time due to a lack of resources.

source: Stack Exchange

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Enol content of acetylacetone and ethyl acetoacetate
Why is the enol content of acetylacetone higher than that of acetoacetic ester? Hydrogen bonding stabilisation is present in enol form of both of them. I cannot figure out which factor makes one predominate over another.

Since the stabilisation by hydrogen bonding is about the same, we might as well just take it away. Try comparing acetone and ethyl acetate, which is more prone to enolisation?

@orthocresol Acetone is more prone to enolisation, but same problem persists, why is it more prone?

To enolise a carbonyl compound, you need to deprotonate the adjacent oxygen and push electrons into the C=O π* orbital, right? (And then you reprotonate on oxygen, but that doesn't make a difference here.) Well, in an ester, there's already one oxygen pushing electrons into that C=O π*. That makes it less interested in enolisation.

@orthocresol Got that! Accept my word of thanks.

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Reduction of sugars
Why reduction of d-glucose is not generally given as compound B of the following figure?
Is the formation of that compound possible? What should be its name?( since I can't find that compound I don't know it's name). Iam sure that it's not l-sorbitol.
Also I would like to know why almost all sources(on google) shows, on reduction hydroxyl group goes to right hand side (in Fischer projections)?

Try to write C-6 in the starting material and C-1/C-6 in the products simply as $\ce{CH2OH}$. Are A and B still different then?

@KlausWarzecha No. Thanks,  I got it, I was highly confused. Should I delete question?

Fisher projection shows stereochemical structure of a compound. When we say stereochemical structure, we specify those carbon/atoms that are chiral and the spacial arrangement of the groups attached to it.
Here you have glucose reduced to sorbitol, in which the aldehyde is converted to a primary alcohol, which is an achiral carbon. Thus there will be no meaning for Fisher projection of that carbon. Thus, both the compounds $A$ and $B$ are effectively the same.


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What is role of palladium in hydrogenation reactions?
Why is palladium used in hydrogenation reactions? Is there any theory about the role of palladium and other catalysts in those reactions?

http://chemistry.stackexchange.com/questions/2710/how-does-palladium-dissociate-h%E2%82%82-so-easily

According to Wikipedia, it appears that the catalyst simply forms temporary bonds with both reactants, enabling said reactants to react with each other much faster than if they were just floating around. Palladium and other expensive metals are used as the catalysts because they are faster and more efficient than other ones which use cheaper materials.

The unsaturated substrate is chemisorbed onto the catalyst, with most sites covered by the substrate. In heterogeneous catalysts, hydrogen forms surface hydrides (M-H) from which hydrogens can be transferred to the chemisorbed substrate. Platinum, palladium, rhodium, and ruthenium form highly active catalysts, which operate at lower temperatures and lower pressures of H2.

The general catalysis cycle of hydrogenation reaction is pretty simple and AFAIK well documented. It consists of for reactions, all well documented in metalloorganic chemistry.

Hydrogen activation. It comes in many flavors, but idea is the same: $\ce{H-H}$ bond is broken and instead $\ce{M-H}$ bonds are formed.
Alkene bonding with a metal center
Alkene insertion in $\ce{M-H}$ bond, forming $\ce{R-M}$ alkyl bond
Elimination of alkane from $\ce{R-M}$ and $\ce{M-H}$ fragments.

This general process can occur on mononuclear or polynuclear complexes and on metal surface. However, when selecting particular metal for the process it is important that the intermediate particles were neither too stable, or they will cover catalyst completely and remain there, nor too unstable or the process won't happen. It so happens, than Ni, Pd, Pt are in the sweet spot with Pd being one of the best. They are not, however, the only hydrogenation catalysts in existence, they are simply the ones with high activity and simple composition. For example, $\ce{Ni}$-promoted $\ce{WS2}$ catalystalso has significant hydrogenation activity.


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Confusion about boiling point and vapor pressure
It is said that a liquid boils when its vapor pressure equals to the external pressure of its environment (e.g. atmospheric pressure). That’s because the bubbles of water vapor that form will be crushed if the vapor pressure is below the atmospheric pressure. However, it’s the fluid’s pressure that actually does the crushing, isn’t it? And the fluid pressure varies with depth, with only the fluid at the top of the liquid being equal to atmospheric pressure. Shouldn’t we say that the liquid boils when its vapor pressure is equal to the greatest fluid pressure in the liquid?

Let's put it this way: true, water at precisely $100^\circ\rm C$ will boil at the top and not deep down below, but as it boils, the upper layers vanish and new layers become free from the liquid pressure and also start to boil. So it goes.

@IvanNeretin So at precisely the boiling point, there won't be any bubbles, because vaporization only occurs at the surface of the liquid?

Strictly speaking, yes. To have bubbles in the entire volume of your kettle, all the way to the bottom, you might need something like $100.05^\circ\rm C$.

@IvanNeretin Did you use C-Clap Eq? For 1 m deep, I got $102.7 ^\circ , \mathrm{C}$ Supposing constant entalphy...; I don't know what happens if you have $1,m$ of water at $T_B$, will it boil because of vessel irregularities? Otherwise, I guess it won't...the system will be hold at 100...

@santimirandarp I didn't use a thing, other than my common sense. Wait, did you just measure the temperature inside a living geyser?

@IvanNeretin No,no; I've just used an equation...(the one of clausius clapeyron). Total pressure $P_{atm} + \rho,g,h $ & your common sense is ok, but the rise depends on height...

Shouldn’t we say that the liquid boils when its vapor pressure is equal to the greatest fluid pressure in the liquid?

No.  If you heat up water at the ocean's surface to 100 °C, it will boil, even though the pressure at the bottom of the ocean is far higher.
As you note, pressure is not guaranteed to be homogeneous in space (i.e. the same everywhere), especially for deep bodies of water.  But neither is nearly any other fluid property.  Temperature can vary with position in poorly-mixed waters.  So can whether the water is boiling or not.
A given small-ish region of water with nearly constant temperature and pressure will boil if and only if the (P, T) condition of the water is at or above the water's vapor pressure at the same temperature.


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Nucleation and Phase Equilibrium at the Melting Temperature
I used to think that at the melting point, the solid and liquid phases are in equilibrium. That’s because the Gibbs free energy change for the transformation is 0, as is characteristic of equilibria.
However, I recently learnt about nucleation in phase transformations, and I’m very confused. My textbook tells me that below the melting temperature the Gibbs free energy change is not 0 and this energy can go into increasing nuclei surfaces and surmounting the nucleation activation energy barrier. It goes on to say that at the melting temperature, no such energy is liberated (since ∆G = 0) and the nucleation activation energy barrier cannot be achieved. Thus, no nucleation occurs at the melting temperature.
If no nucleation occurs, then how can the solid and liquid phases be in equilibrium? I know about supercooling and that the principle behind it is that it takes time for such an equilibrium to be achieved. However, if nucleation simply doesn’t occur at the melting temperature, then equilibrium will never be achieved...

Nucleation is a kinetics issue, not a thermodynamics issue. Note that nucleation of the melt is very easy (it is very very hard to superheat a solid), while nucleation of the solid can be difficult (supercooling by 100s of K is not that difficult under the right conditions). But, again, that is a kinetics issue. Wait a very long time at the melting point, and it will eventually melt/solidify.

@JonCuster Can you further explain the kinetics issue by addressing what I mentioned about activation energy? Where is the activation energy supplied from if there’s no Gibbs free energy released?

The activation energy required for, e.g. nucleation, has little to do with the release of Gibbs free energy. It is the energy of a transitional state required configurationally to get from A to B. Again, kinetics vs thermodynamics.

@JonCuster I see. Going back to your first comment, my lecture notes tell me that at the melting temperature, nucleation rate is 0. Yet you say that “wait a very long time at the melting point and it will eventually melt/solidify.” How long are you talking about? In the limit $t\rightarrow \infty$?

The ease of melting the solid (and why it is hard to superheat) is the presence of surfaces. The atoms on the surface are not fully bonded (as they would be on the interior), and so they start to disorder below the bulk melting temperature, so called 'pre-melting'.  There is not similar phenomena upon cooling (well, except for impurities/oxides remaining on the outside of a liquid drop, so fluxes are often used to prevent heterogeneous nucleation).

@JonCuster Perhaps I’ve made myself unclear by referring to the “melting point” thus far, but I’m actually more interested in the freezing process.

Well, the bulk freezing and melting temperatures are the same... So, turn it around. You have some amount of liquid. You take a little energy out - the temperature goes down. You keep doing that until you reach the melting point. The liquid is quite happy to stay liquid at the melting point. Now you take the next bit of heat out - the temperature drops slightly below the melting point. Solid wants to nucleate now, and will eventually. Enthalpy is released, until the temperature is the melting temperature. Pull a little more heat out, and the solid fraction increases.

You probably know that water can be supercooled down to well below -30°C. And it has probably been suggested (if not outright stated) that at those temperatures, the water is in a "metastable" liquid state and that "eventually" it will spontaneously solidify - even if "eventually" may require trillions of years. Since thermodynamics is empirical, you could wonder how that has been empirically observed...but let's not go there. You've been taught that ΔG is a measure of the spontaneity of a process at constant temperature and pressure and at equilibrium. (How a "process" happens when the system is at equilibrium (steady-state) is often ignored. It's fair to say that nothing "happens" at equilibrium, rather when we say "equilibrium process" we mean "a process occurring sufficiently close to equilibrium". But I think I'm getting off topic, again. In considering nucleation, it is vital you distinguish between homogeneous and heterogeneous nucleation. This is the crux of the issue. Water is almost never without contaminates which provide nucleation sites, even if those sites are only on the surface of the container holding the liquid (getting the water this pure is an enormously difficult feat, and how you create a liquid phase without a container I leave to your imagination (hint:rain)). So, when reading about nucleation, you need to distinguish between the situations where there is preexisting centers (contaminants) or whether the water is assumed to be absolutely pure and centers must form spontaneously. Perhaps to answer your question (it would have been most useful if you had copy/pasted the exact text from your book). Once centers (seeds?) exist, the continued growth is pretty much identical whether their origin was homo- or hetero-. Obviously there is no equilibrium between a dust particle (solid) and liquid water. So how can nuclei exist at the MP? Obvious, right? They're hetero-, no formation is required. (Here I avoid the issue of whether FP is (or is not) identical to MP). Wikipedia has a short article on Classical Nucleation Theory which may help. Consider a solid. At its surface, the atoms (or molecules) are different than in the bulk (for most solids, bulk properties aren't reached for top 2 or 3 (or more) layers) but just consider an ideal crystal. Obviously, the surface has fewer bonds and hence has higher energy (is less stable). We call that surface tension (or exactly equivalently, surface energy). The classical theory states it's always positive. This is the energy which needs to be "overcome" to get nuclei, and it's proportional to the surface area. You should note here that our discussion has gone down to the microscopic level, where classical thermodynamic quantities become less and less useful. In other words, the macroscopic classical world thermodynamics needs to be treated carefully here. Statistical thermodynamics is most suitable transition from the classical to the quantum world. These types of processes can be studied using either, but fundamentally the world behaves according to the rules of (statistical) quantum mechanics when you're concerned with atoms depositing on or leaving a surface. The perspective of Classical Thermodynamics is of limited use. (In this case, I think it is probably just as useful, if not more useful, than the quantum mechanical approach. But you're unlikely to get many "fundamental" insights using it.) I'll point out one other apparent flaw in the "activation energy" approach: if the energy barrier is symmetric, then at its peak there should be no preferential direction: the odds of the system moving towards the solid or towards the liquid should be exactly the same (i.e. 50:50), yet the Classical Nucleation Theory requires a fudge factor for this probability - it is an indication that there is more going on here than this approach can handle, that the nucleation activation energy is, at best an approximation. Meaning don't take it too seriously; unless it can be derived from the quantum mechanics, it has no fundamental basis. I don't know much about this area of chemistry. I don't know whether or not it can be derived from the QM, but given that the Classical Nucleation Theory article in wikipedia is speaking about radius as if it is continuous, then I have my doubts. In some systems you begin to see bulk properties at N=100 to N=1000 (N=number of particles), so you'd have to be above this threshold for "classical" behavior to emerge.


