Precision Balance (1mg) vs. Analytical Balance (0.1mg) I'm in the lucky position to building my own home lab and I'm looking to get a nice balance, but I don't want to waste money on something I won't use though.

I'm currently about to start second year chemistry (Organic) and I plan to take at least a few courses beyond this as time allows (I already have my Comp Sci degree and a full time job and I've been in and out of grad school for Mathematics and Computer Science). So this is just a hobby.  As far as what I'll be doing I honestly don't know chemistry well enough to say exactly, but I think it'd be fun to synthesize various molecules you come across and see what kind of purity I could get after running the experiment a few times.

I probably won't ever get much past 4th year undergrad or 1st year graduate level knowledge since I can only take one night class at a time.

I understand that analysis and p chem have a decent amount of use for sub-mg level accuracy, but how often do these situations come up in practice?

Should I spend the extra on a 0.1mg balance or just go for a 1mg balance?  This decision won't affect future lab purchases much more likely just the timeline of the purchases. Eventually I'd like to get additional useful instruments as I start needing them like a centrifuge and/or autoclave. But I probably won't ever get a FT-IR for example.

Either of these balances will require some care to set up. It will need to be on a solid surface away from any drafts, etc. However, neither would have the more expensive setup and maintenance requirements of say a $\pu{0.001mg}$ balance. And in either case you can take care of the cleaning and calibration yourself, particularly as it's just a hobby with no certification requirements.

I would seriously consider looking at used equipment from a place like LabX (I have no affiliation whatsoever). I would hesitate to purchase a used $\pu{0.001mg}$ balance because that is a very sensitive piece of equipment, but a $\pu{0.1mg}$ analytical balance; my lab always bought that sort of thing used.

You can probably get a good used $\pu{0.1mg}$ Mettler for example for the same or less than a new $\pu{1mg}$ balance and there is really no downside. It's easy to imagine you wanting to weigh out a few mg of some reagent and having the ability to do that with the extra accuracy of at least a $\pu{0.1mg}$ balance would probably be desirable.


Is Alkanol an official part of chemical Nomenclature? There are several references to alkanols here and here but neither of these sources are too reliable. The IUPAC gold book seems not to recognise the term and a quick google search reveals no really good sources.Whilst still not a super reliable source the fact there are no results on Wikipedia but simply a redirect to alcohol also furthers concerns and so the question is: Is the term Alkanol a 'real', 'proper', and 'official' term in chemistry?

IUPAC gold book doesn't recognize it, that's a pretty solid answer. I've never heard anyone say 'alkanol' before either, I've only heard people refer to these as simply alcohols. However, there are people who still use the word 'ethylene' to describe the compound 'ethene' so it could just happen to be some common language I haven't been exposed to.

To be clear, no it's not official, but yes it might be real terminology used by some organic chemists.


Does the amount of solute affect rate of solvation? I know that in order to increase the rate of solvation, you need to increase how often solute and solvent particles come into contact.

You can do it by:

agitating the mixture
increasing the surface area of the solute
heating up the solvent

But my question is: Does adding more solute (like adding more salt to water) increase the rate of solvation since the contact surface area is larger?

Edit: The original question in the book describes dissolving a certain amount of $\ce{NaCl}$ into water and asks, “What can be done to increase the rate of dissolution without affecting solubility?” I assume the rate is measured by how fast the mass of $\ce{NaCl}$ is consumed, which is $-\frac{dW}{dt}$.

If the answers are to be exclusive, then obvious choice is agitating the mixture.

If the answers are not to be exclusive, then agitating the mixture + increased surface by adding more solute increases $\frac{\mathrm{d}c}{\mathrm{d}t}$

Adding more solute does not increase rate of solvation,  but as the absolute surface increases, it increases $\frac{\mathrm{d}c}{\mathrm{d}t}$.

Note also without affecting solubility from your comment, so increasing temperature is out.


Why does the corrosion of a steel nail develop iron ions in the ends of the nail and hydroxide in the middle of the nail? We did an experiment in school where we put a steel nail in a solution consisting of:

1 g agar;
100 ml water;
100 ml indicator solution (which gets blue whenever $\ce{Fe^2+}$ is created and red when $\ce{OH-}$ develops).

We let it rest for a few days and this is what happened to the nail:

My hypothesis was that the corrosion would create two ends of the nail where reduction takes place in one end and oxidation on the other end. Why isn't this the case?

My teacher gave us a hint and told us it could have something to do with both:

The length of the nail;
The ion concentration of solution.

