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What happens to a radioactive carbon dioxide molecule when its carbon-14 atom decays?
When carbon-14 decays, the decay products are nitrogen-14 and an electron (and an electron antineutrino, but that's chemically irrelevant*):
$$\ce{^14_6C -> ^14_7N + e- + \overline{v_e}}$$
Let's assume that the carbon atom in question is part of a carbon dioxide molecule in the atmosphere. What would happen to the molecule when the atom decays into nitrogen? Will it be converted into a $\ce{NO2}$ molecule, or will it split apart? Will the electron created in the decay have sufficient energy to escape the molecule and form a positive ion?
Here's a somewhat related question dealing with the formation of radioactive carbon dioxide.
* Of course, not all the energy from the defect will be transferred to the beta particle's kinetic energy, so this is in fact relevant for the rate. See Loong's answer for details.

The molecule will probably split apart, together with a good many other molecules that will happen nearby. Nuclear energies are way above anything in chemistry.

The electrons are ejected with maximum kinetic energy of $156\ \mathrm{keV}$, which is on the order of $\mathrm{15\ MJ/mol}$. However, its only $\mathrm{2.5\times 10^{-20}\ J}$ per particle, which is not a lot but enough to get that electron traveling a maximum distance of 22 cm in air. You more likely have $\ce{NO2+}$ before it starts falling apart.

@IvanNeretin: The problem is how much of that energy is carried away by the ejected electron. You have simultaneous conservation of energy and impulse, in what's now basically a two-body system. Impulse conservation is easiest: since both bodies have opposite relative impulse, the electron since it's several thousand times lighter, must move several thousand times faster. But since kinetic energy goes up with the square of speed, the electron carries millions times more energy. There's less than 1eV left for NO2+, insufficient to split it. Any other molecule hit by the electron is fair game.

@MSalters: it's thousands, not millions (the electron is faster all right, but also lighter).

@MSalters: Your argument doesn't work, because your estimate of 1 eV is only a bound on the energy tied up in the center-of-mass motion of the recoiling nitrogen nucleus. It isn't a bound on the amount of energy that can go into other forms of energy, such as breaking the bond.

@BenCrowell: "breaking the bond" requires a physical process, such as one of the atoms recoiling at high enough speeds. The bonds themselves are electrons in orbit, which aren't directly impacted by an event in the nucleus. And as Loong points out in his answer, the new electron (beta particle) leaves at relativistic speeds, so it's unlikely to break those bonds either.

And as Loong points out in his answer, the new electron (beta particle) leaves at relativistic speeds, so it's unlikely to break those bonds either. This doesn't seem like a valid logical step to me. It seems perfectly reasonable to imagine that the electron would interact electromagnetically with the other electrons on its way out of the molecule. In fact, doesn't your argument claim to prove something that contradicts the experimental results described in the Snell paper? The time scale for the electron to exit is short, but its EM radiation is intense due to the rapid time variation.

@MSalters: But since kinetic energy goes up with the square of speed, the electron carries millions times more energy. Actually the betas are normally ultrarelativistic, and the neutrinos are always ultrarelativistic. Therefore the proportionality is closer to $KE\propto v$ for them, but $\propto v^2$ for the nucleus. The 14C example is unusual because of the very low decay energy, which makes the electron (but not the antineutrino) approximately nonrelativistic.

There seem to be three processes to consider here:

As the high-energy electron exits, it creates a rapidly varying electrical current, which produces intense electromagnetic fields for a short time. This may act on the other charged particles that are present, possibly creating further ionization and/or breaking the chemical bond.
The nitrogen nucleus recoils, and this recoil could break the bond and/or create further ionization.
The charge of the nucleus changes, so even if effects 1 and 2 didn't exist, the state of the electrons is no longer the ground state. It is now a linear combination of different states of the new element, which means there is some probability of excitation or ionization.

A paper by Oksyuk and Gerasimenko says that effect 2 is normally the important one.
From Linear Christmas's answer, we know that in the case of $\ce{^14C}$ decay, which has an unusually low energy, there is a fairly small but still appreciable probability of breaking the bond, and this is presumably caused by one or more of the above processes.
For process 1, it certainly isn't safe to assume that it's negligible because the electron leaves the molecule so quickly. If this were true, then beta particles wouldn't produce ionization when they came along from the outside and hit atoms. Although the time-scale for the electron to exit is short, its electromagnetic fields are intense. By the way, the motion of this electron is not insanely fast compared to the velocities of the other electrons. It has a typical velocity in this decay of about $0.5c$, which could be compared to an estimate of $Zc/137\approx 0.04c$ for an inner-shell electron in carbon.
To estimate process 1, let's use the mean rate of energy loss for beta particles in a solid. For $\pu{0.1 MeV}$ betas in a solid made of fairly light elements, this is about $(dE/dx)/\rho\approx 0.3$ MeV.m2/kg. Taking $\rho$ to be the density of water, and $\Delta x=0.2$ nm, we find $\Delta E=0.06$ eV, which seems to be a couple of orders of magnitude too low to break a bond. However, energy loss of betas is a process that has a lot of random variation about the mean, so it doesn't seem unreasonable to me to imagine that there is something like a 1% chance that it deposits 100 times this energy in the parent atom on its way out. This puts us in the right ballpark for this mechanism to contribute significantly to the observed probability of breaking a bond in NO.
So now let's consider process 2. Let $Q$ be the energy released in the decay and $M$ the mass of the recoiling nitrogen nucleus. We'll see that for kinematical reasons, almost all of the energy goes into the electron and antineutrino, not the nucleus. Suppose we want to find the maximum energy of the recoiling nucleus. This is attained in the case where the electron gets almost 100% of the energy, because for a fixed energy, a more massive particle carries more momentum. If the electron and neutrino were to share the energy, then their momentum vectors could also partially cancel, further reducing the recoil.
The decay energy of $\ce{^14C}$ is unusually low, but in most beta decays the beta is much more relativistic. Let's do the ultrarelativistic case first, both because the math is simpler and because it's a better guide to our intuition about what happens in general.
In the approximation that the beta is ultrarelativistic, its momentum (in the case where it carries all the energy in this example) is $p\approx Q/c$, and by conservation of momentum, this is also the momentum of the recoiling nucleus. Since the nucleus is nonrelativistic, its kinetic energy is $K=p^2/2M\approx Q^2/2Mc^2$.
As a typical example, let's take $\ce{^40K}$, which is the strongest source of naturally occurring beta radioactivity in our environment. It has an 89% probability of decaying to $\ce{^40Ca}$ plus an electron and an antineutrino. The energy for this decay mode is $\pu{1.33 MeV}$, which is almost $10$ times that of $\ce{^14C}$. The ultrarelativistic approximation for the electron is not too ridiculous; in the case where it gets almost all the energy (none to the neutrino), its velocity is about $0.96c$. The maximum kinetic energy of the recoiling calcium nucleus in this approximation is about $\pu{24 eV}$, which is clearly plenty of energy to break a chemical bond.
Without the ultrarelativistic approximation, the momentum of the beta, in the maximum-recoil case, isn't $Q/c$ but rather $\sqrt{(x+Q)^2-x^2}/c$, where $x=mc^2$, and $m$ is the mass of the electron. Even in the case of $\ce{^40K}$, it turns out that the ultrarelativistic approximation is not so great. The actual maximum recoil energy for the calcium is $\pu{41.7 eV}$, so although the approximation gives the right order of magnitude, it's off by almost a factor of $2$.
In the example of $\ce{^14C}$, the result for the maximum energy of the recoiling nitrogen is $\pu{7.0 eV}$, in agreement with Loong's answer. This is plenty of energy to break the bond. Another interesting case with a very low energy is $\ce{^3H}$ , in which the recoiling $\ce{^3He}$  has a maximum energy of only about $\pu{3 eV}$. People have used this to try to study the chemistry of helium.
So the take-away here is that in almost all cases, beta decay is very likely to break up the molecule in which it occurs.

An article by Snell and Pleasanton, 'The Atomic and Molecular Consequenses of Radioactive Decay', (J. Phys. Chem., 62 (11), pp 1377–1382, $1958$) supports Ben Norris's comment.

It is clear ... that $\ce{^{14}CO2}$ remains predominantly bound as $\ce{NO2+}$, a result that is perhaps not surprising. [This occurs in] $81$% of the decays. In $\ce{^{14}CO2 -> NO2^+}$ dissociation yielding $\ce{NO+}$, $\ce{O+}$ and $\ce{N+}$ follows [in], respectively, $8.4$, $5.9$, and $3.6$% of the decays.

A table summarising the results is given.
$$\begin{array}{|c|c|}
\hline
\mathbf{Ion} & \mathbf{\%\  abundance} \\
\hline
\ce{NO2+} & 81.4(16) \\
\ce{NO+} & 8.4(4) \\
\ce{O+} & 5.9(6)  \\
\ce{N+} & 3.6(4) \\
\ce{NO2^{2+}} & 0.40(06)\\
\hline
\end{array}$$

Nice find! I guess the main point is that the electron goes flying off - the 'kick' to the nucleus is small enough that the (remaining) bonds are OK with it.

@Jon Custer As Loong shows in their answer, the recoil energy transferred to the nucleus does in fact appear to (potentially) be able to break the bonds. The question then is whether the other ions result from this process or from the instability of $\ce{NO_2^+}$.

@Marcel - of course the recoil of the nucleus could break bonds. Whether it has to seems to be a no, which is just nice to see. I would presume there would be a fairly complex and hard to calculate dependence on dissociation and the direction of nuclear recoil with respect to the molecular bonds.

The recoil of the nucleus can be sufficient to break bonds if almost the maximum possible recoil energy is transferred to the nucleus. However, considering the shape of the beta spectrum, such events are relatively rare. The average recoil of the nucleus is only about 2 eV and thus insufficient to break a bond.

@Marel $NO_2^+$ is a stable ion.  $N_2O_5$ is $NO_2^+ NO_3^-$ in the solid state.  Other stable salts of $NO_2^+$ are known, such as $NO_2SbF_6$

@Demi I know that the $\ce{NO_2^+}$ ion is stable as part of several salts, but I doubt this would be the case with an unstabilised gaseous ion.

@Marcel It is stable in the sense of "won't break apart spontaneously".  It will nevertheless react with the nearest available reducing agent and/or nucleophile (probably water).

The 14C decay has some idiosyncratic features that may be a little misleading. Many beta decays proceed through multiple processes, including e+, e-, and internal conversion. The Q value for 14C decay is unusually small, only 156 keV. Many beta decays have multiple branches to excited states in the daughter, so there is x-ray or gamma emission following the beta decay, but 14C goes 100% to the g.s. There can be a long chain of α, β, or γ emissions, but 14N is stable. This answer shows that decay of CO2 sometimes results in bond breaking. Most examples would have much higher probabilities.