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How much Citric Acid to neutralise bicarbonate?
I live in an area with very hard water, where the bicarbonate in the water is 357  mg/L. I would like to add some citric to the tap water to bring it down to around 50mg/L, for the curious, this is for coffee brewing.
How much citric acid would I need to add per liter? If it would be a very tiny quantity, then it would be impossible to weigh it accurately since I have a scale that is only sensitive to ~0.1g. In this case, could I make some solution which I could say add some milileters to a liter of tap water?

Coffee brewing includes boiling, which should remove the bicarbonates anyway.

could you clarify? I don't boil it for long periods, just bring the water to a boil.

That should be enough. After boiling (if even briefly), don't you observe white precipitate in your water, or maybe a deposit on the walls of the vessel?

Yes that is true. But I don't think all of the carbonate would be neutralized? Also I would like to avoid having to descale the kettle as well


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Is sedimentation coefficient constant and why?
Biochemistry by Voet writes the sedimentation rate of a macromolecule in a  centrifuge this way:
$$s=\frac{v}{\omega^2r}=\frac{M(1-\bar{V}_\rho)}{Nf}$$
where $\omega$ is the angular velocity of the rotor, $r$ is the position of the macromolecule with respect to the rotational axis, $M$ is its molar mass, $\bar{V}_\rho$ its partial specific volume, $N$ the number of moles, and $f$ the frictional coefficient.
The book further discusses the uses of $s=M(1-\bar{V}_\rho)/(Nf)$ and the information it reveals. Since they haven’t discussed any changes to $s$, it seems to me that they are holding it constant, but why should that be? If $s$ is a function of $r$, and $r$ changes, then $s$ should change too. Is my reasoning wrong? Is $s$ considered to be constant and, if so, why?

Sedimentation due to Gravity
The sedimentation coefficient of a particle falling in a fluid due to gravity is easier to understand. It is the ratio of the settling velocity ($v_{t}$) to the acceleration causing the settling($a_{g}$).
$$s=\frac{v_{t}}{a_{g}}$$
In fluids on earth $a_{g}$ is acceleration due to gravity which is known to be constant.
$v_{t}$ , which is also referred to as terminal velocity, is constant.  It is constant because the force driving the acceleration of the particle through a fluid ($F_{g}$) is equal in magnitude and in an opposite direction to the drag force ($F_{d}$).  Since these two forces cancel out the net force ($F_{n}$) is 0.
$$F_{g} - F_{d} = F_{n} = 0$$
There is no net acceleration $a_{n}$ since $F_{n}=m_{p}a_{n}=0$.  The particle mass ($m_{p}$) cannot be zero so there is no net acceleration of the particle, thus the terminal velocity is constant.
A ratio of two constants are constant so the sedimentation coefficient is constant.
Sedimentation due to Centrifugal Acceleration
Radial acceleration, $a_{r}$, equals the angular velocity $\omega$ squared times the radius, $r$.
$$a_{r} =\omega^{2}r$$
If the angular velocity is constant, which it is for most of centrifugation, and the radius is constant, then the radial acceleration, like the acceleration due to gravity, is constant.
The sedimentation velocity in centrifugation, just as in gravity settling, is constant.  It is constant because the drag force of the particle in the fluid is balanced by the centrifugal forces generated by the centrifuge. A centrifuge can generate forces several times higher than gravity so settling occurs much faster during centrifugation.
Again, the ratio of two constants are constant so the sedimentation coefficient is constant.
The other equation looks like it is a derivative of Stoke's law, which is used to determine the drag force of a particle in a fluid.

But why is the radius constant? You have a particle moving radially outward as it sediments. This does not suggest to me that the radius of its motion about the rotor axis is constant.

The radius is approximately costant, the flows are approximatly laminar, the particle is approximatly smooth, the particle interference is approximately zero. These are the simplifying assumptions going into the model. Models which are simple work better. The United States Naval laboratory discovered this phenomena and developed the KISS principle to help people remember this.

If centrifuges had really, really long tubes then the radius change would need to be in the model.

How long would such tubes need to be?

This is a different question

Could you give an example nonetheless? Just some figures to show that indeed the length of a typical test tube is not sufficient to alter the value of the sedimentation coefficient appreciably. I’m still not entirely convinced by your claim; how long are you talking about?

No.  You are confusing a claim with a simplifying assumption.  A claim is asserted as true; however, a simplifying assumption is assumed to be close enough to true to get a model working.  A model should not be used in practice if any of the simplifying assumptions are highly questionable.  This happens frequently in practice and should be common sense.

When a particle of mass m moves through a fluid under the action of a force, such as gravity, then a steady or terminal velocity $dz/dt$ can be reached as the frictional effect of the fluid balances the effect of the applied force. An example could be dropping a ball-bearing into a tall column of thick oil. In the case of a molecule a far greater force that that due to gravity is needed and this can be found in a centrifuge.
The force on a particle is reduced from simply mass times acceleration as in a fluid the particle will have some buoyancy. The force due to acceleration $a$ becomes
 $F=(1-\rho/\rho_m)ma$
 where $\rho$ is the fluid's density and
$\rho_m = M/V$
is the density of the particle where M is its molar mass and V the partial molar volume.
This force is balanced by the frictional force $f(dz/dt)$ where
$f = 6\pi\eta r_0$
is the (Stokes) friction for a sphere of radius $r_0$ in  a fluid of viscosity $\eta$ and has units $\pu {Pa\cdot s\cdot m}$. The friction is related to the diffusion coefficient ($\pu{m^2s^{-1}}$) as
 $D=kT/f$.
Equating the two forces produces
$$ f\frac{dz}{dt}=(1-\rho/\rho_m)ma   $$
In a centrifuge the rotational motion produces a force far greater than that due to gravity alone and on a mass m produces
$F=(1-\rho/\rho_m)m\omega^2 r$
where r is the distance from the centre of rotation.
$$ f\frac{dz}{dt}=(1-\rho/\rho_m)m\omega^2 r   \tag 1$$
The sedimentation constant is defined as
$$ s= \left(\frac{1}{\omega^2 r}\right )\frac{dz}{dt}$$
and has units of $\pu{10^{-13}s^{-1}}$ (Svedberg's). This is a constant because $dz/dt$ is the steady terminal velocity measure at positoin r. Typical values for proteins are $2$ to $200\cdot \pu{10^{-13}s^{-1}}$
Equation $(1)$  can be re-written as
$$f=(1-\rho/\rho_m)m/s$$
which in not particularly useful but substituting for the diffusion coefficient produces
$$ m=\frac{skT}{D(1-\rho/\rho_m)}$$
from which the mass can be obtained.
There is a second method used in centrifugation which is the sedimentation equilibrium method. When equilibrium is reached the diffusion inwards caused by the concentration gradient balances the centrifugal force pushing molecules outwards. If a solute of low molecular weight is centrifuged there will be at equilibrium a concentration gradient, and hence density gradient along the tube.
If a molecule of high molecular mass is now added it will  move outwards until its buoyant density is matched by the density of the solution. If a mixture of molecules is added then several bands will appear as each reaches equilibrium.


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When does phenolphthalein containing solution change colour?
We were estimating the amount of dissolved $\ce{CO2}$ in water using the American Public Health Association method. It was a Titrimetric Method using phenolphthalein indicator.
Titrant used was $\ce{NaOH}$ and analyte was sample water.
The following questions came to my mind:

Why does the phenolphthalein $\ce{[HIn]}$ solution change colour at pH 8.2-10 when addition of a little titrant $\ce{NaOH}$ could result into production of some $\ce{In−}$ ions, the ions that give the colour to the solution. Why does it show colour change in only  when there's $10^{-5.8}$ to $10^{-4}$ free $\ce{OH-}$ in the water?
Is the source of this resulting basic pH, the titrant $\ce{NaOH}$? I think so because all the $\ce{OH-}$ formed from the reaction of titrant $\ce{NaOH}$ and analyte $\ce{CO2}$ reacts with the $\ce{H+}$ ions of phenolpthalien. So its only possible that the shift from pH 7 to pH 8.2 is due to the $\ce{OH-}$.

I am not absolutely sure I understood the question completely, but I will try to shed some light on the colour change of an indicator.
As a rule of thumb the human eye can make out a change in colour when there is a 10:1 ratio of the components. Since phenolphtalein, $\ce{HIn}$, is without colour, and its deprotonated form $\ce{In-}$ is pink, it means that a solution with

$\displaystyle \frac{n(\ce{HIn})}{n(\ce{In-})} = \frac{10}{1}$ will appear clear, while
$\displaystyle \frac{n(\ce{HIn})}{n(\ce{In-})} = \frac{1}{10}$ will appear pink.

Since phenolphtalein is a weak acid, you will see, that the colour change will theoretically happen around equilibrium constant, i.e. $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$. (Which is also a simplification, because phenolphtalein has multiple acidic protons.) At that point you will probably not notice significant changes. Though phenolphtalein is a bit of an exception here, because it has only one colour. The human eye can probably see a smaller surplus of the colourful component. (It is much more difficult for an indicator like methyl red.)
If you substitute the above fractions into the Henderson-Hasselbalch equation, you can approximate the range in which the colour change will happen:
\begin{align}
\mathrm{pH}
  &= \mathrm{p}K_\mathrm{a} - \lg\left(\frac{n(\ce{HIn})}{n(\ce{In-})}\right)\\
  &=%
  \begin{cases}
     \mathrm{p}K_\mathrm{a} - \lg\left(\frac{10}{1}\right); & \text{clear}\\
     \mathrm{p}K_\mathrm{a} - \lg\left(\frac{1}{10}\right); & \text{pink}  \end{cases}\\
\mathrm{pH}
  &= \mathrm{p}K_\mathrm{a} \pm 1
\end{align}

When you are estimating the amount of carbon dioxide dissolved in water, then the following equilibrium will be present:
$$\ce{CO2 + H2O <=> HCO3- + H+ <=> H2CO3}$$
Adding hydroxide ions you are shifting this equilibrium to the right, essentially deprotonating the acid. Once you have fully (in first approximation) converted it to carbonate, you start deprotonating the indicator until you notice the colour change.


text:
Why do we say there's no carbon dioxide present?
Were were estimating dissolved $\ce{CO2}$ in water by  American Public Health Association method. It was a titrimetric method using phenolphthalein indicator.
Titrant used was $\ce{NaOH}$ and analyte was sample water.
Our instructor said that on adding phenolpthalein if the water sample turns pink prior to the experiment (addition of $\ce{NaOH}$) then it can be concluded that no free $\ce{CO2}$ is present . But how's it so?