Very good question and thought provoking too. It appears that the nail somehow knows or is it is able to decide which part will be a cathode and which part will be an anode! Why is that so? Why one end was not a cathode and the other end an anode, just the one shown in the cartoon taken from here Imaging metal corrosion

So the fact you should keep in mind is that corrosion is superficially a random process just like cancer. Corrosion comes from corrodere "to gnaw to bits, wear away," it can start from any defect on the metal (scratch, impurity, etc.). However once it starts, just like cancer, it decides its propagation (i.e., the flow of electrons), hence the direction of cathode or an anode. It is likely that in your case the nail tip and the head had more defects than the center, so corrosion started at the tips. The pattern you are seeing is not universal or a fundamentally symmetric phenomenon.

Let us now bend a nail, create stress or defect in the center. Now the corrosion also starts in the center. Have a look at this gallery. Corrosion gallery


How does a volcano for a science fair work? In this video it shows me how to make a Volcano but what is the science behind it? http://www.youtube.com/watch?v=9b_gltKtERY

The food color-dyed water is just there to add volume and color.

The household chemicals used are acetic acid (vinegar, $\ce{CH3COOH}$) and sodium bicarbonate (baking soda, $\ce{NaHCO3}$).

The acetic acid and sodium bicarbonate react to form carbon dioxide gas ($\ce{CO2}$), dissolved sodium acetate ($\ce{CH3COONa}$) and water ($\ce{H2O}$).  It is the liberation of $\ce{CO2}$ gas that causes the foaming and bubbling.

The chemical reaction is written as:

$$\ce{CH3COOH(l) + NaHCO3(s) -> CH3COO-(aq) + Na+(aq) + CO2(g) + H2O(l)}$$

Where:
l = liquid
s = solid
aq = aqueous (dissolved)
g = gas (vapor)


Z-Matrix Generator for Gaussian I am looking for a way to convert XYZ coordinates into z-matrix for Gaussian input files. I need to specify certain coordinates in the matrix, therefore, the automatic built-in function from gaussview isn't of much help. Do you know of a program that can help me?
Thanks in advance for your help!

Try Open Babel, the windows GUI has a lot of features that may be of help depending on what are those specifics you need in your Z-matrix.


Are there any amphoteric substances other than water and ammonia? I have read about the Brønsted-Lowry and Arrhenius concepts to define a substance as an acid or a base. Our teacher told us that there are substances which behave both as an acid and a base, using water and ammonia as examples. He also asked us to find other substances with the same behavior. Does anyone know any more substances which are amphoteric in nature?

First, it may be worth elucidating the nomenclature a bit and distinguishing between amphiprotic and amphoteric  substances. The former are those that can react as either acids or bases in the Brønsted-Lowry sense, while the latter are those that react in either manner per the broader Lewis definition. As you mention, water and ammonia are amphiprotic (and thus also amphoteric by definition), which is evident even from the fact that both undergo auto-ionization:

$$
\ce{H2O + H2O <=> H3O+ + OH-}\\
\ce{NH3 + NH3 <=> NH4+ + NH2-}
$$

In fact, a large number of protic substances exhibit this type of behavior to an appreciable degree, including concentrated solutions of various acids (e.g., sulfuric, hydrofluoric, etc.). Another canonical example of an extremely common amphiprotic molecule prevalent in nature is bicarbonate, which you find in countless consumer products, as well as within the human body as a physiologic buffering agent.

Under the right conditions, a very wide array of molecules of diverse types can be amphoteric. The two necessary and sufficient criteria for amphoterism are:

The molecule possesses lone pairs, so that it can act as a base.
The molecule can accept electron pairs, so that it can act an acid.

Those are not rare properties, hence amphoterism isn't really an exotic phenomenon.


Melting point of objects and food What is the melting point of egg yolk?
Well, generally, when we break an egg, the yolk is in liquid state. After we pour it on a pan, the egg starts to cook and it hardens. My question is, can an egg melt/boil ?
What are the conditions to be imposed?

As Poutnik said, the melting point is well defined for pure substances. In the case of egg yolk, it is not a pure substance and also when you heat it, some molecules decompose. So we can't measure or even define a melting point for it. Also, the temperature is not constant, when the egg is melting.
A substances' phase doesn't get influenced only by temperature. Maybe it is strange that heating the egg yolk in a pan, makes it to behave like a solid. Actually, some complex changes in intermolecular forces are getting done.
If you want to cause an egg yolk to melt, I think that the only work to do is to heat it under the normal pressure.


What is the correct Lewis structure for calcium phosphate? In calcium phosphate, $\ce{Ca3(PO4)2}$, since the calcium and phosphate share an ionic bond, and the phosphorus and oxygen share a covalent bond, should the Lewis structure be like the following?