The antineutrino is not completely irrelevant. The decay energy is shared between the beta particle and the neutrino. Therefore, the beta particles appear with an energy distribution that ranges from zero to the maximum beta energy. The maximum beta energy for the decay of $\ce{^14C}$ is $E_\text{max}=0.1565\ \mathrm{MeV}$; the average beta energy is $E_\text{avg}=0.0495\ \mathrm{MeV}$.

Nevertheless, even beta particles with a relatively low beta energy have a much larger kinetic energy than any chemical bond dissociation energy or ionization energy. Thus, the beta particle cannot be captured in the electron shell of the affected atom. The beta particle leaves the atom with almost relativistic speed before the rest of the molecule understands what went wrong. This process is too fast for immediate chemical reactions. The daughter nuclide remains in the same chemical structure. Thus, the beta decay of a $\ce{^14C}$ atom in $\ce{CO2}$ simply leaves a $\ce{NO2+}$ ion in an excited state. This primary product is unstable and can dissociate slightly later. The resulting secondary products can be radicals or ions, which tend to react with almost any other molecule or atom.
However, the energy released in beta decay is not only distributed between the neutrino and the beta particle. A small amount is also allocated to the recoil of the nucleus. Since the mass of the nucleus is much larger than the mass of the beta particle, the recoil energy is much smaller than the energy of the neutrino and the beta particle and can usually be neglected in a first approximation.
For the beta decay of $\ce{^14C}$, the maximum recoil energy of the $\ce{^14N}$ nucleus can be calculated as $E_\text{recoil, max}=7.08\ \mathrm{eV}$. This value is much smaller than the total decay energy of $0.1565\ \mathrm{MeV}$ and could thus be neglected, but it is larger than the energy of a chemical bond. By way of comparison, the bond dissociation energy in $\ce{NO2}$ is only about $306\ \mathrm{kJ/mol}$ or $3.17\ \mathrm{eV}$. Therefore, the recoil of the $\ce{^14N}$ nucleus can be sufficient to break a bond immediately during the beta decay of $\ce{^14C}$.

Thanks, great explanation! The graph you provided is interesting - could you explain why it has that shape? Also, how did you reach the calculated value for $E_{recoil}$? Did you apply conservation of momentum and assumed the electron's velocity to be directed orthogonally away from the nucleus? Lastly, how would the situation be if we assume that there is a large amount of $\ce{C^14O_2}$ gas instead of an isolated molecule? Would it be possible for the $\ce{NO_2+}$ ions and the electrons to "pair up", forming stable $\ce{NO_2}$?

I suspect my incorrect assumption is the result of too many "assume spherical cow"-type high school physics exercises.

@Marcel 1) Analytical equations that can approximately describe beta spectra are usually complicated since you have to take into account the conservation of energy and momentum for three particles, relativistic effects and Coulomb interaction between the nucleus and the beta particle. In this case, the plot is simply based on tabulated data and only serves to illustrate the continuous shape of the energy distribution. 2) Yes, the maximum recoil is calculated using conservation of momentum, taking into account the relativistic momentum of the beta particle.

@Marcel 3) The beta particle will lose its energy in the surrounding gas until it is slow enough to be captured by another molecule. The maximum range in $\ce{CO2}$ at $p=1\ \mathrm{bar}$ and $T=25\ \mathrm{^\circ C}$ is about $19\ \mathrm{cm}$. If the excited $\ce{NO2+}$ ion is able to deexcite without dissociating, it could attract an electron from somewhere to form $\ce{NO2+}$. However, considering the short lifetime of the $\ce{NO2+}$ ion and the low specific activity of $\ce{^14C}$, it is extremely unlikely that this electron is the beta particle from another $\ce{^14C}$ decay.


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Calculation of CO2 mass in 330mls can coke
How can I Calculate mass of CO2 in a 330 ml's of can coke?

As in this question, it seems that there's a little over 2g of carbon dioxide in a can of Coke.
Do you mean 'calculate', or 'measure'?
If you mean 'measure', many electronic balances will weigh to this sort of precision, so you could just weigh the can before opening, then let it go flat, then weigh it again.
To calculate it, you'd need to know how whether they saturate the solution, what the can pressure is, and the solubility of carbon dioxide in Coke at that pressure.  This may be significantly different from the solubility in water at the same pressure, and not easy to find out, since the Coke recipe is secret.

The coke recipe is not exactly a secret. I takes a reasonably well-equipped laboratory half a day to find out exactly what's in there, and be assured that Pepsi and all other competitors have done  that, and probably do it as a weekly routine. ;-)

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text:
How can dish-soap make glass less misty?
Warm glass that face a cold temperature becomes misty - but becomes less misty if sprayed with dish-soap. Moreover, dish-soap sprayed on glass makes the glass less misty, if it face a drastic change in temperature - at least makes it harder for the glass to become misty.

Why is it so?

The fogging is caused by tiny droplets of water. These condense onto the glass when hot humid air comes in contact with a cold surface . The dish-soap acts as surfactant. The dish soap prevents droplets from forming. It forces any water molecules that condense onto the surface to be evenly spread across the surface. Thus, the water molecules form a thin film instead. This film is usually thin enough to not impede vision.

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What types of glycosidic bonds are in oligosaccharides that occurs naturally not in the lab?
An oligosaccharide is a carbohydrate formed by two or more monosaccharides.
If both monosaccharides are pyranose types, they can bond through ($\alpha$ or $\beta$)1-4 or ($\alpha$ or $\beta$)1-6. If one monosaccharide of them is of furanose type, they can bond through ($\alpha$ or $\beta$)1-2.
Examples:

or

Is any other type of glycosidic bond type between monosaccharides in oligosaccharides that occurs naturally not in the lab?

If you have $n$ hydroxy groups on one sugar, you can form $4n$ glycosidic bonds with the anomeric centre of the other sugar. I’m willing to bet that examples can be found for almost each ‘variant’.


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Reaction of aldoxime and a ketoxime with acetic anhydride
In mt book I saw these two reactions:

I don't know what is the mechanism for these reactions & that's why I am unable to understand how these products are formed & why they are so different. Please explain the mechanism in detail.

This might sound obvious, but since the reagent and the reaction conditions are identical, the reason for the different outcome of the reactions must be related to the starting materials.
In the case of the aldoxime, the carbon atom of the former $\ce{CHO}$ group still bears a hydrogen atom.

 Together with the $\ce{OH}$ group on the adjacent nitrogen atom, this allows (formally) for the elimination of water from the aldoxime under suitable conditions. High temperatures and the presence of a dehydrating agent facilitate exactly that.

The formation of a nitile from an aldoxime in boiling acetic anhydride is (formally) a dehydration reaction.
For obvious reacons, this pathway is not possible in the case of the ketoxime.
What is the role of acetic anhydride here?
I mentioned that the formation of nitriles from aldoximes is formally a dehydration. In reality, however, it is more likely that acetic anhydride actually reacts with both oximes.

Both the aldoxime and the ketoxime are

 acylated. For the ketoxime (R' = alkyl), the story ends here. In the case of the aldoxime (R' = H), a subsequent elimination (of acetic acid) eventually furnishes the nitrile.


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How to identify which species will undergo Oxidation and which will go Reduction
I have been given $\ce{B2}$, $\ce{N2}$, $\ce{(O2)-}$  and $\ce{O2}$.
I have to short out  which will undergo oxidation, reduction or both.
How can I tell it when reactants are not given?

$\ce{B2}$ or $\ce{Br2}$?

@KlausWarzecha $\ce{B2} $

You have to look at the possible oxidation states of the atoms involved.
For instance, if you were given the molecule $\ce{F2}$, you should consider that fluorine has two possible oxidation states: 0 (in the $\ce{F2}$ molecule) and -1.
Hence, $\ce{F2}$ can only undergo reduction.
For the other molecules, you should use the same strategy: try yourself!

To find out which substance has gone oxidation or reduction, you have to check their oxidation nos. If the oxidation increases, the substance is oxidised, if it decreases, the substance is reduced. For eg. . Here, Sulfur is getting oxidised while Bromine is getting reduced.


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Do moles of an acid and base have to be exactly equal for neutralization to occur?
I'm supposed to prove whether a statement is true or false:

5 moles of sodium hydroxide will neutralize $\ce{250 cm^3}$ of hydrochloric acid of concentration $\ce{10 mol/dm^3}$.

So far, I've balanced the equation: $\ce{HCl + NaOH -> NaCl + H2O}$. I've also found the moles of $\ce{HCl}: 10/1000 \times 250 = \pu{2.5 moles}$.
Since the ratio of $\ce{NaOH}$ and $\ce{HCl}$ is $1:1$, I don't have to multiply anything, but I don't know whether the statement's true or false. I thought since $5>2$. $5$, the statement's true, but I was told that in the answer key, the statement's false.
So, does the moles of the substance have to be exactly equal in order for neutralization to occur?

The answer to this question lies in the concept of stoichiometry. The number of moles required of each reactant must be in proportion to it's stoichiometric coefficient in the balanced chemical equation.
So accordingly, the statement should be false.


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How is ozone formation form oxygen **spontaneous**?

DeltaH for process is positive and deltaS is negative. Then by thermodynamics,deltaG for process must be positive at all temperatures. Then how is ozone formation spontaneous?

Why do you think it is spontaneous.

'...a slow dry stream of oxygen is passed through an [electrical] discharge...'   (the picture cut off at the end of the line).   If an electrical current is required then the process isn't spontaneous at all, you're putting energy in

Do you regularly notice oxygen in the air spontaneously converting into ozone?

How long would it take you to either type out the contents of the text in your image or re-image it so the rest of us can see it?  I will then probably spend at least that much time working towards an answer for you.  I cannot promise that I will come up with an answer good enough for posting, but I'll try ;-)

It seems reasonable to suspect that some O2 molecules will dissociate in the discharge. Some of the resulting atoms will bond to other O2 to form O3 instead of recombining. To evaluate the thermodynamics of this process, one should therefore start at atomic oxygen.


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What does the IUPAC definition of the emf of a galvanic cell mean?
IUPAC's Gold Book defines the emf of a galvanic cell in the following way:
The limiting value of E(cell) for zero current flowing through the cell, all local charge transfer and chemical equilibria being established, was formerly called emf (electromotive force). The name electromotive force and the symbol emf are no longer recommended, since a potential difference is not a force.
Check: https://goldbook.iupac.org/E01934.html
I understand that zero current is required to find the emf since passing a current through the cell decreases its voltage. Moreover, we want to find the emf of the cell at some specific composition of the electrolytes, also ensuring a reversible reaction is taking place.
But what does it mean by "all local charge transfer and chemical equilibria being established"? Isn't the emf of the cell ZERO when the cell reaction attains equilibrium? What "equilibrium" does it refer to? And what does "local charge transfer" mean?

Note that the Gold Book is only a collection of terminology from various other sources. Sometimes, you might find more information in the original sources (here: Green Book, 3rd ed., p. 71 and Pure Appl. Chem. 1996, 68 (4), 971).