Phenolphtalein's color change does appears at around pH=8.2.
If you just bubble $\ce{CO2}$ in water, the most of the $\ce{CO2}$ is dissolved into the water and only a small fraction converts to $\ce{H2CO3}$. But if the solution is very basic then all the $\ce{CO2}$ will react via a series of reactions and there is no free $\ce{CO2}$ left in water.
$\ce{CO2(aq) + H2O <=> H2CO3}$
$\ce{H2CO3 + 2OH^- <=> CO3^2- + 2H2O}$

Well more truthfully there is an insignificantly small amount of $\ce{CO2}$ in basic solutions since we discussing chemical equilibria. So $\ce{[CO3^{2-}] >> [CO2]}$

You're doing a titration of the carbonic acid $\ce{H_2CO_3}$ which is the product of the dissolution of $\ce{CO_2}$ in water  by a base here the sodium hydroxide. According to wikipedia for carbonic acid, $\ce{pKa_1=3.6}$, $\ce{pKa_2=6.3}$ and $\ce{pKa_3=10.32}$.
When the carbon dioxyde is in water, the solution is acid then your phenolphtalein is colorless. During your titration while not all the carbonic acid as react then the solution will still be colorless because phenolphtalein's color change appears at around $\ce{pH=8.2}$ (Ref). Then when you finish you're titration the color of the solution will change.

You need to make some prediction to find the theorical pH at the equivalence.
In this case at the equivalence $\ce{[H^+]=\sqrt{Ka_1Ka_2}}=1.26 \times 10^{-10}$ then $$\ce{pH_{eq}=-\log_{10}(1.26 \times 10^{-10})=9.9}$$
So at this point the phenolphtalein must be pink and then no more acid (so carbone dioxyde) will be in your solution. You'll have only carbonate ion which will not react with sodium hydroxyde.

I don't get this sentence - During your titration while not all the carbonic acid as react then the solution will still be colorless because phenolphtalein's color change appears at around pH=8.2. Carbonic acid reacts with what?

@SanjuktaGhosh I add some calculation for you. I let you find the result by yourself if you're doing chemistry it would be not too hard. :-)

I confused. How many reactions occur in the conical flask when we add 1)phenolpthalein to water 2)add NaOH to phenolpthalein containing water until equivalence 3)add NaOH beyond equivalence until end pt? Until you pointed that CO2 could form H2CO3 which inturn reacted with OH- of NaOH to form H2O I knew that only the two following reactions occurred in the flask: 2Na2OH + CO2 = Na2CO3, Na2CO3 + H2O + CO2 =NaHCO3. Now I have another question. If CO2 gets convertd into H2CO3 and the equilibrium b/w H2CO3 and its ions shift rightwards depleting the CO2 then how on earth will CO2 react with NaOH?

@SanjuktaGhosh wow this is much more than one question haha. I can answer you but it will take time and I need to sleep soon. Also I don't understand what "b/w" means ?

Its alright. I'm glad to receive a little consolation after being so much confused. b/w means between. Reply whenever you are free. Thank you.

I'm confused. How can a diprotic acid have 3 pka's !?!

@MaxW look at wikipedia. I have not detailed everything on my answer it can be a little tricky haha. But you know carbon dioxyde it quite special ! :)


text:
Am I using the Nernst equation correctly?
From my understanding the Nernst equation is
$$E = E^\circ - \frac{RT}{nF}\ln Q$$
Where $Q$ is the reaction quotient.
So for a Daniell cell with copper and zinc (valence of 2) shouldn't the coefficient in front of $\ln Q$ be $0.0128$ at $25\ ^\circ\mathrm{C}$? Everywhere I'm looking the coefficient seems to be $0.0295$ but I can't work out where this number comes from. Could somebody explain this please?

I will show you why $\ce{0.295}$ is correct. :-)

Nernst equation can be written in different ways, I let you choose which you prefer. So for any reaction such as $$\ce{\nu_{Ox}Ox + n e^- \rightleftharpoons  \nu_{Red}Red}$$
One can show that at $\ce{25°C}$,
$$E_{Ox/Red} = E_{Ox/Red}^{°} + \frac{RT}{nF}\ln\left(\frac{a_{Ox}^{\nu_{Ox}}}{a_{Red}^{\nu_{Red}}}\right) \tag 1$$
Which can also be written as, see after how,
$$E_{Ox/Red} = E_{Ox/Red}^{°} + \frac{0.06}{n}\log\left(\frac{a_{Ox}^{\nu_{Ox}}}{a_{Red}^{\nu_{Red}}}\right) \tag 2$$
Where $\log$ is the logarithm in base $\ce{10}$.

So we have $$\frac{RT}{F}\ce{=\frac{8.314 \times 298}{96500}}$$
And we know that $\log(x)=\ln(x)/\ln(10)$ then the coefficient next to $\log$ is $$\frac{RT}{F}\times \ln(10)=\frac{8.314 \times 298}{96500} \times \ln(10) = 0.0591 \approx 0.06$$
To be a little more perfect I sometimes use $\ce{0.059}$ instead of $\ce{0.06}$

So for two electrons exchange the factor coefficient is $\ce{0.059/2}=0.295$


text:
How to determine the equivalent weight of CO2
We were estimating dissolved $\ce{CO2}$ in water by American Public Health Association method. It was a titrimetric method using phenolpthalein indicator.
Titrant used was $\ce{NaOH}$ and analyte was sample water.
After the end point was reached and the volume of titrant estimated the ultimate calculation was done to find out the amount of $\ce{CO2}$ present in the sample per litre. The formula mentioned in the book was
$$\ce{CO2} \mathrm{(mg/l)} =\frac{\text{[Volume of NaOH} \times \text{Conc. of NaOH (in Normality)}\times 22 \times 1000]}{\text{Volume of sample water}}$$
$22$ here is the equivalent weight of $\ce{CO2}$. My question is how's it so?
NOTE:
Our teacher said that since 2 molecules of $\ce{NaOH}$ react with 1 molecule of $\ce{CO2}$ to form $\ce{Na2CO3}$ so $n=2$ here.
Is it right? I doubt it because I have always known $n =$ the number of $\ce{H+}$ or $\ce{OH-}$ released by an acid /base or the number of electrons released or accepted in a redox reaction.
The reaction here is
$$\ce{2NaOH + CO2 -> Na2CO3 + H2O}.$$
But as $\ce{CO2}$ is not an acid or base and even doesn't undergo redox so how do we determine its equivalent weight?
P.S. I'm a Zoology student and have learned chemistry for XII + 2 years.

This is the danger of using formulae from books without fully understanding where the formulae came from.

So let's take it step by step.
We have a titration of an unknown concentration solution with a known (standardised) solution.
The reaction $$\ce{2 NaOH + CO2 -> Na2CO3 + H2O}$$ gives us the stoichiometric relationship between $\ce{NaOH}$ and $\ce{CO2}$, i.e. 2 moles of $\ce{NaOH}$ react with 1 mole of $\ce{CO2}$.
How much $\ce{CO2}$ Was used? Half the amount of $\ce{NaOH}$!
How much $\ce{NaOH}$ was used?
$$\begin{align}
c(\ce{NaOH})  &= \frac{\text{amount of substance}}{volume}\\
              &= \frac{n_{\ce{NaOH}}}{V}\\
n_{\ce{NaOH}} &= c(\ce{NaOH})\times V\\
n_{\ce{CO2}}  &= \frac{1}{2} n_{\ce{NaOH}} \\
              &= \frac{1}{2} c(\ce{NaOH})\times V\\
\end{align}$$
What is the concentration of $\ce{CO2}$ in sample?
$$c(\ce{CO2}) = \frac{n_{\ce{CO2}}}{\text{Volume of sample}} = \frac{1}{2}\frac{c(\ce{NaOH})\times V}{\text{Volume of sample}}$$
Concentration is also expressed as $\mathrm{mg/L}$. 1 mole of $\ce{CO2}$ weighs $44~\mathrm{g}$ or $44\times 1000$ mg so sticking this into the above
$$c(\ce{CO2})\mathrm{(mg/L)} = \frac{1}{2}\frac{c(\ce{NaOH})\times V}{\text{Volume of sample}} \times 44 \times 1000 = \frac{c(\ce{NaOH})\times V\times 22\times 1000}{\text{Volume of sample}}$$
That's where the $22$ came from.
This may not address exactly where the equivalent weight of $\ce{CO2}$ comes from but at least it shows you the derivation of how the $22$ got into the formula.

I have a follow up question: While calculating the number of moles of NaOH required to perform the reaction, We took the conc. of NaOH (in M) and volume of the titrant (which is usually in ml). Shouldn't we convert the $ml$ into $l$?

Simple answer is yes. This is relatively straightforward but is a common stumbling block for many. Molarity, M, is the number of moles of substance in 1 litre. So a 0.5 M solution of NaOH has 0.5 moles in 1 litre of solution. So 1 litre is 1000 mL, so we have 0.5 moles in 1000 mL, or 0.0005 moles in 1 mL. If the volume of titration was 19.5 mL, the number of moles used is 0.0005 moles/mL x 19.5 mL = 0.00975 moles. So make sure units are similar and they should cancel out to give the right unit for the variable you are looking for.

If your end point is a consistent light pink, around a pH of 8.3 with phenolphthalein then this calculation is wrong. While it does take $\ce{2NaOH}$ to strip both hydrogens off of $\ce{H2CO3}$, at a pH of 8.3 there is little to no carbonate formed. Instead it is nearly all bicarbonate. The reaction from $\ce{NaOH}$ and $\ce{H2CO3}$ to bicarbonate ($\ce{HCO3}$) is 1 to 1. So instead of $\pu{22g/mol}$ you would just use $\pu{44.01g/mol}$ (the molar mass of $\ce{CO2}$)

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text:
Is iron pyrite dangerous to handle?
I have a cube of pyrite similar to:

(Source: Britannica)
Can pyrite turn into sulfuric acid?
Is this dangerous for me to touch, especially if my hands are sweaty?

No, pyrite is not dangerous to touch. "Can turn into" is an awfully broad term, it does not mean anything, unless you specify the conditions. Table salt can turn into a poisonous gas (if electrolyzed, that is). Water can turn into an explosive gas. So what? Nothing.