While I was checking other web pages I noticed there was another Lewis structure for the same molecule:

Are they both possible Lewis structures, or is the first structure the only "correct" version of the molecule?

Calcium phosphate crystallises in a number of different modifications of which β-$\ce{Ca3(PO4)2}$ is the stable room temperature and ambient pressure modification. It has a rather complex structure not easily described, consisting of chains and rings with phosphate and calcium ions in specific patterns. It is orthorhombic, but $c \gg b \approx a$.

Any attempt to simplify its structure by drawing it in pseudocovalent depictions like your lower scheme is doomed to fail. Better just use the upper scheme; unless you are expected to know a certain salt’s crystal structure that one is safe.


In cracking, why does C form double bonds? I've been reading about cracking and I've found various images and animations that show a process like this:

But I haven't been able to find out the sequence of events to get to this, and why this would occur.

I understand that cracking involves thermal decomposition, where a CC bond is cut. But that's as far as I can truthfully say I understand. To get this reaction below, you need:

(1) Two cuts. One cut would be between two Cs, and the other cut between a C and a H. Why does a H have to be cut off as well? Is this example I found actually one of many random and messy ones that just happens to 'give us what we want in the end'?

(2) The free H would then have to join the alkane. I get why this happens.

(3) A double bond is made. Why?

When we heat hydrocarbons to high temperatures (somewhat lower temperatures if we use a catalyst) they crack.  That means that carbon-carbon bonds are broken and smaller fragments are formed.  In the top part of the following drawing, I've picked n-heptane as an example.  You can see that there are 3

different bonds we can break and a number of different fragments will result.  Each of the fragments that result from breaking a C-C bond has an unpaired electron (there were two electrons in the bond we broke, each fragment retains one electron).  These fragments with an unpaired electron are called radicals.  The carbon with the unpaired electron does not have a stable octet configuration, therefor radicals are extremely reactive species and will react rapidly in a variety of ways to regain their stable octet configuration.

I've picked one of the possible radicals generated in the top drawing and show how it may react further in the bottom half of the figure.  Sometimes it will break a C-H bond and lose a hydrogen atom and form a stable compound with a double bond (an olefin), sometimes it will remove (abstract) a hydrogen from a hydrocarbon (heptane in our example) to generate a new alkane (pentane) and a new radical (perhaps a radical not formed in the initial cracking step).  Many of the radicals formed can rearrange to more stable radicals, this adds further complexity to the overall process.

So starting with just heptane we can get a large variety of saturated hydrocarbons and unsaturated hydrocarbons.  The situation becomes exponentially more complex if, as usual, we do have many more hydrocarbons present in addition to heptane.

The key here is that radicals are very reactive species - they want to stabilize by regaining the octet of electrons around the carbon atom that has the radical.  They can get that octet and become stable by adding a hydrogen atom (alkane formation) or breaking a C-H bond and eliminating a hydrogen atom to form an alkene.


Is it possible to calculate the 'overall' redox potential of a system? For a solution where many redox reactions occur (e.g. cytosol, soil solution), can we consider anything as 'overall' redox potential of a system? Is there a formula to quantify it in relation to the amount of each of the constituents?

Is it possible to calculate the 'overall' redox potential of a system?

No, it is impossible. You can have a situation where a cellular compartment contains a high ratio of $\ce{NAD+:NADH}$ (i.e. favoring oxidation) and a low ratio of $\ce{NADP+:NADPH}$ (i.e favoring reduction). This is advantageous for the cell because it can run oxidation of some metabolites in parallel with reduction of other metabolites, just by employing enzymes with distinct co-factor requirements.

This is different from pH. A solution is either acidic or basic, not both. The difference arises because acid/base reactions are usually fast, and water acts as acid or base, "communicating" acid/base status across the solution.


Why doesn't frozen sugar solution taste sweet? I'll try to make this as brief as possible:

Dissolved two teaspoons of table sugar (sucrose) in about 250ml water. Sipped it, and as expected it tasted sweet. I let the rest of it sit in the freezer overnight. Next day, I took out the frozen sugar solution and, well, licked it.

Surprisingly, I could barely taste any sugar in it. It was almost as though I was licking regular ice.

Why is it that I'm not able to perceive any sweetness here?

I was under the impression that since the solution, being a homogeneous mixture of sugar and water, was sweet, the "popsicle" I made ought to taste sweet too (since the sugar would be evenly distributed over the volume of the ice).

Where is the sugar?