OK I'll take a look

related http://chemistry.stackexchange.com/questions/65911/how-is-the-standard-electrode-potential-measured-without-allowing-the-electrolyt

We think of electrochemical processes happening immediately in one step, but that's not really true. In a galvanic cell, you have to have the dissolved ions move from the bulk to the surface, adsorb to the surface, and possibly undergo an activation step before they can reduce or oxidize. The last step is for the electron to actually move from the reduced to the oxidized species. This specific step is the "local charge transfer" reaction.
You are right that when the cell reaches equilibrium, the $E_{cell}$ would be zero. However, here we're interested in  the equilibrium of each local process separately: eg. diffusion of ions from the bulk electrolyte to the surface will reach an equilibrium.
When you first submerge the electrode into the electrolyte, all of these steps are able to reach an equilibrium, and things like concentration gradients between the electrolyte bulk and electrolyte surface can effect the potential experienced by the electrode. In fact, some charge transfer even occurs initially, but because the half-cells are not connected, electrostatic forces build up and the concentrations are such that the chemical potential is quickly opposed and a steady-state is achieved.

Electrochemical equilibria are just like chemical equilibria. The forward reactions and the reverse reactions happen at their own rates and when the two are equal, the system is in equilibrium. This is true regardless of the nature of the cell potential (electrolytic or galvanic).
If, you short circuit a galvanic cell externally, then, it will pass a current and go into a non-equilibrium condition until the current stops and the potential goes to zero. This however, does not change the fact that before the external short, there was an equilibrium inside the cell, whose potential could be measured.
The "local charge transfer" probably refers to any side reactions inside each half cell that may exist due to the initial reagents being prepared in a non-equilibrium concentration ratio.


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Dehydration of amide with PCl5
Can $\ce{PCl5}$, phosphorus pentachloride, react with an amide?
I saw a reaction where the corresponding cyanide was formed in this process. What other reagents could do the same?

Dunno about PCl5, but POCl3 gives a chloroiminium ion (see the Vilsmeier reaction)

There's a couple of newer methods, using

zinc + N-Methyl-N-(trimethylsilyl)trifluoroacetamide (MSTFA)
copper + N-Methyl-N-(trimethylsilyl)trifluoroacetamide (MSTFA)
diethyl chlorophosphate $\ce{(EtO)2POCl}$
$\ce{PdCl2}$ in $\ce{CH3CN/H2O}$
$\ce{(COCl)2}$ with DMSO at -78 °C (Swern-type conditions)

Thionyl chloride, $\ce{SOCl2}$ is almost a classic here.
Under these conditions, the side products of the amide dehydration are $\ce{SO2}$ and $\ce{HCl}$.
$\ce{R-CONH2 + SOCl2 -> R-C#N + SO2 ^ + 2 HCl ^}$
On the first view, $\ce{PCl5}$ looks like leave quite a mess. Thinking in the workup, that's probably the last thing I would try.

Thanks a lot sir.!!...btw pcl5 is considered as a dehydrating reagent ,right?

Yes, it is. But it can be used to convert carboxylic acids to their chlorides too.

As someone who has done lots of amide dehydrations, there are two methods I'd recommend.  The cheapest is thionyl chloride in dry DMF.  You need to dry the latter thoroughly by stirring overnight over calcium hydride (this evolves hydrogen, so don't stopper the vessel).  Then distil under vacuum.  Then suspend your amide in the dry DMF and stir in an ice bath, and drop in the thionyl chloride.  Let it stir and room to warm temperature, then pour onto crushed ice.  I also used to purify the thionyl chloride first by reflux over triphenyl phosphite and then distilling.
A more expensive method, but just as effective, is to use trifluoroacetic anhydride / pyridine as the agent, and sodium-dry dioxane as the solvent.  This is incredibly mild, specific, and leads to a very clean work up and high purity product.

Welcome to Chemistry SE! We hope you continue to share your expertise, on this site.

Thank you!  Always nice to have a warm welcome!

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text:
Can a strong nucleophile carry out an SN1 reaction?
For example, if I take a tertiary halide in water + ethanol solution and add I- ions to it (which is a good nucleophile), then shouldn't the reaction be SN1; because SN2 is not possible as the substrate is tertiary. Please explain.

http://11452-presscdn-0-51.pagely.netdna-cdn.com/wp-content/uploads/2012/08/compare-sn1sn2.png

You've pretty much said it all already. SN2 is ruled out at the tertiary centre. Since the attack of the nucleophile in the SN1 mechanism is not rate-determining, it doesn't matter whether the nucleophile is weak or strong; iodide works fine.
In this case, it's more likely to get intercepted by the solvent itself, which is present in much larger quantities and is also nucleophilic, but that's another story.

So basically it would be SN1 but it would form alcohol instead of iodide.

@DHMO Yes, indeed. It'll either give the alcohol or the ethyl ether.


text:
An **atom** is the smallest constituent unit of ordinary matter that has the properties of a chemical element
An atom is the smallest constituent unit of ordinary matter that has the properties of a chemical element.
my Question:ordinary matter = chemical element???
A chemical compound is an entity consisting of two or more atoms, at least two from different elements, which associate via chemical bonds .
my Question: Is a single molecule can be a compound? ??
A chemical element  is a species of atoms having the same number of protons in their atomic nuclei .
my Question: Is a single atom can be a compound ???

Yes, in simplest terms, ordinary matter is the elements. It's what we see and interact with everyday, although it only makes up 4% of the universe.
In many cases, a molecule can be a compound, as long as it includes two different atoms. Something like H2 is a molecule but not a compound because it only has one unique element.
No, a compound must have two different atoms.

Hope that helps!


text:
Why does copper(II) sulfate react with ammonia?
We've done preparation of Tetraamminecopper(II) sulphate $\ce{[Cu(NH_3)_4(H_2O)_n]SO_4}$ using copper sulphate and 50% ammonia solution.
Initially, a pale blue color of $\ce{Cu(OH)_2}$ appeared with turbidity then after a dark blue color solution is obtained.
It is a substitution reaction Initially,
why formation of $\ce{Cu(OH)_2}$ is taking place ??
Is initially NH3 acting as a base and ripping off the proton from the complex $\ce{[Cu(H_2O)_6]^2+}$?
Then on the further increment of the concentration of $\ce{NH_3}$, it starts to act as Nucleophile??
My high school teacher taught me that acid-base reactions are always fast, so in this example is everything is about kinetics? What is the role of the thermodynamics? From where can I get a quantitative explanation with proper data?

Related: http://chemistry.stackexchange.com/questions/29126

http://chemistry.stackexchange.com/questions/66938/adding-non-stoichiometric-amount-of-nh3-to-copper-solution


text:
Why does neighbouring group participation happen?
Can someone please explain why neighbouring group participation happens?
Does it stabilise the carbocation? If so then is NGP dominant over an intermolecular SN2 reaction (especially in aprotic medium)?

There may be some confusion over what neighbouring group participation (NGP, also sometimes referred to as anchimeric assistance) is. Carey puts it quite succinctly:

When a molecule that can react by nucleophilic substitution also contains a substituent group that can act as a nucleophile, it is often observed that the rate and stereochemistry of the nucleophilic substitution are strongly affected. The involvement of nearby nucleophilic substituents in a substitution process is called neighboring- group participation.
Source: Advanced Organic Chemistry (Part A). Francis Carey.

Rather than a discrete carbocation forming, which is then stabilised, NGP usually (though not exclusively) occurs via two consecutive SN2 processes, with overall retention of stereochemistry.
The reason why the NGP happens, rather than an external nucleophile just doing the substitution is generally due to the fact that intramolecular reactions happen at a faster rate than intermolecular ones, since this removes the necessity of encounter between nucleophile and electrophile (the two are essentially tethered together). The initial internal attack forms a ring (usually 3 or 5 membered for reasons that should be familiar to you) which may then be opened by the external nucleophile.
A classical example is of a sulfide displacing a halide, with subsequent opening of the reactive 3-membered intermediate:

Source: Organic Chemistry. Clayden/Warren.

Thank you NotCorey. Your answer was very helpful. But I'm sorry,  i dont know why it can't be a four or six membered ring. Can you please explain?

Thats a whole new question. You'll find a clear explanation in any organic textbook – if you still don't get it, ask a new question.

@PranoyDe A six-membered ring is another common ring size (favored over 4 and 7 in general), but not as fast as 3 and 5. I agree that the rationale merits another question.


text:
Why is AlF3 soluble in anhydrous HF only in the presence of KF?
I am pondering over that what must be the reason that $\ce{AlF3}$ is insoluble or less soluble in anhydrous $\ce{HF}$? And pageants it is soluble to a good extent only in the presence of $\ce{KF}$? Is there any other reagent in presence of which $\ce{AlF3}$ is soluble in $\ce{HF}$?
I guess a probable answer for the cause of insolubility could be:
"the nonavailability of $\ce{F-}$ ions because of the presence of intermolecular hydrogen bonding in $\ce{HF}$, which makes $\ce{AlF3}$ insoluble in $\ce{HF}$."

Can someone elaborate the cause and mention other such reagents like $\ce{KF}$ which can be used for the same purpose?

Could you clarify the conditions? I find that $\ce{AlF3}$ is okayish soluble in water, see Wikipedia, so I am curious under what conditions it is insoluble in HF, because if we talk about an aqueous solution, then I doubt that this is true. Otherwise the question is more or less why $\ce{AlF3}$ is so much more soluble in the presence of $\ce{KF}$.

Do you mean pure hydrogen fluoride, or a concentrated aqueous solution?

I mean pure  HF.

$\ce{AlF3}$ is insoluble in the anhydrous $\ce{HF}$ because the $\ce{F-}$ ions are not available in intermolecular hydrogen bonded $\ce{HF}$, but it becomes soluble in the presence of $\ce{KF}$ due to the formation of soluble complex, $\ce{K3[AlF6]}$.
$$\ce{AlF3 + 3KF -> K3[AlF6]}$$

Hmm, it looks like too strong simplification.

From the reaction equation it looks like you need a massive amount of KF. And while we are at it, what is a little amount?

Probably there is more than one complex ion.  But yes, complex ion formation is the key.  It's like an aqueous system where zinc or aluminum hydroxide is almost insoluble in just water but dissolved easily when you add NaOH.

Since it is pure $\ce{HF}$ we are talking about, there wouldn't be any hydrogen bonding in the first place. Therefore, there wouldn't be any dissociation to form $\ce{H+}$ and $\ce{F-}$ ions.
Whereas for the $\ce{KF}$ solution, it would easy dissociate to form $\ce{K+}$ and $\ce{F-}$ ions since it is dissolved in water. I believe the crux to this question is the water; since you need it to produce $\ce{F-}$ ions which can form the $\ce{[AlF6]^3-}$ complex. This is compounded by the fact that $\ce{AlF3}$ is ionic in character since fluorine is very electronegative, the $\ce{Al-F}$ is relatively weaker and will easily break in the presence of water to form the complex.


text:
Oxidation state of tungsten in W3O8
Recently I came across this equation:
$$\ce{WO_3 +SnCl_2 +HCl\rightarrow W_3O_8 + H_2SnCl_6 +H_2O}$$
The compound $\ce{W_3O_8}$ that appears in the right hand side of the equation seems to be mixed oxide. Normal calculation of the oxidation state of W in the compound yields $\ce{\frac{16}{3}}$ which is a fraction. A google search on the compound does not give any good results too.
So, is $\ce{W_3O_8}$ a mixed oxide? What are the oxidation states of the different tungsten atoms?