Thanks for the response Ivan. Good point. By touch I just mean to play around - roll it around in my hands etc.

Yes, I got that. No, not dangerous.

Also I should probably have been clearer with the conditions you mentioned. By sweaty hands I was referring to the pyrite getting water on it. I heard pyrite on mine sites turns into acid mine drainage when it gets water on it and oxidises. So I was worried the sweat from my hands would turn my pyrite into sulphuric acid.

OK, if you'd wear it like a jewel in constant contact with your skin for years, then maybe it would be somewhat dangerous for the stone.

Take it for granite, it's OK. Just don't eat it with St. John's bread, or you'll get pyrites of the carob bean.

Hahah, took me a bit of google ("Showing results for pirates of the carob,
Search instead for pyrites of the carob") for that to joke to click. Also thanks Ivan, I can roll it in my hands to my hearts content.

Pyrite does not turn into sulfuric acid. You can burn it with oxygen to give iron oxide and $\ce{SO2}$.
$\ce{4 FeS2 + 11 O2 -> 2Fe2O3 + 8 SO2}$
$\ce{SO2}$ with water gives sulfurous acid. To really get sulfuric acid, $\ce{SO2}$ needs to be oxidised to $\ce{SO3}$ (Contact process), and then reacted with water. Actually with more sulfuric acid to give oleum, which decomposes into two equivalents of sulfuric acid upon dilution with water.
The surface of the pyrite will react very slowly in the acidic conditions in your hand. Very, very slowly. There might be a faint smell of $\ce{H2S}$ recognisable if you handle the pyrite for a while in your sweaty hands.
(Also with water and oxygen from the air, pyrite will slowly react, to finally give sulfuric acid, which will speed up further reaction/degradation. But this typically needs decades of very bad ventilation.)

Actually the decomposition of pyrite specimens is a well known problem in the mineral collecting community. It isn't dangerous to the collector, but the specimen can be destroyed. See left hand column on page 2 of the NPS pdf https://www.nps.gov/Museum/publications/conserveogram/11-02.pdf

A nice crystal as shown in the picture above is not likely to be susceptible to decomposition by a bit of humid air, i guess.

You guess wrong...

@MaxW right, but on what timescale?

Evidently if the pyrite specimen is kept in a very dry case then it will last a very long time. In humid climates perhaps just centuries. The problem became known because museum mineral collections date back over several centuries.

This is my pyrite right here. I got it from a jeweler. It is not dangerous to handle. However touching to much could cause it to lose its lust and shine. You could polish it with a polish rag to a tool that could remove rust. Again safe to handle just don't handle it to much

It's not a safety concern.  However you should be aware that it will oxidize slowly when surfaced and exposed.  This can happen in ores (e.g. at gold mines) and is know as "pyrite rot" in fossil specimens.

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text:
Why do different metal ions release different colors of light?
For example, potassium chloride released a purple colored flame?
Does it have to do with the energy level of the valence electrons in the metal cation that is being burned?

You are right, it depends on the energy level of valance electrons, but more precisely it depends on the difference between the highest occupied and lowest unoccupied levels' energies. If the energy difference is $\Delta E$ then the light emitted by the excited atom has a frequency $\nu = \frac{\Delta E}{h}$ so a wavelength $\lambda = \frac{c}{\nu}=\frac{h c}{\Delta E}$

How does this relate to the energy level of valence electrons? Lithium's valence electron energy level is 2, and it creates a red pinkish color. Barium , with a valence electron energy level of 6, creates a light green flame color. Potassium on the other hand, with a valence electron energy level of 4, burns a purple color, meaning that it has a high frequency. So is there even a relation between the valence electron energy level and the frequency of the wavelengths of the flames?

The energy level can be expressed in two different ways; the one, you are using is denoting the principal quantum number $n$ that only enumerates the energy levels but it doesn't say anything (directly) about their energy-value measured in Js or in eVs. When you solve the Schrődinger equation for an atom or ion you get the possible energies (eigenvalues) of your system and these are quantized so you can denote them with whole numbers ($n$, $l$, $m$ and $s$). However these are only serial numbers, not energies, so their difference is meaningless in this context.


text:
Kinetics of molecule reacting with another of itself
For the following reaction:
$$ \ce{O3 + O3 -> 3O2} $$
The rule (which makes sense for combination of two different moluecules) states that the rate constant R should be:
$$ R \propto \ce{[O3][O3]} = \ce{[O_3]^2} $$
But in the case of a combination of the same molecule this seems counter-intuitive. If you had double the amount of $\ce{O3}$, wouldn't the rate simply double (instead of increasing as the square)?

If you had double the amount of $\ce{O3}$, wouldn't the rate simply double (instead of increasing as the square)?

No. You seem to be really close to understanding why this would not be the case, though. To help bridge the gap, let's consider the kinetics of a general bimolecular reaction:
$$\ce{A + B -> C}$$
We know that: $$\ce{\frac{d[\ce{P}]}{dt}} \propto \ce{[A][B]}$$
and that doubling $\ce{[A]}$ doubles $\ce{\frac{d[\ce{P}]}{dt}}$ because when we substitute $[2\ce{A}]$ for $\ce{[A]}$, we get: $$\ce{\frac{d[\ce{P}]}{dt}} \propto [2\ce{A][B]}=2\ce{[A][B]}$$
If $\ce{A}=\ce{B}$, however, then it follows that we must also substitute $[2\ce{A}]$ for $\ce{[B]}$, which then gives us: $$\ce{\frac{d[\ce{P}]}{dt}} \propto [2\ce{A}][2\ce{A}]=4\ce{[A][A]}=4\ce{[A][B]}$$


text:
what ways of TS searches are available in Gaussian09
I have a system for which I could not find the transition state using Gaussian09 corresponding to the experimental product formation. I have tried
opt=modredundant

route command fixing the two atoms in the (C-C) bond formation. It worked well showing the vibration in the bond formation in this direction (one -ve frequency of about -200), however, when i release the constraint and do a TS search, it gave a TS for a remote bond rotation (no longer along C-C bond formation).
I have tried to do a PES scan along the C-C bond forming direction varying the bond length from 3 Angstrom to 1.2 Angstrom, there is no maxima (but a minimum) within this range; the energy of the system only increases when the C-C bond approaches too close to each other.
What other ways can I try?

I'm interested in this, can you explain how you want to find a TS using frozen coordinates? Every time you set a coordinate to a non-minimum configuration you will get at least one negative frequence, this doesn't say anything about if that's a TS.  And did you do berny afterwards?

yes, you are right, i think. when you freeze the atoms for which you expect the bond formation and do an opt=modredundant, you are checking if you could get a TS after you release the constraint. indeed a berny optimisation for TS search is done after you have confirmed the existence of a negative vibrational frequency in the direction of the desired bond formation.

You could also try the Synchronous Transit-Guided Quasi-Newton (STQN) method requested by Opt=QST2 or Opt=QST3 in the route section. It is described in many different places, including the Gaussian official web-site.
Note that

QST2 requires two molecule specifications, for the reactant and
  product, as its input, while QST3 requires three molecule
  specifications: the reactant, the product, and an initial structure
  for the transition state, in that order.

Another important thing to whatch for is that

The order of the atoms must be identical within all molecule
  specifications.

thanks Wildcat! yes indeed! this is one method that I forgot! will try and update if it works for my system!

it did not work for my system; it gave a C-C rotation outside the C-C bond-forming carbon atoms. This is the same as the remote bond rotation that I obtained when I release my constraint in TS search starting from the modredundant results..

@XZhang, without details it would be difficult to suggest something.


text:
Problem digesting sample for Ti analysis using ICP-OES
I'm having trouble digesting a paper sample for $\ce{Ti}$ analysis using our ICP-OES. I've been digesting my ash samples with 50% $\ce{H2SO4}$, but I'm still getting lower $\ce{TiO2}$ results than the target. What is the best way to digest paper sample with $\ce{TiO2}$?

When you say "trouble digesting" what is your trouble? You say the analyses come out other than expected, but do you mean "the solution is cloudy". Is this a fact? If the solution is clear it can still be a digestion problem for certain but if an analysis is off target it could be other issues. In short, why do you assume it is digestion?

@StianYttervik I have converted your answer to a comment since it would have probably been deleted anyway. You are close to being allowed to comment anywhere, good luck!

@StianYttervik: Hi Stian, trouble digesting because my sample doesn't get dissolved at all. I've added H2O2 and it works. My problem before about getting low results was actually because the process in our mill had a retention problem which we found out few weeks after I've done the test. All good now. Cheers mate

If you want to stay with a sulfuric acid type of digestion, there is this(1) paper that describes a microwave digestion using copper(II) and hydrogen peroxide to facilitate the digestion.
From my personal experience, if your lab is properly equiped, my first choice would be an $\ce{HF}$ digestion or a nitric/perchloric acid digestion.
1)"A rapid and simple microwave-assisted digestion procedure for
spectrophotometric determination of titanium dioxide
photocatalyst on activated carbon", ScienceDirect, Talanta 71 (2007) 1867–1872

EMPHASIS ON "PROPERLY EQUIPPED"!! If you don't immediately know how nasty $\ce{HF}$ is, research it thoroughly before you even consider beginning work with it. Same thing with perchlorate: research first, then decide whether it's feasible to use. In both of these cases, specific safety equipment/apparatus is usually required.


text:
What kind of unit is this [A*m^2/mol]?
I found a term on this website that gives a unit of [A*m^2/mol]. The equation is:
$k_d^{eff}=\frac{2RT\kappa^{eff}}{F}(t^+-1)\left(1+\frac{\text{d}\ln(f)}{\text{d}\ln(c)}\right)$
where R is the gas constant [J/mol/K], K is ionic conductivity [S/m], T is the temperature [K], t+ is unitless, f is unitless and c is the concentration [mol/m^3], F is the Faraday constant [C/mol].
This equation gives the unit $\frac{A \cdot m^2}{mol}$. What kind of unit is this one? I can see that they define it as electrolyte diffusional conductivity, but I can't find this unit and better definition anywhere in the literature.
Can anyone help me with this?

First thing to say, I don't know much about electrochemistry. But concluding from thermodynamics, rewriting your equation a bit and making dimensional analysis there are some things to say:

It seems that you interchange the words dimension and unit. So an example: Length is a dimension and meter, inches... are units that measure length. You are probably asking for the physical interpretation of the electrolyte diffusional conductivity and not for the specific unit it is measured in.
The logarithm can be applied only to dimensionless numbers. In order to calculate it, you have to divide by some $c_0$ usually this is given in mol/l. In addition both terms in brackets have to be dimensionless, because there is a sum with a dimensionless number and in sums the dimensions have to match. So the only term relevant for the dimension, is the fraction.
The resulting dimension of the fraction is in SI units [A / m] and not [A m$^2$ / mol]

Thank you for your answer.