When you freeze a dilute aqueous sugar solution pure water freezes first, leaving a more concentrated solution until you reach a high concentration of sugar called the eutectic concentration.  Now you have the pure water that's frozen out, called proeutectic water, and the concentrated eutectic sugar solution from which the sugar is finally ready to freeze along with the water.  Upon freezing this eutectic composition forms a two-phase eutectic mixture, in which the sugar may appear as veins or lamellae (like veins of some ores among Earth's rocks, though these typically form form a different process).  If that structure is in the interior of the ice cube, likely since you cooled the solution from the outside, then licking the outside you got only the pure water proeutectic component.

See https://en.wikipedia.org/wiki/Eutectic_system for more about this process.

Addendum:  I tried this with store-bought fruit juice which was red in color.  Poured it into an ice tray and froze it overnight in my household freezer.  It appeared to be a homogeneous red mass and tasted sweet, but was also mushy implying some liquid was still present (after overnight freezing for an ice cube sized sample).  The juice was roughly 10% sugar by weight.


Is there any difference between balancing redox reactions in acidic and basic medium? For balancing redox reactions in an acidic medium I follow this procedure:
1. Writing the half reaction
2. Balance the atoms other than $\ce{O}$ and $\ce{H}$
3. Balance the oxygen by adding water
4. Balance the hydrogen by adding protons
5. Make the number of electrons equal and add the half reaction.

Is there any difference if I change the medium if so what are the differences ?

The five points to pay attention are correct; and I would like to add a sixth one: depending on the reaction condition, namely if the reaction occurs in acidic, neutral, or basic solution, the outcome and the reaction product(s) of the half reaction may differ.

Take potassium permanganate, for example.  Under aqueous acidic conditions, this is a very powerful oxidiser, and formally $\ce{Mn^{7+}}$ is reduced all the way down to $\ce{Mn^{2+}}$.  This half reaction, starting from purple coloured solution of $\ce{KMnO4}$ to practical colourless $\ce{MnSO4}$ (assuming $\ce{H2SO4}$ as the acid present) may be written as

$$ \ce{MnO4^-  + 5e^- + 8H^+ ->  Mn^{2+} + 4H2O} $$

Under neutral conditions, the product however is manganese dioxide (formally $\ce{Mn^{4+}}$), which is poorly soluble in water and precipitates

$$ \ce{MnO4^-  + 3e^-  + 2H2O  ->  MnO2 + 4OH^-} $$

Eventually under (very) basic conditions, the reaction will stop already at a green colour in form of oxidation state $\ce{Mn^{6+}}$:

$$ \ce{MnO4^-  + e^- ->   MnO4^{2-}} $$

Paying attention to this has practical, preparative relevance, too.  It offers one way to attune the oxidation power of potassium permanganate.  A moderated reactivity often correlates with selectivity of the reactions.

This pH-dependence of oxidation-reduction reactions correlates with electrode potentials which often are already tabulated, like here or in the CRC Handbook of Chemical and Physics (keyword Nernst equation).

As a side note: These equations sometimes do not tell the complete story.  $\ce{MnO2}$, for example, may act / may be used itself as oxidiser. The simple sum formula however hides well that the oxidising power of this material depends how this material was prepared, too.


Easiest/Fastest way to obtain a good quantity of fine Iron powder? I am wanting to make a reasonable quantity (10-100g) of iron powder that I can use to visualise magnetic fields (by putting the powder in a thin layer between two transparent materials).
The finer the powder and more consistent the size the better. Ideally I would like a particle size 1 micron or smaller.
I've used a very fine grit sandpaper of about 2000 grit and this had a great result but took an eternity to produce.
I also made iron powder via electrolysis with an iron anode/cathode, in an aqueous solution of iron sulphate. I used an old repurposed PC PSU to supply 3V and however many amps the PSU could output for the electrolysis.
This produced some iron flakes which seemed a bit too large for what I wanted but it made a lot of iron particles quickly and with little effort.
Main question:
Do you know of any good processes I could use to make fine iron particles using electrolysis or another chemical process?
Related side questions:
What's a good liquid medium for suspending iron particles?
Is iron the best material to use for a viewing magnetic fields? How does it compare to other materials like Cobalt,Nickel,magnetite,Tungsten (Paramagnetic), NdFeB dust, etc?
A magnetic field viewer with silvery white particles (as opposed to black particles) could look nice. What might be a good way to turn the magnetic powder more silvery instead of black?
Is there a way to produce a good quantity of very fine, needle-like rods of iron? This could also work well for visualising magnetic fields as I believe the 'needles' would want to align to the magnetic fields.
I appreciate any help you can offer me!