This might be helpful. https://www.researchgate.net/profile/Jun_Li36/publication/7378383_On_the_Structure_and_Chemical_Bonding_of_Tri-Tungsten_Oxide_Clusters_W_3_O_n_-_and_W_3_O_n_n_7-10_W_3_O_8_As_A_Potential_Molecular_Model_for_O-Deficient_Defect_Sites_in_Tungsten_Oxides/links/09e4150a75dbf3d339000000.pdf?origin=publication_detail

this http://dx.doi.org/10.1063/1.3505689 article claims that some 'extra' electrons on W are localized on specific atoms. So it seems that the compound may be classified as mixed oxide. However, it belongs to a weird family of Magneli phases, which are ... somewhat special. They are not non-stechiometric per se, but close to it. I'm not qualified enough to give a full conclusion, though.

TL; DR: If you are forced to list the formal oxidation states, then with a high level of confidence it's tungsten(V,VI) octaoxide $\ce{W^{VI}W^V2O8}$. However, the real product of this reaction would be a number of non-soichimetric tungsten(IV,V,VI) oxides. Formation of molecular oxoclusters or peroxo compounds is very unlikely.

Reduced tungsten oxides are generally non-stoiciometric. Glemser et al. [1] reduced tungsten(VI) oxide hydrates $\ce{WO3 * H2O}$ and $\ce{WO3 * 2 H2O}$ with $\ce{Zn}$ in $\ce{HCl}$ to the blue tungsten oxide hydrates of the general formula $\ce{WO_{3‐x}(OH)_x}$ $(x → 0.5)$ with the lowest reducing degree of $2.65$, which is very close to the one in $\ce{W3O8}$ $(8/3 = 2.6(6))$.
I would expect that $\ce{SnCl2}$ in acidic media would have comparable reducing effect. With the moderately mild reducing agent such as tin(II) chloride there is very little chance of ending up exclusively with $\ce{W3O8}$. Also, a presence of a reducing agent also eliminates a possibility of the formation of the peroxo-compounds, so that with a high level of confidence we are dealing with oxide $\ce{O^2-}$ ligand.
I find it equally improbable that any cluster formation takes place. For the existence of a polyoxotungstate cluster the terminal $\ce{W=O}$ bonds must be "disarmed" in order to prevent further olation/oxolation and aggregation into bulk metal oxide. This is often achieved via the template effect, however even reduced by 1 to 6 electrons forms of 12-heteropolytungstates $\ce{[(P,Si)W12O40]^3-}$, the phosphate- or silicate-templated Keggin structure — one of the most stable — only exist briefly in solution [2, pp. 341–343]. In the presented system there is no suitable anion to preserve the three-membered $\ce{W3O8}$ ring, and it's very likely going to hydrolyze quickly.
To sum these thoughts up, the more appropriate way to write the reaction would be
$$\ce{n WO3 + SnCl2 + 4 HCl → W_nO_{3n-1} + H2SnCl6 + H2O}$$
where $\ce{W_nO_{3n-1}}$ is homologous series analogous to the mixed-valence tungsten blues. On crystallographic level this can be seen as withdrawal of oxygen atoms from $\ce{WO3}$, which is isostructural to $\ce{ReO3}$ (Fig. 1a). New point defects are eliminated by introducing a crystallographic shear (CS, Fig. 1b). As the reduction proceeds, the octahedra $\ce{[WO6]}$ in this structure changing from corner-sharing to edge-sharing motif (Fig. 1c). (Caption and adapted illustration from [3, p. 1086].)

Figure 1. Crystallographic shear in $\ce{WO3}$ (a) the idealized structure
of $\ce{WO3}$, drawn as corner-shared $\ce{WO6}$ octahedra, squares in projection (b) {102} CS planes (arrowed), consisting of blocks of four edge-shared octahedra (c) {103} CS planes (arrowed), consisting of blocks of six edge-shared octahedra.

It's impossible to deduce exact oxidation number for such structure with floating stoichiometry, but sure enough an average oxidation number of tungsten can be found trivially from the $\ce{W_nO_{3n-1}}$ formula:
$$\text{O.N.}(\ce{W})\cdot n - 2(3n - 1) = 0 \qquad\to\qquad \text{O.N.}(\ce{W}) = 6 - 2/n,$$
e.g. for $\ce{W3O8}$ $\text{O.N.}(\ce{W}) = 6 - 2/3 = 5.3(3)$. Boundary case $n = 1$ dictates the lowest $\text{O.N.}(\ce{W}) = 4$, so that $\ce{W_nO_{3n-1}}$ phase consists of a mixed-valence oxide with various ratios of $\ce{W(VI)}:\ce{W(V)}:\ce{W(IV)}$. Exact ratio is to be determined from XPS/XANES analysis for a given specimen; spectrophotometry would be difficult to implement as tungsten oxides have negligible solubility in most solvents unless some peroxide or $\ce{HF}$ is added. Exact location of the reduced tungsten centers can be averaged by finding CS areas with EM/PXRD analysis.
Now, to the $\ce{W3O8}$. In fact, it's already been synthesized by Zakharov et al. as follows [4]:

The sample examined was prepared by partial reduction of $\ce{WO3}$ with carbon at high temperature in combination with high pressure. The reaction occurred between the graphite container material and a pressed tablet of $\ce{WO3}$ in a closed system ($T = \pu{1773 K}$, $P = \pu{50e5 kPa}$). [...] The product obtained was a mass containing minute dark-red crystallites. The colour suggested that considerable reduction had taken place.

and two phases were characterized with EM/PXRD. Authors also suggest that $\ce{W3O8}$ can hardly be obtained under ambient conditions:

The two phases, both of composition $\ce{W3O8}$ $(\ce{WO_{2.667}})$, do not appear in the binary $\ce{W-O}$ system at ambient pressure. $\ce{W3O8(I)}$ has a structure of the $\ce{U3O8}$ type (Andresen, 1958), which is denser than that of $\ce{W3O8(II)}$. The latter has a new type of structure, although some features are shared with other previously known tungsten oxides ({102} CS structures).

Figure 2. Idealized polyhedral representation of $\ce{W3O8(I)}$

Figure 3. Idealized polyhedral representation of $\ce{W3O8(II)}$
So, there are two possible combinations for the formal $\text{O.N.}$s:

$1~\ce{W(VI)} + 2~\ce{W(V)}$;
$2~\ce{W(VI)} + 1~\ce{W(IV)}$.

Unfortunately, oxidation states are not mentioned in the article and I haven't found any citeable sources with the determination of oxidation numbers for $\ce{W3O8}$, but the XPS studies of uranium octaoxide $\ce{U3O8}$ indicated the presence of two oxidation states, $\ce{U(VI)}$ and $\ce{U(V)}$, in a $1:2$ ratio, respectively. Since phase $\ce{W3O8(I)}$ is isostructural, there is a high chance that the scenario 1, $1~\ce{W(VI)} + 2~\ce{W(V)}$, is quite plausible.
Having the CIFs of both structures at hands, I've been trying to perform bond-valence analysis, but with the common tabulated empirical parameters I haven't reached any meaningful results. Mixed-valenced tungsten and molybdenum oxides often contain delocalized $\mathrm{d}$ electrons, making bond-valence analysis alone less meaningful [5].
References

Glemser, O.; Weidelt, J.; Freund, F. Genotypische Oxidhydrate des Wolframs. Zur Frage der Wolframblauverbindungen. Zeitschrift für anorganische und allgemeine Chemie 1964, 332 (5–6), 299–313. https://doi.org/10.1002/zaac.19643320511.
Advances in Inorganic Chemistry and Radiochemistry; Emeléus, H. J., Sharpe, A. G., Eds.; Advances in inorganic chemistry and radiochemistry; Academic Press: New York, 1967; Vol. 10. ISBN 978-0-08-057859-0.
Encyclopedia of Inorganic Chemistry, 2nd ed.; King, R. B., Ed.; Wiley: Chichester, West Sussex, England ; Hoboken, NJ, 2005. ISBN 978-0-470-86078-6.
Sundberg, M.; Zakharov, N. D.; Zibrov, I. P.; Barabanenkov, Y. A.; Filonenko, V. P.; Werner, P. Two High-Pressure Tungsten Oxide Structures of $\ce{W3O8}$ Stoichiometry Deduced from High-Resolution Electron Microscopy Images. Acta Crystallographica Section B Structural Science 1993, 49 (6), 951–958. https://doi.org/10.1107/S0108768193005701.
Domenge`s, B.; McGuire, N. K.; O’Keeffe, M. Bond Lengths and Valences in Tungsten Oxides. Journal of Solid State Chemistry 1985, 56 (1), 94–100. https://doi.org/10.1016/0022-4596(85)90256-7.

Note that $\ce{W3O8}$ is a non-stoichiometric compound and it is not that simple compound you may think. It has a very complex structure as such its oxidation state is very difficult to calculate. I am not going in details about the structure of the compound and I will only mention it a non-stoichiometric compound actually a oxygen-deficient cluster.
Please give a good read to Xin Huang et.al., J. Phys. Chem. A 2006, 110, 85-92, the link that @JM97 has posted. It says that the $\ce{O-O}$ is much longer than normal oxide bond and so it is assumed to be a peroxo bond $\ce{O2^2-}$. Now, peroxide has oxidation state of -1. So, assuming $\ce{W3O8}$ to be a simple compound (no stoichiometric defect or any kind of defect), the oxidation state is:
$$3x + 8(-1) = 0$$
$$x = +{8/3}$$

There is no way a molecular cluster of such composition is going to be produced by the OP's reaction, it only exists in the gaseous phase and is very unstable, not to mention that it's a reducing media (no peroxides). The non-stoichimetric oxides are the so-called Wadsley and Magneli phases $\ce{M_nO_{3n-2}}$ and $\ce{M_nO_{3n-1}}$, whereas $\ce{W3O8}$ polymorphic phases preserve its stoichiometry all right.


text:
Why do shampoo ingredient labels feature the term "Aqua"?
I keep seeing the term "Aqua" in the ingredient labels on several shampoo varieties, but I really don't see why it should be there in the first place.

I mean, if the manufacturers just wanted to say it contains water, couldn't they've printed out "Water" instead?
Or could it be that "Aqua" is slang for purified water (or water that's been treated in some godforsaken way) in the shampoo industry? Well, I guess there could always be the possibility that the manufacturers think "Aqua" sounds a lot fancier than plain ol' "Water".
So why's the term "Aqua" mentioned there?