Yes. I would like to know the interpretation of that conductivity.
I agree, but there is no $c_0$ in that equation. $c_e$ is solved using Fick's law, so it is in [mol/L]. If you look at Eq. 3.15, the term is multiplied with the logarithm of concentration again.
I agree. I get [A/m] if I ignore the term with the logarithm. But then the units in 3.15 don't match if that term is given in [A/m]. Or am I not getting something fundamental?

Units are not carried out from inside the logarithm or square radicals. Yes, this may lead to strange things like ln(m) unit that have no reasonable physical meaning. Furthermore, if an equation contains addition of values with different units, something is disastrously wrong. So I would cross-check the formula, as (1+k) member should by all common sense be fully unitless. I also see suspicious $d$, which may be a differentiation symbol, and than things become even more interesting.

That said, not all units have names and finding a weir unit somewhere in a table is not surprising. Especially in borderline areas.

I agree with everything you said. It is a weird notation, indeed. I have artificially assumed that 1 would have the unit [m3/mol], so that the units would be homogeneous. It is definitely incorrect, but I have the feeling that there is something weird about those equations, and it's not the first time I see it... And then there is the differential and the logarithm... So weird


text:
Order of Filling Orbitals vs. Order of Increasing Energy?
Is the order of filling orbitals the same as the order of increasing energy in orbitals, or is it different?
Please tell me if I am right or wrong:
Order of filling orbitals:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, etc...
Orbitals in order of increasing energy:
1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f, 5s, 5p, 5d, 5f, 6s, 6p, 6d, 6f, etc...
I am pretty sure I got the second one wrong but I am not being able to find an answer anywhere

Why would electrons fill in any order except that which minimises energy? The two lists should be (and unless I'm recalling and reading entirely wrongly) and are the same.

Energy is the currency of chemistry. If the order in which orbitals fill up and the order in which the energy increases were different, it would mean that that the atom has more energy than it needs. This can happen (the atom is then said to be "excited"), but typically the extra energy distributes into the environment quickly - e.g. by radiating the energy as light - and the atom gets back to the ground state.
That said, the orbital energies are not necessarily the same for all atoms. Getting the correct order is a tricky business, but if you stick to the Aufbau principle  (the first ordering you mentioned) you'll be pretty close with the guess. Don't bet on it for d and f elements though.


text:
Determination of the integrated rate equation for a reaction
We were given this equation to determine its integrated rate equation:
$$\ce {A + B <=>[$k_1$][$k_2$] C +D ->[$k_3$] E + F}$$
We were given these assumptions to simplify the derivation:

$\displaystyle \frac{\mathrm d[\ce{A}]}{\mathrm dt} = \frac{\mathrm d[\ce{B}]}{\mathrm dt}$, and $\displaystyle \frac{\mathrm d[\ce{C}]}{\mathrm dt} = \frac{\mathrm d[\ce{D}]}{\mathrm dt}$
At any time, $[\ce{A}]=[\ce{B}]$, and $[\ce{C}]=[\ce{D}]$, and $[\ce{E}] = [\ce{A}]+[\ce{B}]$
$\ce{F}$ is continuously removed from the system such that $[\ce{F}] = 0$ at any given time

The answer given to us was:
$$\ln[\ce{A}] = \frac{1}{2} \ln \left( \frac{k_2+k_3}{k_1}\right) + \ln\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2 [\ce{A}]\right)$$
However, I refuse to believe that an integrated rate equation can be without the time variable. Our teacher argued that the time variable was removed by using steady state approximation, and that $\frac{\mathrm d[\ce{C}]}{\mathrm dt} =0$. However, I think there is something wrong with using the rate of formation of an intermediate in deriving the integrated rate equation for a certain reaction. Can anyone please help me?
Edit: Here is how our professor got the answer:
$$\begin{align}
\frac{\mathrm d[\ce{C}]}{\mathrm dt}=\frac{\mathrm d[\ce{D}]}{\mathrm dt} &= k_1[\ce{A}][\ce{B}] - k_2[\ce{C}][\ce{D}] - k_3[\ce{C}][\ce{D}] \\
\frac{\mathrm d[\ce{C}]}{dt} &= k_1[\ce{A}][\ce{B}] - (k_2 +k_3 )[\ce{C}][\ce{D}]
\end{align}$$
$\ce{A}$ and $\ce{B}$ are the only species initially present such that:
$$[\ce{A}]_0 + [\ce{B}]_0 = [\ce{A}] + [\ce{B}] + [\ce{C}] + [\ce{D}] + [\ce{E}] + [\ce{F}] $$
Since $[\ce{C}] = [\ce{D}]$ and $[\ce{A}] = [\ce{B}]$ and $[\ce{F}] = 0$ and $[\ce{E}] = [\ce{A}] + [\ce{B}]$,
$$\begin{align}
[\ce{A}]_0 + [\ce{B}]_0 &= 4[\ce{A}] + 2[\ce{C}] \\
[\ce{A}]_0 + [\ce{B}]_0 - 4[\ce{A}] &= 2[\ce{C}] \\
\frac{[\ce{A}]_0 + [\ce{B}]_0 - 4[\ce{A}]}{2} &= [\ce{C}]
\end{align}$$
$$\begin{align}
\frac{\mathrm d[\ce{C}]}{\mathrm dt} &= k_1[\ce{A}]^2 - (k_2 +k_3 )[\ce{C}]^2 \\
\frac{\mathrm d[\ce{C}]}{\mathrm dt} &= k_1[\ce{A}]^2 - (k_2 +k_3 )\left(\frac{[\ce{A}]_o}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right)^2
\end{align}$$
By steady state approximation (SSA), $\frac{\mathrm d[\ce{C}]}{\mathrm dt} = 0$
$$\begin{align}
\frac{\mathrm d[\ce{C}]}{\mathrm dt} &= k_1[\ce{A}]^2 - (k_2 +k_3 )\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right)^2 = 0 \\
k_1[\ce{A}]^2 &= (k_2 +k_3)\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right)^2
\end{align}$$
Applying natural logarithms to both sides,
$$\begin{align}
2 \ln(k_1) + 2 \ln[\ce{A}] &= \ln(k_2 +k_3)+ 2\ln\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right) \\
2 \ln[\ce{A}] &= \ln(k_2 +k_3 ) - \ln(k_1)+ 2\ln\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right) \\
\ln[\ce{A}] &= \frac{1}{2}\ln\left(\frac{k_2 +k_3}{k_1}\right) + \ln\left(\frac{[\ce{A}]_0}{2} + \frac{[\ce{B}]_0}{2} - 2[\ce{A}]\right)
\end{align}$$
As you can see from the solution, he was able to eliminate time from the final equation by using the steady state approximation. Is this a valid move to get the integrated rate equation? Please help.

The steady state only applies in the 'middle' of the reaction, and not in the initial or final stages. You can see that initially the equilibrium has to be established (assuming only A+B present initially) and that at long times all the species will end up as E & F. In the steady state /equilibrium phase  concentrations do not change with time.

If we are not to use SSA in the derivation of the integrated rate equation? How then would the rate equation be derived?

The simple answer is with difficulty :) However, its a lot easier to do this numerically that algebraically. There is lots of code written in free languages such as python with which to do this. To solve any set of differential equations the initial conditions have to be stated.  If you want to do this algebraically then this is much harder and you need to consult some standard texts first. Your scheme can be solved; have a look at the equations for pyrene excimers or the Lindemann scheme as its very similar to this. If you have no luck in a day or two I can post an answer .

@porphyrin Hello, I have posted what I think is the answer, and I want to know what you think of it, thank you :)

After consulting the Lindemann mechanism, here is my attempt for an answer:
$$\frac{d[\ce{C}]}{dt}= k_1[\ce{A}]^2 - (k_2 +k_3 )[\ce{C}]^2 = 0$$
$$[\ce{C}]^2=\frac{k_1}{k_2 +k_3}[\ce{A}]^2$$
$$\frac{d[\ce{A}]}{dt}= -k_1[\ce{A}]^2 +k_2[\ce{C}]^2$$
Substituting $[\ce{C}]^2$,
$$\frac{d[\ce{A}]}{dt}= -k_1[\ce{A}]^2 +\frac{k_2k_1}{k_2 +k_3}[\ce{A}]^2$$
$$\frac{d[\ce{A}]}{dt}= \frac{-k_2k_1[\ce{A}]^2 + -k_3k_1[\ce{A}]^2 +k_2k_1[\ce{A}]^2}{k_2 +k_3}$$
$$\frac{d[\ce{A}]}{dt}= \frac{-k_3k_1[\ce{A}]^2}{k_2 +k_3}$$
Let $k'=\frac{-k_3k_1}{k_2+k_3}$, then
$$\frac{d[\ce{A}]}{dt}= k'[\ce{A}]^2$$
$$\frac{d[\ce{A}]}{[\ce{A}]^2}= k'dt$$
$$\int_{[\ce{A}]_o}^{[\ce{A}]} \frac{d[\ce{A}]}{[\ce{A}]^2}= \int_0^t k'dt$$
$$-\left(\frac{1}{[\ce{A}]}-\frac{1}{[\ce{A}]_o}\right) = k't$$
$$\frac{1}{[\ce{A}]}-\frac{1}{[\ce{A}]_o} = -k't$$
$$\frac{1}{[\ce{A}]}= -k't +\frac{1}{[\ce{A}]_o} $$
I wonder how is it correct when I also used the steady state assumption in my solution. How is it different from the answer posted by my professor? Which is more correct and corresponds to the principles of chemical kinetics? Why would the other one be wrong and the other correct? (That is if my current derivation is correct :) ).

Your answer seems to be ok once the equilibrium is established. In your original question the derivation stops at your third equation above and substitutions were made. You now assume that [A] is still changing with time and but you now assume that [C] and [A] are in a fixed ratio. (your second equation) .  As you realise some simple chemical  kinetics equations are very hard to solve and this is why we often try to use the steady state approximation. This particular scheme does seem to be one that is very difficult to obtain a full time-dependent solution.