A less laborious path to very fine iron dust (in fact, likely  pyrophoric iron) perhaps follows from first purchasing or preparing iron(II) oxalate (a path is detailed in a reference provided below). Next, thermally decompose the salt and finally, stabilize the nano-iron product with say mineral oil.
The details on the thermal decomposition in an atmosphere of hydrogen is discussed in this paper: Pyrophoric Nanoparticles and Nanoporous Foils for Defense Applications. See also this discussion which includes a suggested preparation for Iron(II) oxalate together with a video here.
Lastly, follow a similar process as is outlined in this US Patent 3,520,676 which claims, to quote:

Pyrophoric powder of metals (such as iron, cobalt, nickel and alloys thereof) is stabilized to prevent its spontaneous ignition. Such stabilization is achieved by Wetting the metal particles with a high-boiling organic liquid (such as an ester of carboxylic acid, mineral oil, silicone oil or fatty acid) and holding the wetted particles in the presence of oxygen until a thin oxide layer is formed on the surfaces thereof.

in order to stabilize the very fine iron product for your intended use.
A particular advantage to this approach is the low cost, and actually, not a great need for a large amount of the nano-particles for your demonstration purposes.


Calculating the potential of an electrochemical cell (Nernst equation) This is the problem:

Calculate the ${E_\mathrm{cell}}$ (not ${E°}$) at 25°C.
I found that the reaction is $$\ce{Cu(s) + Fe^2+(aq) -> Cu^2+(aq) + Fe(s)}$$ $\ce{Cu}$ gets oxidized and $\ce{Fe^2+}$ gets reduced.
I added the ${E°_\mathrm {Cu^{2+}/Cu}}$ and ${E°_\mathrm{Fe^{2+}/Fe}}$ to get ${E°_\mathrm{cell} = -.101}$ . But I'm not even sure if I'm supposed to add them or subtract them because I've seen both done, which is really confusing me. In what situations do you add, and which situations do you subtract half reaction potential values?
I then plugged this calculated value into the Nernst equation.
I said Z = 2 and plugged in the correct constants. However, I am not sure if I calculated Q correctly. I did [anode]/[cathode], but is it supposed to be [products]/[reactants]? It is unclear to me because my instructor did the latter, but I keep seeing the former everywhere else because that is how you would calculate it in an equilibrium problem. I know it doesn't make a difference in this specific problem, but I want to know for future problems when concentration of all components are given (such as if they are all aqueous).
From these steps, I calculated that ${E_\mathrm{cell} = -.164 V}$
How to approach this question?

If you are not sure about the use of the parameter Q, go back to Nernst's law for each electrode, namely $$\ce{E_{M^{2+}/M} = \pu{ E°_{M^{2+}/M}} + \pu{0.0296 V} \times  \log[M^{2+}]}$$ Here the redox potentials of the two half-cells are, according to Nernst's law :$$\ce{E_{Cu} = \pu{+ 0.339 V} + \pu{0.0296 V} \times  \log0.10} = \pu{0.309 V}$$  $$\ce{E_{Fe} = \pu{-0.44 V} + \pu{0.0296 V} \times  \log0.0030} = \pu{- 0.515 V}$$
So the overall performance of the cell is $$\ce{E_{cell} = E_{Cu} - E_{Fe}}=  \pu{0.309 V} - \pu{(-0.515 V)  =  0.824 V}$$


Optical techniques for kinetics studies What optical methods can be used to measure the concentration of reacting gases in a reactor during an exothermic and fast reaction? I need to monitor and measure the local concentration of my gases (in the methane production reaction) along a reactor bed in order to be able to clarify the kinetics of this reaction. Which optical techniques are able to make such measurement?

There are a number of techniques but what you would use depends on how you plan to start the reaction. What use to be called Flash-Photolysis and now called Pump-Probe Spectroscopy is the most general and can be used from the femtosecond to multi-second time range. This is a general method as it measured the absorption of all species produced. It works by using a suitable duration and wavelength flash of light, usually from a laser, to initiate the reaction and then at various time delays after this,  the absorption spectrum of any species produced is measured. Absorption can be in the uv, visible or ir ranges. Sometimes Raman scattering or fluorescence can be used instead of absorption. What is used depends on equipment available and the species you need to measure.

If you cannot start the reaction with light then mixing methods can be used, but these are relatively slow, longer than microsecond time scale and more like milliseconds. In the Stopped-Flow method the reactant gases   are mixed as rapidly as possible in the chamber so as to start the reaction then this is monitored with absorption etc. as above. Additionally, there are flow methods whereby the gaseous reactants are mixed at the end of a long flow tube and are monitored spectroscopically at various points as they flow along a tube. The gases are pumped to cause the flow and so distance along the tube is an alias for reaction time.  Each of these methods can be used with gases or liquids.