While I agree with the answers, I suspect it's also used to "hide" the fact that what you've just bought, is actually mainly water...

@BaardKopperud this is a very unfair remark. What you've bought is a finished product, not a bunch of ingredients. There is an optimal range for water content in a shampoo, and if there was less water in it, it would not be a good shampoo. So, you would have a product that is worth less, not more, if they were to start putting less water in it. By your logic, anybody buying a $200 CPU has just paid $200 for 2 grams of sand - true on some level, but misleading, and based on bad assumptions like "the whole is the same as the sum of its parts" and "market value is determined by material cost".

@rumtscho, It is a fair remark/conjecture. Ultimately, the manufacturers use "aqua" because they believe they will make more money using that term rather than "water". There are, probably, multiple contributing reasons.  One of which is that some percentage of people will not recognize that "aqua" is "water". What that percentage is, and how that lack of understanding actually affects buying choices is something that such companies should have taken into account when making the choice to use "aqua". Their marketing department would not be doing their job properly if it was not considered.

@Makyen Yes, it is the marketing's department job to sell more units, but 1) deriving that this must be why they use "aqua" is a logical leap, and in this case, wrong - read the answers below, there are regulations requiring them to use "aqua" and they would not be allowed to use "water" even if they wanted. 2) my comment was not about why they use the term, it was about the implicit suggestion that a shampoo made up of mostly water is somehow worth less than what you paid for it.

This post could benefit from some freehand red circles.

Water is a food ingredient. I suspect that the term, "Aqua" is used to distinguish it in non-food products. I can't even get a satisfactory answer out of my local (UK) trading standards department.

In most countries, cosmetic product labels use the International Nomenclature of Cosmetic Ingredients (INCI) for listing ingredients.
The INCI name “AQUA” indeed just describes water (which is used as a solvent).

So, why does INCI name water aqua instead of just using English terms like in most other cases?

Indeed. The question asks "why", so the answer should explain why, not simply restate that it is so.

@MaxRied then they don't need to change the list to sell in another country.

@Tim That is a much better answer than any of the other ones.

@MrLister I phoned up Original Source (a UK brand) to find out why. Effectively it is due to regulations, which are in place so its multi lingual.

@MaxRied when you are creating what is in essence a dictionary of controlled vocabulary, "just using English terms" is low on your list of priorities. If you ask "why it is low priority", I suggest you read a bit about the basics of information architecture and data curation - it is a fascinating topic, and one which many people today could profit from in their professional lives, but way too much information  to be put in a single Stack Exchange answer, and straying too far away from chemistry.

@rumtscho Of course your conclusion is arguably true, and, well, that would make the question off topic. But the question "why are some of the words on shampoo bottles in Latin, but most of them in English,  even in non-English-speaking countries?" (and I just checked and it's true here on my own shampoo bottles, somewhere in Europe) is still a very interesting one, and why shouldn't it be asked here. Or, if it shouldn't, where on SE should it be asked?

@MrLister Nowhere on SE. Questions of the type "Why is the world not the way I would have designed it" rarely have an answer which satisfies the asker's brain's desire for meaning. If it is a matter of human design, like Google choosing 4 colors for their logo instead of 2, or the INCI people choosing a mixture of Latin and English instead of something else, the story is usually "because somebody had the job to do it and this is the first idea which came to his mind without the boss opposing it". Asking it for natural phenomena ("why didn't pigs evolve wings") is similar, it is (cont.)

(cont.) just the consequence of random events interacting. The worst thing about such questions is when eager answerers start connecting their knowledge into a sensible-sounding story, become convinced of it and write it up. These stories are usually unprovable, but because they sound logical, they appeal to people, get upvoted, and turn into myths. Which makes us a source of desinformation.

Aqua means water in many languages, so maybe it`s selected as a more universally recognised term?

There are three reasons which I can see:

labelling regulation and a common nomenclature of ingredients,
ubiquity of the term aqua,
the romantic ring to the word.

I shall expand upon the first as the others are self-explanatory.

Overview of laws and regulations
The United States Food and Drug Administration (FDA) summarises the Fair Packaging and Labeling Act (FPLA).

If a cosmetic product is marketed on a retail basis to consumers—such as in stores, online, or “person to person”—the ingredients must be listed by their common or usual names, generally in descending order of predominance.

For such consumer products, the FPLA actually prohibits the use of the word aqua on its own.

Cosmetic companies sometimes ask FDA about identifying botanicals only by their Latin names ... such as “Aqua” ... instead of “Water”. Under the FPLA, however, ingredients must be listed by their “common or usual names,” and FDA does not accept these alternatives as substitutes.

They do, however, allow simultaneous use which might be something that makes uniform international labelling easier. Specifically, writing

water (aqua)

is allowed. Often times people buy their product directly from professionals, and in this case simply aqua might still end up on your bathroom counter.

[FPLA does not apply] to products distributed only

as free samples,
for professional use, or
for institutional use, such as at schools, hospitals, or the workplace.

In the European Union, the INCI is followed but indirectly as far as I can tell. Indeed, regulation no 1223/2009 (available as a documentation here) states

The Commission shall compile and update a glossary of common ingredient names. To this end, the Commission shall take account of internationally recognised nomenclatures including the International Nomenclature of Cosmetic Ingredients (INCI).

Consult your local EU lawyer for further information. :-)

I live in the EU and, although I don't have a direct reference at hand, I think the correct reason is indeed regulations. I remember that a couple of decades ago products in Italy were labeled using Italian ingredient names. Then I remember having heard about some kind of EU directive that imposed uniform labeling for ingredients of cosmetics.

@LorenzoDonati I am also in the EU (for a little longer).  I thought that use of Latin names was an option rather than a requirement.  A while ago, a visiting Danish friend went to a pharmacy for a treatment (I forget for what condition).  I was disappointed that she was sold a homeopathic product whose principle ingredient was "allium" - Latin for garlic.  She had just bought some expensive garlic tablets.

Latin names are used to simplify identifying allergens.
The reason for this nomenclature is that it should be easy for everyone, wherever they are (even abroad), to identify substances they are allergic or otherwise sensitive to.
For water, this might not make much sense, but it does make a lot of sense for plants and chemicals, whose names vary widely between languages, so in this respect it's customer-friendly to have standardized terms for them.
The way by which this goal is achieved is by using the INCI nomenclature, as the other answers say. The downside of this is that ingredients unwanted by some persons (e.g. non-vegan or mineral oil based ingredients etc.) can still be hidden with rather cryptic names, which can make those labels harder to read than if they were stated in the local language.

It seems to me that most of the responses here are not really getting at the heart of the issue.  Saying that the term "aqua" is used instead of "water" because of regulatory reasons is merely pushing the question back one step, leaving us still to wonder why the people who created those regulatory rules decided that the term "aqua" should be used to indicate the common substance water (H2O) on cosmetic product ingredient lists. User bot47 alluded to this in his/her question above.
What was the primary reason for regulations requiring the use of "aqua" to mean water on cosmetic ingredient lists? I don't think that the primary motivating factor behind that regulatory rule was clarity or understanding across different regions/languages, because the Latin word "aqua" is not a word commonly used for "water" in any living language that I know of -- certainly not in any of the world's most common languages, such as English, Chinese, Spanish, Arabic, etc.
In English, the term "aqua" only means "water" in a technical usage (like on cosmetic labels), and otherwise "aqua" refers to a blue-green color (like "aquamarine") or is found in water-related words like "aquatic" or "aquarium." I think that if you asked a person on the street, "What is aqua?" without any other context, they would mostly likely be confused about your meaning. They might guess that you are referring to something water-related, or they might have no idea what you mean.
Some people might think of Latin as a universal language of science, and therefore universally understood, but most people do not speak Latin no matter which part of the world you are in. The language spoken by the most people in the world, Mandarin Chinese, bears no meaningful relation to Latin.  While Spanish "agua" sounds similar to "aqua," one cannot require the consumer to assume that a similarity in spelling/sound reliably indicates an equivalent meaning. If "sugar" were labeled using its Latin translation "saccharum," then consumers might mistakenly assume that the product contained the artificial sweetener saccharin instead of actual sugar.
If clarity were the main reason for the regulation, using English "water" or the chemical formula "H2O" would, I suspect, be better understood than "aqua."  No one language is going to be perfectly understood in every situation, but English is arguably the single most widely understood language in the world, and "water," being a common word, is reasonably likely to be understood even by someone with only a scant knowledge of English. This is why many terms on the INCI list are in English or use the scientific name for a plant species (in Latin per scientific convention) followed by English terms providing further information. Examples: The INCI term for "beeswax" is "Beeswax," for "raspberry extract" is "Rubus Idaeus (Raspberry) Fruit Extract," and for "sunflower oil" is " Helianthus Annuus (Sunflower) Seed Oil." The only reason Latin is used in the names of these substances is that this is the standard way to refer to a specific species of plant unambiguously. For instance, Rubus ideaus refers to the species also called "red raspberry" or "European red raspberry," which is different from Rubus strigosus (American red raspberry), Rubus crataegifolius (Asian raspberry), Rubus occidentalis (black raspberry), Rubus ellipticus (yellow Himalayan raspberry), or many other raspberry species.  In order to accurately and precisely label the plant species used in an ingredient, such that there is no ambiguity, it is most practical to use the scientific name for the plant, which is in Latin by convention. (Thank you, Carl Linnaeus.) If we extend the same spirit of employing the specific, unambiguous labels used in scientific writing to the topic of interest here, we should be using the term that is typically used in science to refer to water, and that term is "water." I have studied and worked in science for years, and I gave only ever heard of water referred to as "water," "H2O," or (humorously) "dihydrogen monoxide" (the last one being a tongue-in-cheek joke about chemistry nomenclature).
I suspect, as the the OP and some early commenters suggested, that the primary reason that "aqua" is used instead of "water" is that people don't like the idea of paying for water in their products, even if the inclusion of water is necessary and justified for scientific reasons. Perhaps this is a surprise for some people, but human beings are irrational creatures, by and large.
Let me digress to emphasize this point a bit: In my home country, the USA, not only are people irrational, but they tend to have a very poor understanding of science. As one example of this, a 2015 survey found that 82% of respondents supported mandatory labels on foods "produced with genetic engineering," which refers to GMO's (genetically modified organisms). In the USA, at least, "GMO" has become a dirty word, and food labels commonly boast their their products are "GMO-free" in order to appeal to consumers. However, in that same survey where 82% wanted mandatory labels on GMO foods, presumably because these people thought GMOs were harmful, 80% (nearly the same number of people) also supported "mandatory labels on foods containing DNA." These people apparently did not understand that DNA is a normal part of all living organisms on earth, and that we eat these organisms (or parts of them) for food. All food item containing plant or animal cells or single-celled organisms would require this ridiculous label warning about "DNA" that 80% of Americans favored -- this label would include everything from apples, oranges, potatoes, broccoli, and carrots, to fish, steak, milk, cheese, and eggs, to wheat, oats, rice, nuts, and soy, and anything made using such ingredients). The only DNA-free items would be highly refined and purified substances such as table salt, table sugar, cooking oil, or water.
My point in bringing up this survey is that the vast majority of Americans know very little about science, and so they are likely not to read any ingredient list with a particularly informed or critical eye, and they are prone to illogical conclusions about what they see. As a result, no matter what the explanation is for why water is the first ingredient on the list of ingredients in a fancy/high-end/expensive product, pretty much nobody will be pleased or impressed by the fact that water is the single largest component of this thing which is supposed to make them feel special, refined, exclusive, etc.
First of all, for the typical consumer who is buying cosmetics, water is cheap, so the consumer might feel like they were cheated in a way, being charged top dollar for something that comes out of their faucet at home. And secondly, this fancy product that they're purchasing is supposed to be special and improve their appearance or fix a cosmetic problem -- in some ways, many consumers will expect it to contain only the rarest, most exotic ingredients that will be different than what they've used before -- and so people will tend to wonder how "water," which they already use on their skin every day, could possibly provide an additional benefit for them. "Water" in a list of ingredients probably comes across to many people as an inert filler which may, if anything, be impeding the action of the other, more important ingredients.
(I realize this line of thinking is not entirely logical, but I think the vast majority of people would tend to have this reaction on some level. It's well-established that irrational thinking influences people's decision making all the time and in many ways.)
At best, people might feel neutral about seeing "water" on a label, recognizing that there are legitimate reasons for water's use in a formulation, even in luxury products with other rare/expensive ingredients. But on average, the disclosure of the aqueous nature of a product, especially those aiming to be seen as high-end or luxury items, will decrease the product's appeal among consumers.  In the same way, no matter how delicious a $100 bowl of soup from the world's best chef tastes, nobody will feel even better about it after being reminded that the main ingredient in that dish was water.
Using "aqua" obscures the fact that water is used in a product, which will be favorable for most products' sales, which is really what cosmetics companies care about. "Aqua" may even sound like something good to some consumers, giving the product some mysterious appeal ("Oooh! 'Aqua'... I wonder what that is! Must be fancy...") If anyone asks a cosmetic company about this term, then various explanations can be rolled out: the notions that "aqua" is universal, that it is less English-centric and therefore more culturally sensitive across many markets, or that it's a scientific term and therefore inherently preferable to use. However, all of these explanations are disingenuous and avoid what is likely the real reason (improving the company's bottom line), but are they just plausible enough to allay suspicion and discourage follow-up questions.