Your derivation is fine, so feel free to accept your own answer (not sure if this is technically possible, though). Your professor's answer is absurd on more than one level.
Now, when solving differential equations with multiple unknown time-dependent functions, it is not uncommon to stumble upon certain relations involving some of these functions and not involving time (think of conservation laws in celestial mechanics, or look at your second equation which links $[\rm A]$ and $[\rm C]$), so this is not necessarily wrong per se. But your professor's answer is an expression with only one unknown, namely $[\rm A]$. So in fact this is a static equation which defines one particular value of $[\rm A]$. What's the meaning of it? This can't be the equilibrium concentration, for there is no equilibrium. As the second reaction is irreversible, it will inevitably drain the system of all $\rm{[C]+[D]}$, and also of all pathways leading there. The final concentration of $[\rm A]$ can't be anything other than $0$ (that is, unless we have insufficient $[\rm B]$, which is not the case). How this wild expression came about is hardly of any importance. It just can't be right.
Still, if we want to investigate further, we'll find the root of all evil: it is the nonsensical condition $\rm{[E]=[A]+[B]}$ at all times. Just how can that be true? As both reactions progress, $[\rm A]$ and $[\rm B]$ go down, while $[\rm E]$ goes up. To maintain the condition, we must use some sort of Maxwell's demon that would monitor the concentrations of $[\rm A]$ and $[\rm B]$ and manipulate $[\rm E]$ accordingly. But wait, any such manipulation would break the conservation law $\rm{[A]_0+[B]_0=[A]+\dots+[F]}$, which we relied upon, so the whole thing can't be true even then.
So it goes.


text:
Is there a systematic approach to determing structural formula from NMR spectra?
I have not been able to find a methodologically consistent approach. If NMR spectroscopy is a rational analysis, it must not be beyond logical description in principle.
Here is what I have surmised so far:

Calculate degrees of unsaturation (I understand this)
Determine functional groups from IR spectra (I understand this)
Figure out chemical shift of each peak (I usually understand this)
Calculate integration of each peak (I usually understand this)
Determine number of neighboring protons (I usually understand this)
"Put the molecule together" (Variable results)

I "usually" understand steps 3-5, but sometimes I can't honestly figure out how many peaks I am looking at. How can I reliably tell the difference between lines within a peak due to coupling and distinct peaks when they are sometimes so close together?
Step 6 is difficult. Sometimes I can deduce an answer, and sometimes I am wondering how to tell the differences between hundreds of isomers. How can I systematically determine the right answer (or even be able to recognize IF there is single valid answer)?

Is there a systematic approach to determing structural formula from NMR spectra? || there is none that would guarantee the result. For example, 2) and 3) in your list may be hindered by unusual group or combination of groups. and the only guaranteed option for 6) is exhaustive search and sometimes it is not enough and resort to 2D NMR is required. X-ray crystallography is superior in this regard, but AFAIK it is also an exhaustive search, only automated.

NMR is not the only technique to figure out the structure of unknown compound. Therefore, it is used in combination to others technique like MS, FTIR to find out the exact structure of a unknown compound.

Many molecules cannot be distinguished from one another with NMR. Usually, other methods are used along NMR. Nevertheless, there is software that can predict NMR spectra (IRC ChemDraw). Maybe you could do some practice in the opposite direction and confirm your answers with said software.

ACD/Labs’ NMR processor includes a structure predictor in the latest versions (or maybe that part hasn’t been published yet, but I definitely know they contacted a neighbour group of mine to ask for complicated NMRs to assign a structure to). It seems to be pretty good. However, unless you join the relevant group over at ACD, I doubt you’ll get hold of the algorithms.

Of course there is a systematic approach to solving structure. The key is that it is an iterative process, and it is not uncommon to have to repeat the cycle several times. For complicated molecular structures (for example, alkaloids with C35+ frameworks) you might get 20 or more possible candidate structures that you have to work through very closely.
Of course, the method you use will depend on the class of compound, but for unknown organic molecules, I would start with a molecular mass and NMR data from 1H, COSY, HSQC, HMBC. Subsequent iterations or refinements might require NOESY, TOCSY and if there is enough material maybe even an ADEQUATE.
The general approach that I teach is as follows. The first step is to identify the molecular formula from mass spec data and calculate the double bond equivalence. Subsequently:
1. Environments

From 1H and 13C data, determine number of carbon environments, and account for all carbon atoms, as these will represent the individual jigsaw pieces
Determine proton environments attached to 13C - tally up CH3, CH2, CH groups.
Tally up integrals, identify symmetry and equivalence.
This should account for almost all of your C and H atoms

2. Chemical shifts

Establish functional group types based on chemical shifts (aromatic, alkene etc).
Look for possibility of heteroatom inclusion that ties in with the molecular formula

3. Couplings and spin systems

From obvious 1D coupling and COSY data, start assembling spin system fragments
From HMBC start assembling C–C connectivities
Consider drawing out a correlation map to show connectivities, especially for complicated systems
You should be reasonably confident about the spin system fragments leading into the next stage

4. Structure hypothesis

Assemble spin system fragments into allowable possible structures, and test the structure against the data - that is rationalise the chemical shifts and couplings for all signals to see that they fit the model. If not, refine and try again.

A number of structure elucidation software packages exist, and they essentially base their method on building up a molecular correlation map, and then doing a chemical shift prediction for the proposed structure to give an overall confidence score. The sort of molecular map you might get looks something like the picture below. On the left are all the explained correlations, and on the right the incorrect correlations. What is important is look closely and the incorrect correlations, and rationalise them. In this case, all of the incorrect correlations would be allowable, and therefore we would have good confidence in this structure (which is understandable, as we know exactly what it is!). If we couldn't rationalise some of the correlations, we need to either reconsider our molecular structure, or re-analyse our data.


text:
Reaction of Grignard reagent with ammonia
What is the exact reaction mechanism of the reaction of Grignard reagent with ammonia?
$$\ce{RMgX + NH_3 -> RH + MgNH3X}$$
[I found the above equation from a book, and I am not completely sure that that reaction actually happens]

Why do you think the reaction might not be taking place?

@EashaanGodbole Because I took it from an unreliable book.

Oh...what I meant to ask was what you think is the chemical/physical reason why such an interaction doesn't take place.

Your reaction isn't balanced. It should read:
$$\ce{RMgX +NH3 -> RH +MgNH2X}$$
This is a simple proton transfer. $\ce{MgX+}$ is a spectator ion, so the reaction equation could also be written:
$$\ce{R- +NH3 -> RH +NH2-}$$
When the carbanion collides with an ammonia hydrogen, the hydrogen is transferred between the two species. The result is a newly formed carbon to hydrogen bond and an amide anion. Carbanions are several orders of magnitude more basic than amide ions, so the reaction definitely occurs, and will essentially go to completion.

Well, it shouldn't be written like that. Grignard compounds don't generate free carbanions.

I'd be hesitant to say that $\ce{MgX+}$ is a spectator ion. It is effectively that way, but the M-C bond is actually pretty covalent.

My point wasn't to imply that it doesn't participate in the reaction or that it isn't closely coordinated to the anions in solution, simply that it exists in the same form on both sides of the equation (though it is much more closely coordinated to one species on one side than the other).


text:
Carbon dioxide sequestration in Limestone
The chemical reaction where dissolved carbon dioxide forms carbonic acid and then reacts with various calcium containing minerals to form calcium carbonate and bicarbonate is rather slow. I know that several years ago there was research into possible ways to speed up the reaction, possibly as a candidate for a truly long term way to scrub $\mathrm{CO}_2$ from the atmosphere. The inspiration for this idea is to speed up the part of the geologic carbon cycle that regulates atmospheric $\mathrm{CO}_2$ concentration.
Does anyone know how the research into catalysts or processes to speed this up are going?

This should give you a start on recent developments in the field: https://scholar.google.com/scholar?as_ylo=2016&q=carbon+sequestration&hl=en&as_sdt=0,39

This process does not rely on catalysts, but it addresses the essence of the question. The process described is to inject liquefied carbon dioxide into basalt formations, even if the basalt is very old. This makes the it plausible that we could scrub the excess $\mathrm{CO}_2$ from the atmosphere by:

growing plants,
burning those plants, capturing the CO2, and
burying that carbon using, partly, the energy from the burned plants.

Yes, but nature has already converted plants to more energy efficient hydrocarbon compounds (oil & gas) via heat and pressure.

Going via plants seems rather an overelaborate way of getting carbon dioxide. Why not just capture it from the atmosphere chemically? Or, just grow plants and don't burn them.

@matt_black Simple - concentration. This process goes this fast because the carbon dioxide is highly concentrated. Doing that mechanically from the atmosphere requires the input of energy. Doing it from an exhaust flue requires less energy because it's more concentrated.

@SeanLake You still don't need to grow plants and burn them as we already burn vast quantities of coal, oil and gas and could put the capture sequestration on their exhausts.

@matt_black Preventing fossil carbon from entering the atmosphere doesn't remove carbon that was dumped there in the past, which is the goal.

@SeanLake But avoiding emissions would have exactly the same effect on the atmosphere as sequestering already emitted gas. The only significant difference is economics. Besides, growing plants already sequesters carbon dioxide and burning them just wastes extra energy.

No, it does not. We're already at the point where we want to reduce CO2, and waiting for the natural geologic cycle while we just stop making the situation worse won't cut it. Also, plants don't sequester carbon in a meaningful way, since it's still part of the biologic carbon cycle. There are only two long-term sequestration strategies that I'd trust: CO2 to limestone, and burying charcoal.

@SeanLake a net increase in growing plants will lead to removal of CO2 from the atmosphere. More forests = less CO2. And the major geological route to remove CO2 is weathering of olivine-rich rocks. Injecting CO2 into such rocks has been proposed as a way to speed this process up to reduce atmospheric cO2 meaningfully on human timescales.

@matt_black "Injecting CO2 into such rocks has been proposed as a way to speed this process up to reduce atmospheric cO2 meaningfully on human timescales." Did you even read the question/answer? That's what I said.
Also, no net increase in plants will be sufficient because atmospheric $\mathrm{CO}_2$ is not the major bottle neck in plant growth.


text:
Why would Ammonium Nitrate spontaneously combust/detonate when damp?
Why would small amounts (1g or less) of Ammonium Nitrate spontaneously combust/detonate when damp? Possible temperature excursions to +70degC. Unfortunately I cannot give you the context for commercial confidentiality reasons.
I was under the impression it was very stable in small quantities.

To my knowledge ammonium nitrate only becomes explosive with certain additives. So perhaps you should ensure that your samples are what you think they are. Be safe!

@BiggChemT Ammonium nitrate is responsible for one of the most famous industrial explosion killing more than 500 people. They though it stable, too. https://en.m.wikipedia.org/wiki/Ammonium_nitrate_disasters

Ammonium nitrate is not, on its own, an explosive,[11] but it readily forms explosive mixtures with varying properties when combined with primary explosives such as azides or with fuels such aluminum powder or fuel oil. Wikipedia


text:
Hydrocarbons and their derivatives
Which alcohol is used in spirit lamp?
Accoding to my text book, the answer given for this question is Ethyl Alcohol but one of my teachers told me that it is Methyl Alcohol.
Please somebody help me with this question.

the methylated spirit is used in spirit lamp.
but methylated spirit contains methyl alcohol & ethyl alcohol both.
I think, it should be Ethyl alcohol.


text:
Osazone crystal structures
What is the reason behind the formation of different shapes of crystals for different osazones? Like glucose osazone- needle shaped, maltose osazone- sunflower shaped, lactose osazone - powderpuff shaped. Does this involves hydrogen bonds only or something else? If so then can I be able to determine the crystal shape of different osazones by simple calculations?


text:
Why can one product in an equilibrium system not be produced without the other?
I performed an experiment where $\ce{CuSO4 (aq)}$ reacted with $\ce{NH3 (aq)}$ in one test tube, and in the other, $\ce{CuSO4 (aq)}$ reacted with $\ce{NH3 (aq)}$ and $\ce{NaOH (aq)}$. I know two products were formed, one being the complex ion $\ce{[Cu(NH3)4]^{2+}}$.
The goal of the experiment was to see if one of the products could be produced without the other, and I know that is not possible due to the solutions being at equilibrium, but why is this?