Why is the polarity of the B-N bonds in borazine the reason for its reduced delocalisation of π electrons? My professor mentioned that borazine is aromatic, but not as much as, say, benzene. The reason he gave for this was that the B-N bonds in borazine, being polar, inhibit resonance. He also mentioned that the same inhibition of resonance occurs in h-BN. Why is this so? What difference does bond polarity make?

Let us understand and try to answer your question with a few simple points.
First it is important to understand why exactly is borazine aromatic. Boron has an empty p-orbital and Nitrogen has a lone pair of electrons.
According to requirements for pi-backbonding, there needs to be a sigma bond between both the atoms and requirement of one empty orbital for allowing the electrons of the side atom to delocalize into it needs to be satisfied. Along with this, the lone pair which needs to be delocalized should be in kept in a pure p-orbital. Boron and Nitrogen both facilitate these requirements and thus the lone pair delocalization takes place.
Thus in a way, we could say that the aromaticity in the borazine moeity is due the pi-backdonation of unpaired electrons.

Understanding that the main reason is sideways pi-backdonation for the cyclic delocalization of electrons, one also needs to keep in mind that Boron and Nitrogen have a significant difference in their electronegativities. The polarity occurs due to the pi-backdonation but keeping in the mind the electronegativity of nitrogen is higher than boron, the delocalisation isn't as pronounced as in the case of benzene where the bonds are non polar. Thus leading to comparatively lesser delocalisation in broazine than in benzene. You could similarly compare h-BN and grpahite; in the former case the delocalisation is less pronounced due to difference in electronegativity than in the later case where a very well pronounced delocalisation is observed.


For a given teabag and dunking duration - will I get more caffeine in a bigger cup or the same amount? When choosing a cup of tea, I can choose a big cup or a small cup. I'm wondering if I get more caffeine with the bigger cup or the same amount of caffeine. (in moles)

My question is: For a given teabag and dunking duration - will I get more caffeine in a bigger cup or the same amount?

Assumptions

Teabag is placed in cup first - and hot water poured directly onto the bag
Same amount of stirring across both cups (assume three spoon swirls)
The main varying parameter is the volume of solvent changing

Question behind the question - trying to express it another way

Is there a kind of 'fixed rate of dissolution' that limits the amount of caffeine that comes out over a period of time - ie for two cups where one is 250ml and one is 300ml - and each have a tea dissolving time of 30s - wouldn't they have the same number of moles of caffeine at the end because of an upper bound on the rate of caffeine dissolving?

From what I remember from Fick's laws, I would say that the factor that limits the rate of dissolution is the concentration of the already-dissolved compound. At first, you don't have any caffeine dissolved so the rate of dissolution is fastest. As time goes by, more and more caffeine becomes solvated and that hinders the dissolution of remaining caffeine from the tea leaves. At any point of time until saturation, the 300 ml cup will have faster dissolution rate because it has more "free space" to solvate the caffeine.

Edit: However, after thinking this for a while, I have to add that in practice the difference might be too small to be detected. The solubility (~ saturation point) of caffeine in water is 20 g/l at 20°C while the concentration of caffeine in tea is 0.1 - 0.4 g/l – huge difference. It might be that from caffeine's point of view, the "free space" is more or less the same in both cups.


Terminology for describing nuclei participating in metallic bonds I've read many times that in metals (for example sodium metal), what constitutes the crystal lattice is an arrange of atoms, since no electron is given or lost, they are just shared between all the atoms within the structure.

In other texts though, it is written that metallic bonding is comprised of cations "surrounded" by free electrons since electrons are not held to the nucleus as in a single metal atom in the gas phase. From a formal point of view, which of the two statements is correct? Does it make a difference to consider cations or atoms when describing metallic bonding?

Which is the more appropriate terminology?

Both your statements mean the same thing, however it is convenient to think of metals as kernel + sea of electrons, as it greatly simplifies visualization, conceptualization and thus calculations. From a purely numeric viewpoint, i prefer the 'sea of free electrons' model or that cations are surrounded by free electrons.


What is the minimum pressure for existence of a liquid solution? The minimum pressure for the existence of liquid solution is?

  $0.6\;\mathrm{atm}$
  $0.8\;\mathrm{atm}$
  $0.77\;\mathrm{atm}$
  $0.9\;\mathrm{atm}$

Although this seems to be an informative question, I want to know the process by which we get this answer.

I know that the critical pressure for a liquid solution is $p={ a\over {27b^2}}$ but what is the value of a and b in that case?

I expect an easy to understand answer for high-school level.