This seems reasonable to me.  And, consistent with the idea that the decision was motivated by marketing considerations, INCI is an industry group, not a governmental regulatory body:  https://en.wikipedia.org/wiki/International_Nomenclature_of_Cosmetic_Ingredients


text:
When you report Einstein units do you capitalize it?
When you report Einstein units do you capitalize it? Since we're using it in the context of a unit (and not a name) is capitalization necessary?
Example:
The measured value was reported as 10 Einstein.
or
The measured value was reported as 10 einstein.

Unit symbols may or may not be capitalised, but unit names are never capitalised, as far as I know.

It should be lowercase:  N = newton,  Pa = pascal, J = joule, for example, both of which are units derived from actual names.           As a slight aside, the unit is highly redundant and non-recommended. I'd also suggest this was a better fit for physics.SE

There is a question about this on English.SE.

See also our recommendations concerning units on meta.

The non-SI unit einstein is not accepted for use with the SI and should be avoided.
Anyway, unit names are normally printed in upright type, and they are treated like ordinary nouns. Names of units are spelled with a lower-case initial in English, except in the beginning of a sentence when a capital initial is used. This also applies to unit names that are derived from a proper name of a person such as kelvin, newton, joule, pascal, watt, hertz, coulomb, volt, farad, ohm, siemens, weber, tesla, henry, and thus also einstein.
For SI units, it is only the unit name degree Celsius that contains a capital letter. In keeping with the rule, the unit degree begins with a lower-case d and the modifier Celsius begins with an upper-case C because it is a proper name.
However, in quantity names containing a person’s name, the person’s name is spelled with a capital initial, for example, the Einstein transition probability for spontaneous emission $A_{jk}$.
These writing rules are in accordance with following references:

ISO/IEC: ISO/IEC 80000 Quantities and units (14 parts)
BIPM: The International System of Units (SI)
IUPAC: Quantities, Units, and Symbols in Physical Chemistry (Green Book)
NIST: Guide for the Use of the International System of Units (SI)
ACS: The ACS Style Guide

Citing NIST seems to be excessive. I recently asked for the Proper reference for the International System of Units, and it turned out that BIPM has priority over NIST as the latter is only a derivative of initial normative brochure from 2006.

From the ACS style guide (a reference book for conventions when publishing):

➤ Do not capitalize surnames that are used as units of measure:
ampere, angstrom, coulomb, curie, dalton, darcy, debye, einstein, erg, faraday, franklin, gauss, gilbert, gray, hartree, henry, hertz, joule, kelvin, langmuir, newton, ohm, pascal, poise, siemens, sievert, stokes, tesla, watt, weber.
Celsius and Fahrenheit are always capitalized. They are not themselves units; they are the names of temperature scales.


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Negative ions affecting galvanic cell voltage
I don't see why negative ions don't affect the voltage. I mean the electric field of a negative ion like $\ce{SO4^{-2}}$ is not like what $\ce{NO3-}$ produces that means that it should affect the overall voltage.
This is the same reason why for example for $\ce{Cu+^2}$ the reduction electrode potential is higher than for $\ce{Zn+^2}$

What reaction? How is the cell constructed? Need some more information here in order to answer.

Well, assuming that you're talking about an aqueous cell, in order to get the anion into solution, you need a salt: eg $\ce{CuSO4}$ or $\ce{Cu(NO3)2}$. Yes, the local electric field around a single $\ce{SO4^2-}$ anion would be stronger than around a single $\ce{NO3^-}$. However, at the same concentration of $\ce{Cu^2+}$ you would have twice as many $\ce{NO3^-}$ ions so the overall solution would still be neutral.
In a microscopic level, each ion is surrounded by a solvation shell of water molecules that all have their partial positive charge on their hydrogens pointing at the anion (or oxygen towards the cation). For this reason they are also shielded from the electrode. Speculation: double charged anions would probably have a thicker solvation shell.
I don't see how this explains the difference in potential of $\ce{Cu^2+/Cu}$ vs $\ce{Zn^2+/Zn}$ since both have the same charge.

Maybe on nano and micro lengths the charge distribution of electrons on both of them play to produce different electric field


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How does salt dough work?
Sorry for coming in here with little chemistry knowledge and a pretty inane question, but my googling skills have completely failed to cut through the enormous combined output of the baking brigade and all the lovely people making lovely things with this stuff.  Unsurprisingly not one of whom has paused to mention what's actually happening on a chemical level.
For the uninitiated, salt dough is made by mixing half a part table salt with one part wheat flour, and one half to one part water, though recipes vary greatly. This is then moulded into the desired shape and baked in an oven at anywhere between 60 and 180 °C, resulting in a hard and surprisingly tough (to me anyway) finished article which is often used as a decoration on Christmas trees. But what is actually happening?
I can guess at something involving the re crystallization of the sodium chloride, but wheat flour, whilst being so incredibly common, is to me a completely incomprehensible mix of proteins, starches and all manner of other stuff. All of which involves chemistry way over my head.
I would greatly appreciate an explanation of what's going on in this hardening process.

What you get is a polymeric (starch) composite, with salt as inorganic filler.
The salt particles give toughness to the resulting composite, while the starch phase, which is just the same as in your bread, minus the unnecessary blowing agent (baking powder, yeast, whatever), connects the particles in an elastic, flexible way. You get a material that is nearly as hard as a piece of rock salt, but much less brittle.
(Elastic here means more elastic than salt. Dry bread is quite flexible, compared to rock salt ;-) .)
See also https://en.wikipedia.org/wiki/Composite_material

Thanks Karl, definitely helpful though your explanation is somewhat at odds with my (very limited) knowledge of how bread works. As I understand it, in bread the starch acts as the filler with some being turned to glucose to "feed" the yeast. The binder in bread is the glutens, which as well as being very long are able to bond with each other, forming the strong stretchy gluten web, which is somehow effected by the presence of salt ions in the water, though at much lower concentrations than here.

The gluten is afaik important during baking, when the leavening agent produces CO2 bubbles. These pop (coalesce) very quickly without he gluten network, which stabilises the dough. It's an effect called strain hardening. Expanding bubbles stretch molecules at their surface. That's why bakers like wheat: High gluten content. While the dough sets during baking, the previously partially dissolved starch macromolecules form a second, stiffer network.

I guess for salt dough, it's not important, and rice flour (gluten-free) would also do nicely.

Thanks again Karl, that's very interesting. The idea of gluten free salt dough is certainly amusing, perhaps the gluten intolerance "epidemic" will spread to Christmas trees... Could you perhaps edit your answer to elaborate on the formation of the starch network? Are we talking about starch gelatinisation? https://en.m.wikipedia.org/wiki/Starch_gelatinization

I'd hope the epidemic completely moves to the Christmas tree. Certanily wouldn't bother me there. ;-) This wp article is quite confusing, i'm not sure what they try to teach. I'd have to read up again on the starch network before expanding the answer.


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Stoichiometry Chemistry Lab: Copper (II) Sulphate Pentahydrate and Iron Single Displacement Reaction?
In my grade 11 chemistry class, we performed a lab in which we added 1.00g of iron to 6.00g of copper (II) sulphate which was dissolved in water. Using stoichiometry, we predicted that if the iron was ferrous iron, 1.14g of copper would be produced and if the iron was ferric iron, 1.71g of copper would be produced. We waited 24 hours for the copper to dry and then we weighed the copper. The copper had a total mass of 1.24g. We then asked our teacher how this could be as we could not have ended up with more copper than 1.14g (as she already told us that the copper was ferrous). She told us that we need to make sure we rinse the copper really well.
So we decided to do a second trial. We were extra careful in our measurements and we made sure to rinse the copper well with distilled water. We waited 24 hours for the copper to dry and weighed the copper. The copper had a total mass of 1.21g. This was an improvement, but we also understood that we cannot have a percentage yield of over 100%, while ours was 106%.
What could have caused the copper to have a mass larger than predicted? Also, what was the reasoning behind my chemistry teacher advising us to rinse the copper before we let it dry?
Note: I can provide photos and calculations, as well as any other information if it is needed. Thank you!