Your question is not quite clear. What is the "one" product, what is the "other", what where your observations, and what is unclear about all that?

In the first experiment, the dissolved $\ce{CuSO4}$ reacted with $\ce{NH3(aq)}$ to form the soluble, deep blue/purple complex via equation 1, $\ce{[Cu(NH3)4]^{2+}}$, just as you said.
1) $\ce{Cu^{2+} + 4NH3 <--> [Cu(NH3)4]^{2+}}$
Depending on the concentration of $\ce{NH3}$, you would also have precipitated some amount of insoluble $\ce{Cu(OH)2}$ from the hydroxide formed by the ammonia reacting with water as in equation 2:
2) $\ce{NH3 + H2O <--> NH4+ + OH-}$
The equilibrium of equation 2 lies far to the left, so when you add more ammonia, you aren't adding an equal amount of hydroxide.
The difference in the second experiment is that you explicitly added excess hydroxide.  In this case the high hydroxide concentration would have pushed the dominant product to $\ce{Cu(OH)2}$ via equation 3.
3) $\ce{Cu^{2+} + 2OH- <--> Cu(OH)2}$ (solid)
Regarding the statement "The goal of the experiment was to see if one of the products could be produced without the other":
The products of both experiments were the same, just to different degrees; in other words, under different pH conditions, different products predominated.  In the first experiment, you had far more ammonia than hydroxide, so equation 1 dominated; lots of $\ce{[Cu(NH3)4]^{2+}}$ and a small amount of $\ce{Cu(OH)2}$.  In the second experiment, you added excess hydroxide so lots of $\ce{Cu(OH)2}$ was formed an very little $\ce{[Cu(NH3)4]^{2+}}$.
So the upshot is that these reactions are in equilibrium, and because of equation 2 you always had both $\ce{NH3}$ and $\ce{OH-}$ in your reaction mixture so that both equations 1 and 3 always produced some small amount of their respective products, $\ce{[Cu(NH3)4]^{2+}}$ and $\ce{Cu(OH)2}$, respectively.


text:
Why does hand sanitizer heat up?
While washing my hands with soap I added some hand sanitizer just because it was right there on the counter and couldn't resist mixing them... Ha!  However, when I added the hand sanitizer it heated up for a moment and then cooled back down.  Why does this happen?

Hand sanitizers are generally ~50% - 70% ethanol or isopropanol. When these alcohols mix with water, they may release heat. Starting at room temperature, mixing ~100% ethanol with an equal volume of water will raise the temperature of the mixture ~8 0C.

It sounds like some type of exothermic reaction happened between the hand sanitizer which is primarily alcohol, and the soap.


text:
Can solvents like 99.9% isopropyl alcohol and/or n-hexane be "carbonated" with CO2?
I'm working on a prototype for a new machine, and I'm trying to get solvents like $99.9\%$ isopropyl alcohol or n-hexane to be carbonated like a soda pop at cold temperatures. I have the machines and everything for it I just want to make sure I'm not going to explode or anything.

Not sure how much $\ce{CO2}$ will dissolve in isopropyl alcohol or n-hexane, or what you are trying to do. The obvious safety concern is that both of those liquids are flammable. So a rupture in a pressurized container could spew a spray of flammable liquid which could ignite from some ignition source in the room. Such a spray would basically create a fuel vapor bomb (https://en.wikipedia.org/wiki/Thermobaric_weapon) or a BLEVE (https://en.wikipedia.org/wiki/Boiling_liquid_expanding_vapor_explosion).

Well, one of the reason sodas behave nicely while carbonated is because the dioxide is in fact carbonic acid and escapes the liquid calmy and orderly (or "effervescence" as they call it) because there is a chemical reaction happening. I am not sure how the CO2 would behave in other solvents, the dissolution can happen quite immediately (as is the case in sodas if you create nucleation sites by shaking or adding fine/microporous particles) and this can, as MaxW hints at, have catastrophic consequences.

The following are published[1] solubilities of $\ce{CO2}$ in several organic solvents at 1 atm and $\mathrm{20^oC}$. Solubilities are given in mL/g:

acetone: 8.2
  ethanol: 3.6
  benzene: 2.7
  methanol: 4.1
  toluene: 3.0
  xylene: 2.3
  heptane: 2.8
  methyl acetate: 7.4
  diethyl ether: 6.3

For reference, the solubility of $\ce{CO2}$ in water is about 0.9 mL/g under the same conditions.
1) Topham S et al; Carbon Dioxide. Ullmann's Encyclopedia of Industrial Chemistry. 7th ed. (1999-2015). New York, NY: John Wiley & Sons


text:
Copper and water heat capacity concern

What mass of copper, originally at $50\ ^{\circ}\text{C}$ must be added to $1\ \mathrm{kg}$ of $10\ ^{\circ}\text{C}$ water to raise the water temperature by $10\ ^{\circ}\text{C}$?

We can work out that the necessary energy gain by the water is $41.8\ \mathrm{kJ}$, but I'm not sure how to work out the mass of copper, since both the mass and the temperature change is unknown in the equation $\Delta E=mc\Delta T$? Or do we assume that the copper cools to $20\ ^{\circ}\text{C}$ also?

There are all elements in your exercise to solve it expect the heat capacities but this is easy to find them.
You are doing an exercise of thermodynamics and this is not mention that the equilibrium is not reached. So if your water heats up only by $10\ \mathrm{^\circ C}$ at the end even in one million years if your system is adiabatic, your solution will still be at $20\ \mathrm{^\circ C}$. And so at the equilibrium the copper must have the same temperature than the water.
So if you use $\Delta H=mc_p\Delta T$, for water $\Delta T=10\ \mathrm{^\circ C}$ and for copper $\Delta T=30\ \mathrm{^\circ C}$.
Are you able to finish now ? :-)


text:
Health Risks Associated with Working in Chemistry Labs
I've been working in a contaminants food quality lab for about 2 years now and was curious what the chronic health effects would be, should I be worried about cancer?

I think this is far too broad. We don't know what chemicals you are exposed to on a daily basis, how much exposure you get, etc. etc. Obviously as chemists we try to minimise such risks but without a lot more details I'm not sure if anybody can really say anything useful.

I work on several different methods, I work with actonitrile, cyclohexane, residual solvents, and volatiles (containing aromatics) quite often

Thinking about safety is the key consideration. Frankly people are terrible at assessing risks. There are 50,000 deaths in car accidents every year. There are not nearly that many chemists dying from lab accidents.  You're in more risk driving to work that you are in the lab. Are you going to stop driving and traveling in a car too?

I would recommend looking at the MSDS information for anything that you work with specifically and/or researching the long term health effects of those specific substances

I have, I just wasnt sure on a timeline of how long it would take to develop chronic illnesses, I'm currently trying to get out of the lab even though i love it.


text:
Extract all structures of Gaussian 09 molecular dynamics calculation using babel?
Is there a way to easily extract all the structures and energies from an ADMP calculation done in Gaussian 09? GaussView can plot them but not export them in a useful way and babel just gives me the last structure in there.
I tried
babel <G09output> all -oxyz <output.xyz>

but I get just
==============================
*** Open Babel Error  in OpenAndSetFormat
  Cannot open all
1 molecule converted
1 errors 16 audit log messages

Or is there a different easy way to do this? Otherwise I guess I have to write my own script or look into cclib, but that one is using babel too if I understand it correct?
Example output file: http://dl.dropboxusercontent.com/u/6111391/Tz_TCO_dyn6.txt
Input:
%nprocshared=16
%mem=56GB
%chk=Tz_TCO_dyn6.chk
# 6-31g(d) scrf=check geom=(connectivity,crowd)admp=(stepsize=2500,maxpoints=400) rm062x IOp(1/44=990210)

[No Title]

0 1
 C                  1.08049600   -0.58720100   -2.07918800
 C                 -1.08049600    0.58720100   -2.07918800
 C                 -0.58945000    0.34949100    0.13978700
 H                 -0.52250100    1.43457800    0.23933400
 C                  0.58945000   -0.34949100    0.13978700
 H                  0.52250100   -1.43457800    0.23933400
 N                 -1.13318600   -0.73433600   -2.39136300
 N                  0.00078600   -1.35005000   -2.39250200
 N                 -0.00078600    1.35005000   -2.39250200
 N                  1.13318600    0.73433600   -2.39136300
 C                  1.84657900    0.28440600    0.66016000
 H                  2.75382900   -0.14516700    0.22100400
 H                  1.84124500    1.35503200    0.42328200
 C                 -1.84657900   -0.28440600    0.66016000
 H                 -2.75382900    0.14516700    0.22100400
 H                 -1.84124500   -1.35503200    0.42328200
 C                  1.87373600    0.08444000    2.19307200
 H                  2.72499800    0.64701600    2.59375900
 H                  2.07520100   -0.97273600    2.41061700
 C                 -1.87373600   -0.08444000    2.19307200
 H                 -2.72499800   -0.64701600    2.59375900
 H                 -2.07520100    0.97273600    2.41061700
 C                 -0.58945000   -0.50859700    2.93023900
 H                 -0.86607100   -0.70606500    3.97132400
 H                 -0.24860000   -1.47389400    2.53270700
 C                  0.58945000    0.50859700    2.93023900
 H                  0.86607100    0.70606500    3.97132400
 H                  0.24860000    1.47389400    2.53270700
 H                  2.02862900   -1.10167200   -1.96939700
 H                 -2.02862900    1.10167200   -1.96939700

 1 8 1.5 10 1.5 29 1.0
 2 7 1.5 9 1.5 30 1.0
 3 4 1.0 5 2.0 14 1.0
 4
 5 6 1.0 11 1.0
 6
 7 8 1.5
 8
 9 10 1.5
 10
 11 12 1.0 13 1.0 17 1.0
 12
 13
 14 15 1.0 16 1.0 20 1.0
 15
 16
 17 18 1.0 19 1.0 26 1.0
 18
 19
 20 21 1.0 22 1.0 23 1.0
 21
 22
 23 24 1.0 25 1.0 26 1.0
 24
 25
 26 27 1.0 28 1.0
 27
 28
 29
 30

Are you constrained to a specific platform?

Do you mean in terms of doing the calculations or for data analysis?
Calculations are run on a supercomputer which runs on centos7 and I can use g09 and nwchem there, but for these calculations I prefer g09.
For data analysis: not really, using Windows and Ubuntu here, and I'm fine with using whatever software or programming language is useful here.

Yes I meant for analysis. I was assuming from your input stats that you are using a supercomputer. I am currently not in the office so I don't have access to all my toys, you might want to try a recent version of ChemCraft. Iirc you can save an optimisation as a xyz file with multiple structures, so it might work for this problem, too. (It is payware though, but probably the best there is, and if you don't already have it, you might want to consider getting it anyway.)