The only things I can figure for $a$ and $b$ in $p=\dfrac{a}{27b^2}$ are the coefficients from the van der Waals equation for nonideal gases:

$$\left(p+\dfrac{n^2a}{V^2}\right)\left(V-nb\right)=nRT$$

These constants have appropriate units to give you pressure: $a\ \left( \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}\right)$ and $b \ \left(\dfrac{\text{L}}{\text{mol}} \right)$.

$$\dfrac{a}{27b^2}\implies\dfrac{\dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}}{27\left(\dfrac{\text{L}}{\text{mol}}\right)^2}=\dfrac{\text{atm}}{27}\implies p$$

However, these constants are empirically derived, and thus different for every substance, and so the critical pressure would be different for every substance.

For example, for argon $a=1.355 \ \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}$ and $b=0.0320\ \dfrac{\text{L}}{\text{mol}}$, so $p$ for argon is

$$p=\dfrac{1.355\ \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}}{27\left( 0.0320\ \dfrac{\text{L}}{\text{mol}} \right)^2}=49.01\ \text{atm}$$

Were you told which substance you should be using? For all substances, $b$ appears to be less than $1$, so we need to find the minimum $a$ and maximum $b$ combination to get the minimum $p$. However, by my calculation, all values of $p$ will be greater than $1$. The value for water is $220.5\text{ atm}$ (which does not make sense given that a vast number of liquid solutions are stable at normal atmosphere pressure).


How does osmosis work? I have (very) basic knowledge in chemistry.

I am curious as to why osmosis works. I read a couple of introductory chemistry books and also searched online but most explanations I found only explained the phenomena (i.e. solvent molecules move through semipermeable membrane from low solute concentration area to high concentration area), but not the reason behind it at the molecular level (i.e. what makes more solvent molecules move towards the high solute concentration area than in the reverse direction).

Wikipedia seems to go a step further:

  ... there is an interaction between the solute and water that counteracts the pressure that otherwise free solute molecules would exert.

And a bit later:

  The virial theorem demonstrates that attraction between the molecules (water and solute) reduces the pressure, and thus the pressure exerted by water molecules on each other in solution is less than in pure water, allowing pure water to "force" the solution until the pressure reaches equilibrium.

I think I understand what it means, but what I don't understand is this: osmotic pressure is said to be a colligative property, i.e. only affected by the concentration of the solute and not by its type. If the reason for osmotic pressure is the attraction between water and solute molecules - wouldn't we expect different osmotic pressures for different kinds of solutes?

According to this document (https://arxiv.org/pdf/physics/0305011v1.pdf)

  "Water molecules can pass through the membranes in either direction, and they do. But because the concentration of water molecules is greater in the pure water than in the solution, there is a net flow from the pure water into the solution."
  Hydraulic equilibrium is achieved when the lower concentration of water in the high pressure side is balanced by the higher pressure and therefore higher energy pV on that side

This is the diffusion model of osmosis which is easy to understand at your level.


Are all homogeneous mixtures just solutions? Or can there be a homogeneous mixture that is not a solution?

Homogeneous and heterogeneous classifications are scale-dependent. What we mean when we say a mixture is homogeneous is that there are no visible phase differences on the scale of interest.

For example, milk (when homogenized) and lotions are both examples of colloidal suspensions. There are no phase differences at the human scale, but neither one is a solution. On a more microscopic scale, you could call the suspended micelles phase boundaries, and in that case might not call the mixture homogeneous.

"Solution" always implies mixing on the molecular level (single-phase).

To summarize:

  Solutions are always homogeneous mixtures, but homogeneous mixtures are not always solutions.


Universal symbols for molar mass, molarity, and molality This is a really basic (perhaps trivial) question, but I just want to confirm:
What are the SI symbols for molar mass, molarity, and molality? For molality and molarity, there is a difference, but I’ve found that the symbol is the same for molarity and molar mass, i.e. M (a big m).
Also, what are the meaningful differences between molarity and molality? Does it just have to do with mass (i.e. difference in mass between solute and solvent) vs. concentration (i.e. within the solution)?
I get the sense that molarity is defunct? I read this here -
https://www.nist.gov/pml/special-publication-811/nist-guide-si-chapter-8

According to IUPAC, the official name for molarity is amount of substance concentration, and its symbol is $c$. The symbol for molality is $m$ or $b$, and the symbol for molar mass is $M$.

what are the meaningful differences between molarity and molality?

Molality does not change with temperature because mass and chemical amount is temperature-independent. The amount of substance concentration (molarity) does because substances expand (mostly) with in increasing temperature. So for most applications where the temperature changes, molality is the better choice.
On the other hand, for room temperature applications where volumetric measurements are convenient, it is common to use molarity as the measure of concentrations.