Your results are actually very good.  The reason for thoroughly washing your copper is to remove any residual water-soluble un-reacted reagents.  If the copper were simply dried by sitting in room-air, the bulk of your extra 6% could still be water.  That's only 70 mg or 70 ul of water.  They way you produced your copper would likely result in a high-surface area product which can be tricky to dry.  Ideally, you would do multiple rinses in hot DI water then dry overnight at 105 C (really 70+C should do it).  If your yield goes down just a couple percent or so, you can still verify the excess water theory by repeating the drying.  If your yield goes down again then you are just having trouble getting all the water out.  Make sure you cool the copper, in a desiccator, before weighing.
If you don't have lab time for this here's what I'd try: take you copper home and put it an oven on low, about 200F or so, for several hours.  While still hot, place in the smallest reasonably air-tight glass jar you can find an get it back to the balance ASAP!  Good luck!  Oh, and NEVER touch the copper with your sweaty, oily fingers ;-)


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Can sensitive nuclei like 1H be exploited to obtain 13C spectra?
Measurement of 13C-NMR is essential for characterisation of compounds in organic synthesis, however, sensitivity can often be an issue, especially as molecules get larger and quantities get smaller, due to the low natural abundance of the NMR active 13C nuclei.
Standard carbon NMR is ran using a single pulsed experiment with heteronuclear decoupling for the 1H nuclei. This decoupling improves sensitivity by focusing the multiplets into a single sharp peak, with added sensitivity gained due to the nOe. Even with this enhancement, it can still often be difficult to obtain satisfactory spectra without resorting to using a stronger magnet (500 MHz/600 MHz is standard, the use of the 800 MHz spectrometer is prohibitively expensive for routine work) or running thousands of scans (there's a limit to how well this works anyway due to natural drift leading to peak broadening over time).
Unlike the standard 13C-NMR experiment, many 2D experiments such as HSQC are able to record data for 13C by observing the 1H nuclei (this is a huge oversimplification, I realise). This does, of course, lose the data for quaternary carbons, but I believe there are ways to get this data back (DEPTQ and various 2D experiments for example).
Why then are we unable to record 1D heteronuclear NMR using a similar technique?

I'm not sure I understand the question. If you detect 1H in a 1D experiment, you'll get proton chemical shifts out of it. This doesn't help you if you're interested in 13C shifts, then you have to measure a 2D and we're at the HSQC/HMQC again.

Well, yes and no. I'm talking about using/exploiting the 1H, not actually just measuring a proton spectrum.

We are able to record 1D heteronuclear experiments using these techniques, and absolutely do make use of this behaviour as a commonplace method for measuring 1D spectra of low sensitivity nuclei, such as $\ce{^13C, ^15N}$ and $\ce{^29Si}$. However, in order to get frequency information for the insensitive nucleus, we still need to use what is called a direct-detect method, where the nucleus of interest is actually the acquisition nucleus.
The most common method for sensitivity enhancement uses the general class of experiments called INEPT experiments; Insensitive nuclei enhanced by polarization transfer. The DEPT series (DEPT45, DEPT90 and DEPT135) are a sub-class of the INEPT experiment.
The signal of some nuclei can be improved by taking advantage of low power 1H decoupling during the relaxation period to allow nOe transfer to occur. This, as you mention, is typically how $\ce{^13C}$ spectra are recorded, and is why the $\ce{CH3}$ peaks ar much bigger than quaternary peaks. This method cannot be used for nuclei that have a negative magnetogyric ratio, such as $\ce{^15N}$ and $\ce{^29Si}$, as the nOe will actually generate a diminished signal. For these nuclei, the INEPT method is the preferred experiment for acquisition.
The INEPT magnetisation transfer step is at the core of most common 2D sequences, such as the HSQC. It provides a much greater signal enhancement than the nOe, although is strongly dependent on correct calibration of pulses.

The theoretical maximum sensitivity gain for nOe is $\displaystyle1+\frac{\gamma_\mathrm{S}}{2\gamma_\mathrm{I}}$
The theoretical maximum sensitivity gain for INEPT is  $\displaystyle\left|\frac{\gamma_\mathrm{S}}{\gamma_\mathrm{I}}\right|$

For $\ce{^13C}$, this equates to $\approx 3\times$ for nOe and $\approx 4\times$ for INEPT. For $\ce{^15N}$, it is $\approx -4\times$ for nOe (yep, negative) and $\approx 25\times$ for INEPT.
Glenn Facey at University of Ottawa has a couple of nice examples of INEPT and DEPT that are well worth looking at.
Both of these classes of experiment (nOE and INEPT) typically rely on interaction with nearby $\ce{^1H}$ signals: nOe is a through space interaction, and INEPT is a through bond interaction. Other nuclear pairs can be used, and many examples such as INEPT using $\ce{^19F-^29Si}$ have been published. The detection of insenstive nucleus environments with no directly attached sensitive nucleus (such as $\ce{^13C}$ quaternary centres) still requires long range coupling, and there are many published long-range INEPT methods.
The 2D HSQC method is an indirect detection method — we get information about the insensitive nucleus by acquiring another nucleus. Equivalent 1D variants of these 2D experiments do exist (so, a 1D HSQC), but they will only give you information about those $\ce{^1H}$ signals that have coupling to $\ce{^13C}$ — they don't provide any information about the $\ce{^13C}$ frequencies, unless you do a selective variant. So, a sel-HSQC will only irradiate a small band of the $\ce{^13C}$ frequency window, and therefore only give $\ce{^1H}$ signals that correlate to a $\ce{^13C}$ nucleus in this band. This is sometimes useful to probe a specfic question of structure, but a very tedious method of stepping through frequency bands of the $\ce{^13C}$ region (which is what a 2D experiment does for you).
For what it is worth, there are a number of cost-effective ways to improve sensitivity without having to fork out the big bucks for an 800. Choice of probe (X-observe, cryoprobe, microprobe) will offer improvements over increase in magnetic field at a fraction of the cost, and Shigemi tubes and micro cell tubes will also provide significant boosts in  sensitivity.

Seems I upvoted this a long time ago, but I'll drop the extra tidbit for future readers that getting a decoupled X spectrum using INEPT only works perfectly on XH groups; XH2 and XH3 peaks don't appear 2x or 3x as strongly. INEPT gets from H magnetisation to antiphase X magnetisation pretty well (giving a coupled spectrum with antiphase doublets), but converting that to inphase X magnetisation for decoupling is somewhat inefficient for multiplicities other than XH. More info in Keeler chpt 12 and also https://science.sciencemag.org/content/280/5362/421 for a way of going beyond INEPT...

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Why the atomic (metallic) radius of gallium is less than that of aluminium?
It is a well known fact that on moving down the group of p-block elements, the atomic radius is expected to increase. However, gallium is an exception. The atomic radius i.e., the metallic radius of gallium ($135~\mathrm{pm}$) is less than that of aluminium ($143~\mathrm{pm}$).
My book provides the following reason:

It is due to the presence of additional 10 d-electrons in gallium which offer poor screening effect for the outer electrons from the increased nuclear charge.

But if that is the case, why the effect is not that same in case of the heavier members?
Moreover, there must be something else which governs this exceptional property of gallium. Maybe the structure, but I am not sure. Can someone elaborate?

I am not sure of your values, but the book explanation leads me to  think that gallium has a fairly low atomic radius because the 4s, 3d and 4p electrons are so close to each other that the 3d are not completely shielding the 4s and p electrons from the nuclear charge.   It is more like one big level with 13 electrons.

First, you have to look at the definition of metallic radius, which is the half distance between two atoms in a lattice. It has a significant dependence on crystal structure.

Tanget relevant to the question and other answers:  Gallium has an orthorhombic crystal structure (CN = 6) whereas as aluminum has a face-centered cubic crystal structure (CN = 12).  This difference in coordination requires a goldschmidt correction to compare values as if gallium were 12-coordinated which makes the gallium's corrected metallic radius actually even smaller compared to aluminum at about $\mathrm{130pm}$.

Gallium exists as a diatomic solid in the lattice, which causes the atoms to be closer together on the  a axis because they are covalently bonded. Gallium's covalent radius is about $\mathrm{122pm}$ which is very much smaller than the metallic radii of either element. This value between the bonded atoms lowers the average distance between atoms giving gallium a lower metallic radius.
If you look at the Van der Waals radii of the elements, (more representative of a monoatomic gaseous atom) you will see that aluminum is actually smaller though not by a lot, giving some credence to the d-electrons explanations, but not a complete explanation in regard to the metallic radius.

With each additional proton that is added to a nucleus, the attraction between the nucleus and the electrons is increased and thus the wave function is contracted. This trend is most obvious when going horizontally along a group: a lithium atom is much larger than a neon atom even though the valence electrons are in the same shell — and it is even true for the difference between boron and neon, if you want to restrict it to a single subshell.
Each time a new shell is opened, the atomic radius jumps upwards since these always (i.e. quantum mechanics calculations say so) have a greater contribution further away from the nucleus with at least one additional lobe. So far the basics.
What happens when going from aluminium to gallium? We should consider the case when going across the periodic table from the corresponding alkaline metals sodium and potassium. From sodium, it is two steps to aluminium but from potassium to gallium is 12 steps — the entire 3d block is wedged in-between. Thus, from a hypothetical starting point, we experience a much greater contraction by the time we reach gallium compared to aluminium.
Note that it is irrelevant that the 3d electrons are there and ‘shielding’. Shielding doesn’t play that great a role as is often said.
Another ‘step’ can be experienced when going from indium to thallium. Here, we suddenly have 4f elements perched in-between and thus the radii of indium and thallium are again rather similar.

As Joseph's comment above alludes to, the 3d electrons in Gallium exhibit poor shielding, which causes a phenomenon known as "d-block contraction" as seen in elements from Ga to Br.  Despite being in the same group as Aluminium, the introduction of the d-orbital means Ga has significantly more protons (31 vs 13) so the positively-charged nucleus has a much greater pull in Ga than in Al.  Because of d-block contraction the nucleus is able to exert much greater pulling power on the outermost s- and p-level electrons thus reducing the atomic radius.  This also cause the ionisation potential of Ga to be higher than that of Al, when the normal trend is for ionisation potential to decrease down a group.


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What kind of materials are plastic syringes made from?
I'm trying to figure out what kind of materials my disposable plastic syringe is made from. I have a Medefil MIS-1130 syringe and when I visited the company website I couldn't find any information on what the disposable syringes were made from (although they did tell me a lot on what's NOT in them).
I'm trying to find a regulation or something that could help me identify what kind of plastic and rubber are used to make the shill, plunger and seal that make up the disposable syringe. Can anyone on here please give me some guidance?

its typicaly polypropylene

The syringes are advertised as “ISO compliant” on the website of the manufacturer. This is suspiciously vague and does not necessarily mean that the product complies with all requirements of all relevant ISO standards. (Also note that, concerning compliance and quality, the website mentions that the manufacturing facility is “ISO certified”, which is not permissible. However, the manufacturer actually has been certified to comply with ISO 9001:2008 and ISO 13485:2003 – that’s more like it.)
Ideally, such prefilled syringes are made in accordance with the applicable standards of the ISO 11040 Prefilled syringes series and other documents that are referenced in these standards (e.g. ISO 7886-1:1993 Sterile hypodermic syringes for single use – Part 1: Syringes for manual use).
However, the normative minimum requirements for materials are very general and not very restrictive. For example for the material of the barrel according to ISO 11040-6:2012 Prefilled syringes – Part 6: Plastic barrels for injectables:

For the manufacturing of the barrels, suitable polymers shall be selected and used, based upon the intended application and processing techniques, which include, for example, the method of sterilization or the method of decontamination. The material shall exhibit the appropriate performance properties, e.g. oxygen and water permeability.

or the material of the plunger stopper according to ISO 11040-5:2012 Prefilled syringes –
Part 5: Plunger stoppers for injectables:

The elastomeric material used shall meet the requirements specified in Clause 7.
  Plunger stoppers shall be made from the elastomeric formulation originally tested and approved by the end user. The manufacturer of the plunger stopper shall ensure the conformance of each delivery with the type sample and compliance with previously agreed functional and compendial requirements.
The elastomeric material shall withstand two sterilization cycles when autoclaving in saturated steam at (121 ± 2) °C for 30 min without impairment of its function under conditions of normal use. If other sterilization methods are used, e.g. irradiation, the suitability of the material shall be evaluated.