Oh thanks, that works at least for Scan and IRC and it's possible to put all the xyz coordinates in one file. But it doesn't work for dynamics. Chemcraft can read out energies and geometries (but it also says that there's an error) and whatever export option I use it's always the same single geometry. As far as I understand it it interprets this data as optimization steps and I haven't found an option to export those. But at least it shows where it found the values, so in worst case I could write a python script which exports everything.

I recently found this again, and noticed there is a utility newzmat that has some capabilities. It's not ideal, but it should at least be available on systems that have G09/16 installed.

If you have the latest copy of cclib installed, it can extract both the energies and geometries and write the geometries to an XYZ trajectory file that you can open with Avogadro or VMD.
To print the HF or DFT energies to the screen,
$ ccget scfenergies admp.out

but be warned that the energies are in eV, not hartree.
To dump geometries to XYZ files, here's a simplified version of a script I use all the time:
#!/usr/bin/env python

"""cclib_extract_geom.py: Extract geometries from a quantum chemical
output file using cclib.
"""

from __future__ import print_function

import os.path

from cclib.io import ccopen, ccwrite

def getargs():
    """Get command-line arguments."""

    import argparse

    parser = argparse.ArgumentParser()

    parser.add_argument('outputfilename', nargs='+')

    parser.add_argument('--trajectory',
                        action='store_true',
                        help="""Extract all possible geometries into a trajectory-like file?""")
    parser.add_argument('--suffix')

    args = parser.parse_args()

    return args

if __name__ == '__main__':

    args = getargs()

    for outputfilename in args.outputfilename:

        job = ccopen(outputfilename)
        data = job.parse()

        stub = os.path.splitext(outputfilename)[0]
        if args.suffix:
            xyzfilename = ''.join([stub, '.', args.suffix, '.xyz'])
        else:
            xyzfilename = ''.join([stub, '.xyz'])

        ccwrite(data, outputdest=xyzfilename, allgeom=args.trajectory)

To write the energies in to a text file in hartree, you might add something like
        from cclib.parser.utils import convertor # put me at the top of the file
        with open(stub + '.energies', 'w') as fh:
            for energy in data.scfenergies:
                fh.write(str(convertor(energy, 'eV', 'hartree')) + '\n')

If you're on Ubuntu, there's a cclib package, but these things are usually horribly out of date, so I'd download the latest copy from GitHub or pip install cclib. Come to chat if you need help.

Oh thanks! I figured that I will write some code myself, which I don't really mind because I can optimize it for whatever I need, but having a working script where I can actually see how to use it is great. I'm also quite new to python so...
And now that you told me I looked and I had only cclib 1.3 installed but got the 1.5 version now. =)

If, like me, you just want to extract the optimized structures from a potential energy scan, you can use this messy script:
#!/usr/bin/env python
# python Extract_Optimized_From_Gaussian.py filename

from __future__ import print_function
import sys, os

def extract_all(text, target):
    linenums = []
    # Start count at 1 because files start at line 1 not 0
    count = 1
    for line in text:
        if (line.find(target)) > -1:
            linenums.append(count)
        count += 1
    return linenums

code = {"1" : "H", "2" : "He", "3" : "Li", "4" : "Be", "5" : "B", \
"6"  : "C", "7"  : "N", "8"  : "O", "9" : "F", "10" : "Ne", \
"11" : "Na" , "12" : "Mg" , "13" : "Al" , "14" : "Si" , "15" : "P", \
"16" : "S"  , "17" : "Cl" , "18" : "Ar" , "19" : "K"  , "20" : "Ca", \
"21" : "Sc" , "22" : "Ti" , "23" : "V"  , "24" : "Cr" , "25" : "Mn", \
"26" : "Fe" , "27" : "Co" , "28" : "Ni" , "29" : "Cu" , "30" : "Zn", \
"31" : "Ga" , "32" : "Ge" , "33" : "As" , "34" : "Se" , "35" : "Br", \
"36" : "Kr" , "37" : "Rb" , "38" : "Sr" , "39" : "Y"  , "40" : "Zr", \
"41" : "Nb" , "42" : "Mo" , "43" : "Tc" , "44" : "Ru" , "45" : "Rh", \
"46" : "Pd" , "47" : "Ag" , "48" : "Cd" , "49" : "In" , "50" : "Sn", \
"51" : "Sb" , "52" : "Te" , "53" : "I"  , "54" : "Xe" , "55" : "Cs", \
"56" : "Ba" , "57" : "La" , "58" : "Ce" , "59" : "Pr" , "60" : "Nd", \
"61" : "Pm" , "62" : "Sm" , "63" : "Eu" , "64" : "Gd" , "65" : "Tb", \
"66" : "Dy" , "67" : "Ho" , "68" : "Er" , "69" : "Tm" , "70" : "Yb", \
"71" : "Lu" , "72" : "Hf" , "73" : "Ta" , "74" : "W"  , "75" : "Re", \
"76" : "Os" , "77" : "Ir" , "78" : "Pt" , "79" : "Au" , "80" : "Hg", \
"81" : "Tl" , "82" : "Pb" , "83" : "Bi" , "84" : "Po" , "85" : "At", \
"86" : "Rn" , "87" : "Fr" , "88" : "Ra" , "89" : "Ac" , "90" : "Th", \
"91" : "Pa" , "92" : "U"  , "93" : "Np" , "94" : "Pu" , "95" : "Am", \
"96" : "Cm" , "97" : "Bk" , "98" : "Cf" , "99" : "Es" ,"100" : "Fm", \
"101": "Md" ,"102" : "No" ,"103" : "Lr" ,"104" : "Rf" ,"105" : "Db", \
"106": "Sg" ,"107" : "Bh" ,"108" : "Hs" ,"109" : "Mt" ,"110" : "Ds", \
"111": "Rg" ,"112" : "Uub","113" : "Uut","114" : "Uuq","115" : "Uup", \
"116": "Uuh","117" : "Uus","118" : "Uuo"}

logfile_fn = os.path.basename(sys.argv[1])
logfile_bn = os.path.splitext(logfile_fn)[0]
logfile_fh = open(logfile_fn, 'r')
text = logfile_fh.readlines()
logfile_fh.close()

# Find all lines that contain "!   Optimized Parameters   !"
opt_param_line_nums = extract_all(text, '!   Optimized Parameters   !')
# Then find all lines that have "Input orientation:"
input_orient_line_nums = extract_all(text, 'Standard orientation')
# And all lines that have the ---- which we will use to find the ends of the coordinate sections
end_coor_line_nums = extract_all(text, '---------')
# Add 5 because the first actual coordinate is +5 lines
input_orient_line_nums = [x+5 for x in input_orient_line_nums]
# Make sure they're sorted
opt_param_line_nums.sort()
input_orient_line_nums.sort()
end_coor_line_nums.sort()

new_input_lns = []
count = 0
for x in reversed(input_orient_line_nums):
    if count > len(opt_param_line_nums)-1:
        break
    if x < list(reversed(opt_param_line_nums))[count]:
        new_input_lns.append(x)
        count += 1

new_input_lns.reverse()

new_end_lns = []
count = 0
for x in end_coor_line_nums:
    if count > len(new_input_lns)-1:
        break
    if x > new_input_lns[count]:
        new_end_lns.append(x)
        count += 1

intervals = zip(new_input_lns, new_end_lns)

# Now we convert each interval into a .xyz
outfile = open(logfile_bn+'.xyz','w')
for intvl in intervals:
    n_atoms = intvl[1] - intvl[0]
    outfile.write(str(n_atoms)+"\n\n")
    for x in range(intvl[0]-1, intvl[1]-1):
        column = text[x].split()
        print(code[column[1]],float(column[3]),float(column[4]),float(column[5]), file=outfile)

outfile.close()

I adapted this from here. I've tested it with Gaussian 09. No idea if it works with other versions. The resulting file loads into VMD. I don't know about other visualization software.
Edit: I checked the original script I had posted here against GaussView, and my script was wrong. Specifically, it was the structures titled "Standard orientation" that came before each "!   Optimized Parameters   !" line that I needed to extract, not the ones that came after. I have updated the script above to do that, and the output now matches that of GaussView. I removed the original script, but I have copied it to Pastbin here in case anyone still wants access to it (they're not too different).

Thanks, I will look into that and see if I get what's going on here:-D Always useful to see how people do this and it seems to work fine

FYI, I realized a bit later today that I had an error in it. See my edit at the bottom. The script here is now correct.


text:
What happens when we connect a external emf (having emf greater than that of emf of a galvanic cell) to a galvanic cell?
I read a statement given in the book and I did not understand it
"If any external battery is connected whose emf is more than the emf of galvanic cell ,then it will act as Electrolytic cell"
Also can you please explain what happens if we connect a external battery having emf less than that of galvanic cell?

Think of a generic redox pair:
$$\ce{Ox + ne- -> Red}$$
and use Nernst equation (and a few simplifications) to calculate its potential:
$$ E = E^\circ + \frac{\mathrm{RT}}{\mathrm{nF}}\mathrm{ln}\left(\frac{[Ox]}{[Red]}\right)$$
So, now you know what potential you would measure on such a half-cell. However, what would happen if you applied some arbitrary potential $E_{w}$ ?
Well, if $E_{w} > E$, then that also means that
$$\frac{\mathrm{RT}}{\mathrm{nF}}\mathrm{ln}\left(\frac{[Ox]_w}{[Red]_w}\right) > \frac{\mathrm{RT}}{\mathrm{nF}}\mathrm{ln}\left(\frac{[Ox]}{[Red]}\right)$$
Finally, you get that this situation implies an increase of $\ce{Ox}$ at the expense of $\ce{Red}$, which is what you would expect from an electrolytic cell - the supply of electrical energy causes electrochemical reactions.
If you apply a potential that is below $E$, the only difference would be that $\ce{Ox}$ would decrease and $\ce{Red}$ would increase, but the overall treatment remains the same.

That means if Ew<E then it still acts like galvanic cell, right ?

The thing that bothers me about sticking to names instead of the physical picture is that it may be very misleading. Galvanic cells are supposed to operate spontaneously, so, as long as nothing happens until you apply some potential, any value of $E_w$ should "turn" this cell into an electrolytic cell.

Truffaut's answer is comprehensive, I just thought I'd build further.
Given that the cell in question has a reversible setup, the result would be as specified in his answer. If the cell is not set up reversibly, (for instance most common galvanic cells (aka 'batteries') are set up for no charging) there is no chemical reaction to make use of the voltage difference, and the amps will increase until the voltage resistance drop into and through the electrolyte in the cell matches the external applied voltage.
$$
E=Eo + E\omega
$$
This will generate heat, and will quite quickly burst the cell in question, and become a samsung note 7 type handgrenade.

You didn't really specify how the batteries would be connected.
If the opposite poles were connected (both + to other -) then both batteries would be short circuited and a massive current flow would result.
If the same poles were connected (+ to + and - to -) of two batteries of different EMF values, then one battery would be discharging and one battery would be in the "charging" configuration.