The mentioned Clause 7 includes minimum requirements on hardness

The hardness agreed between the manufacturer and the user shall not differ from the nominal value by more than ±5 Shore A when tested in accordance with ISO 7619‑1 on a special test specimen. Alternatively, the hardness can be tested on the plunger stoppers in accordance with ISO 48. If tested according to ISO 48, the microhardness shall not differ by more than ±5 IRHD from the type sample.

resistance to ageing

The maximum time between the date of manufacture of the plunger stoppers and their pharmaceutical use should be agreed upon between the manufacturer and the user.
The plunger stoppers shall maintain their performance characteristics throughout the entire shelf-life of the medicinal product, which is tested as part of the stability test by the user.

chemical requirements

The aqueous extractables of the plunger stopper material shall not exceed the limits specified for elastomeric parts defined in ISO 8871‑1.

and biological requirements

The requirements in ISO 8871‑4 shall apply.

In addition to the normative minimum requirements, ISO 11040-6:2012 also includes informative guidance on materials:

Materials used in the construction of syringe barrels for use in pre-filled syringes should be suitable for the process to be used, the sterilization method to be applied and for the specific requirements as primary packaging for injectable drug products. Materials used in the construction of pre-filled syringes should not cause the syringes to be detrimentally affected, physically or chemically, by the normal use of injectable preparations and should be safe for the intended use.
Certain grades of polypropylene (homo-polymer and/or co-polymer polypropylene), cyclo-olefin copolymer and cyclo-olefin polymer have been used for the barrels of pre-filled syringes in applications with a variety of injectable products (drugs and biologics).
The organization responsible for the marketing approval of the injectable drug product is required to demonstrate the appropriateness and compatibility of the selected material and the suitability of the prefillable syringe as a container closure system. This includes, among others, the verification of drug/container interactions (leachable, extractable and sorption assessments) in reference to regulatory guidance (e.g. FDA and EMA), pharmacopoeias and other guidelines (e.g. ICH).
To develop the necessary data as outlined above, a collaborative approach is recommended between the pharmaceutical company that holds the marketing authorization of the injectable product and the supplier of the packaging system, with reference to ISO 15378.

Furthermore, ISO 11040-6:2012 includes a list of polymer materials codification:

Cyclo-olefin copolymer (COC)
Cyclo-olefin polymer (COP)
Polypropylene (PP)
Other polymer (OTH)

Remarkably, the manufacturer mentions the materials glass, COP, and COC (without further context) for prefilled syringes on another page, whereas the package insert mentions “The luer lock syringes are made using polypropylene.”
For reliable information, you should directly inquire from the manufacturer of the prefilled syringes. Apparently, this company does not make the empty syringes but only fills syringes that are obtained from another supplier.


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What constitutes something being designated as big, bulky in terms of SN2 vs E2
When deciding between $\mathrm{S_N2}$ and $\mathrm{E2}$ for primary and secondary alkyl halides, is there a particular method or molecular size or any other threshold, which allows us to designate something as big and bulky?

I think you are referring to the fact that bulky nucleophiles hinder $\mathrm{S_N2}$, which is something you could have pointed out next time.

http://chemistry.stackexchange.com/questions/35071/can-bulky-nature-of-organic-compound-be-quantified http://chemistry.stackexchange.com/questions/35013/are-bulkier-molecules-more-likely-to-react


text:
Effective chemical for protecting hide glue from microbial attack
One problem with hide glue is that it can be attacked and degraded by bacteria and fungi.
To protect hide glue from this kind of deterioration I am interested in identifying an effective anti-microbial. Toxicity against beetle larvae would be a plus. Note that hide glue is typically prepared at 140F and so the chemical must be stable at that temperature. Also, any kind of volatility is out of the question because it needs to last for a long time (100 years).
The main chemical I am considering right now is triclosan, but would be interested in any other ideas.

I don't know where you are located, what your regional law and regulations are regarding triclosan, or what quantities you would be using, but it's worth noting that in the US it is an Environmental Protection Agency regulated pesticide.

@airhuff LOL I can see the headlines now: EPA raids rare book library looking for triclosan.

Silver salts are used to kill bacteria, algae and fungi.  Different types of clothing are 'laced' with silver to kill the bacteria that cause body odor.  Silver spoons were suspected historically to have self-sanitizing properties.  I ran a small analytical chemistry lab that was exempt from most emission oversight, but we could not dispose of silver waste into the sewer as it would destroy the 'good' bacteria at the water-treatment plant.  Different silver compounds, often phosphates, are used to strike a balance between required toxicity and durability/longevity.  If I were starting your project, I would at least start with a good literature search along these lines and some small scale proof-of-concept experiments if it seemed at all feasible.
I do want to be clear however, that I know of no reason to expect silver to be toxic to beetle larvae, at least not at the concentrations that would provide anti-microbial protection. Maybe you would need a second additive for that purpose

Benzalkonium chloride is commonly used wide range antimicrobal and antifungal agent. Whereas it's allways better to combine different reagents. In GxP rules, rotation of antimicrobals (regular changing) is required to prevent resistant bacterias or fungi.


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Thermal decomposition of magnesium bicarbonate
I was trying to solve a problem from a regional contest where it was a mixture of bicarbonates of $\ce{Na}$ and $\ce{Mg}$ which was put at high temperature. What are the reactions of decomposition?
I've thought that the reactions are:
$$\ce{2NaHCO3 -> Na2CO3 + H2O + CO2}$$
and $$\ce{Mg(HCO3)2 -> MgCO3 + H2O + CO2}$$
But, I found out that the second reaction is wrong. Actually, the reaction is
$$\ce{Mg(HCO3)2 -> MgO + H2O + 2CO2}$$
I searched on the web and I found the decomposition of other bicarbonates, like $\ce{Ca(HCO3)2}$ is similar to the first reaction. Is there any rule? When is carbonate and when is oxide? Or is $\ce{Mg}$ an exception?
Thanks!

In the future please don't use MathJax in Chem.SE titles, thanks. I've edited this one out for you. Anyway, it probably depends on how strongly you heat. MgCO3 itself is susceptible to decomposition to MgO + CO2. Quick google search indicates that MgCO3 decomposes at ~350 deg C whereas Na2CO3 itself decomposes (to Na2O + CO2) at ~400 deg C. I wouldn't fault you for giving the answer you gave.

Then, how can I know which of them has a higher melting point? If I heat them at the same temperature, can I know which carbonate descomposes in an oxide? And thanks for the edit

So there isn't anyting special for Mg. Thanks!

The usual factor controlling the decomposition temperature of carbonates is the charge density of the metal ion.

To my knowledge, group 2 bicarbonates are not stable outside of aqueous solutions. Seems like a nonsense question.

Highly related: https://chemistry.stackexchange.com/questions/124516/does-magnesium-carbonate-exist-what-makes-magnesium-bicarbonate-different-from/124519#124519

I can quite easily remember from my class XI studies that carbonates of alkali metals decompose on heating to give carbon dioxide and the corresponding metal oxide.
Moreover the thermal stability of the alkali metal carbonates increases with increasing cationic size because carbonate ion is big in size and increased cationic size leads to better bonding and hence greater stability.
In this regard the beryllium carbonate is the least stable alkali carbonate or in other words, an unstable alkali carbonate and readily decomposes into $\ce{BeO}$ and $\ce{CO2}$. $\ce{BeCO3}$ is so unstable that it can only be kept in an atmosphere of $\ce{CO2}$. $\ce{MgCO3}$ is also a bit unstable taking into consideration the above fact.
So the reaction you mentioned basically gives $\ce{MgCO3}$ as the product but $\ce{MgCO3}$ decomposes readily to give $\ce{MgO}$ and $\ce{CO2}$.
$$\ce{Mg(HCO3)2 -> MgCO3 + H2O + CO2}$$
and
$$\ce{MgCO3 -> MgO + CO2}$$
giving
$$\ce{Mg(HCO3)2 -> MgO + H2O + 2CO2}$$

The title of this thread is perhaps more accurately the thermal decomposition of aqueous magnesium bicarbonate, as the dry salt does not exist.
Per my personal experience, solutions of the salt can decompose on standing in the course of days or upon warming. To quote  Wikipedia on Magnesium bicarbonate, to quote:

Magnesium bicarbonate exists only in aqueous solution. Magnesium does not form solid bicarbonate as like Lithium. To produce it, a suspension of magnesium hydroxide is treated with pressurized carbon dioxide, producing a solution of magnesium bicarbonate:
$\ce{Mg(OH)2 + 2 CO2 → Mg(HCO3)2}$
Drying the resulting solution causes the magnesium bicarbonate to decompose, yielding magnesium carbonate, carbon dioxide, and water:
$\ce{Mg++ + 2 HCO3− → MgCO3 + CO2 + H2O}$

Further heating of the dry magnesium carbonate can result in its decomposition as noted above.
$$\ce{MgCO3 (s) -> MgO (s) + CO2 (g)}$$
Per Wikipedia on magnesium carbonate, its decomposition temperature is 350 °C.


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Why is finding [H+] when given [OH-]= 3.57e-5 M wrong when I use -log[OH-]?
Question/Problem
I know that [H+][OH-]=$10^{-14}$
And the problem asks to find out the concentration of H+ when the concentration of OH- is $3.57*10^{-5}$M
By using the equation above i find that the concentration of H+ is 2.67E-10
But if I use a different method as follows:
-log[OH]=pOH so by substitution, -log[$3.57*10^{-5}$]
then that gives me 4.425.......
then i subtract 14 from that number to get 9.574......
which that is pH
then i set 9.574=-log[H+]
so then you "e" everything which the answer comes to be  6.95E-5
why is this wrong?

Do you know the difference between $$9.574 = -\log x$$ and $$9.574 = -\ln x?$$

Oh no wonder, i get it now

Well You have to note that you cannot simply "e" everything because the base of log in chemistry is base 10 while you assumed that it was all base e. Ie ln(x). in order to get x, we use e, while if we say log(x) we have to assume that it is base 10 which is the general norm