What is the molecularity of a reversible reaction? I know that the definition of molecularity of a reaction is number of species reacting in an elementary step. But considering the theory of microscopic reversibility for elementary reactions, each reaction can be assumed to be reversible. In that case, if $$\ce{A <=> B + C}$$ then what is the molecularity of the reaction? Is it 1 or 2?

For the reaction $$\ce {A <=> B + C}$$
its molecularity is unimolecular. If we were given $$\ce{B + C <=> A}$$
Its molecularity would be bimolecular. Thus we see molecularity is dependent on the number of reactant particles, even if it's in equilibrium.


Dissolving to Create a Solution with Ethanol If at 95% Ethanol can dissolve chemical X at 1mg/1mL then theoretically if you have 500mg of chemical X you can dissolve it in 500mL of $\ce{EtOH}$ to get 1mg/mL correct?

My question is, if you have 75.5% $\ce{EtOH}$ for example in a 750mL solution - how many mL will you need to create the 1 mg/mL concentration of quantity of 500mg chemical X?

Is it just 19.5% more of the 500mL volume or do you have to figure out the molarity of each and then go from there?

Knowing that the material (solute)  has a solubility of 1 mg/ml in 95% ethanol does not help you predict its solubility in 75% ethanol. It could be more soluble or less soluble, depending on the properties of the solute.


Formulations for wax thermostatic element for effect in alternative temperature range Wax thermostatic elements are used automobile cooling systems and anti-scald mixing valves to control flow rates or to adjust mixing of flow inputs. These devices operate on the principle of thermal expansion of "waxes" to convert heat energy to mechanical energy (Wikipedia article).  A review of thermostatic mixing valves for sale indicates that the common formulation for the "wax" within the element in those devices expands in the range 20 °C to 50 °C.
Is it possible to formulate a "wax" that effectively changes volume in a lower temperature range such that it could operate a mixing valve, beginning at 5 °C, for instance, or are the materials typically used in this application not effective at temperatures below 20 °C?

Although typical applications for wax elements are centered around 40 °C, it is possible to create formulations for wax elements that change volume in the lower range of 5 °C to 20 °C.  The example below, although not linear in the lowest temperature extreme of the range of interest, behaves roughly linearly in the majority of the range.


Naming ionic compounds of group B When naming an ionic compound which includes a transition metal, the oxidation state is written between parantheses, yet some metals show an exception. Which elements represent this exception, and what is the reason?

The relevant document including the rules of formal IUPAC nomenclature is the Nomenclature of Inorganic Chemistry — red book from 2005. Section IR-5.4.2.2 (Use of charge and oxidation numbers) is relevant for the cases you discuss. It reads:

[…]
[…]
The oxidation number (see Sections IR-4.6.1 and IR-9.1.2.8) of an element is indicated by a Roman numeral placed in parentheses immediately following the name (modified by the ending ‘ate’ if necessary) of the element to which it refers. The oxidation number may be positive, negative or zero (represented by the numeral $0$). An oxidation number is always non-negative unless the minus sign is explicitly used (the positive sign is never used). Non-integral oxidation numbers are not used for nomenclature purposes.
Several conventions are observed for inferring oxidation numbers, the use of which is particularly common in the names of compounds of transition elements. Hydrogen is considered positive (oxidation number $\mathrm I$) in combination with non-metallic elements and negative (oxidation number $\mathrm{-I}$) in combination with metallic elements. Organic groups combined with metal atoms are treated sometimes as anions (for example, a methyl ligand is usually considered to be a methanide ion, $\ce{CH3}$), sometimes as neutral (e.g. carbon monooxide). Bonds between atoms of the same species make no contribution to oxidation number.
Note that oxidation numbers are no longer recommended when naming homopolyatomic ions. This is to avoid ambiguity. Oxidation numbers refer to the individual atoms of the element in question, even if they are appended to a name containing a multiplicative prefix, cf. Example 12 above. To conform to this practice, dimercury($2+$) (see Section IR-5.3.2.3) would have to be named dimercury($\mathrm{I}$); dioxide($2-$) (see Section IR-5.3.3.3) would be dioxide($\mathrm{-I}$); and ions such as pentabismuth($4+$) (see Section IR-5.3.2.3) and dioxide($1-$) (see Section IR-5.3.3.3), with fractional formal oxidation numbers, could not be named at all.

While the section in itself does not mention the possibility of omitting oxidation numbers, the examples obviously do:

$\ce{Na2[Fe(CO)4]}$   sodium tetracarbonylferrate($\mathrm{-II}$), or sodium tetracarbonylferrate($2-$)

The generally accepted rule is that oxidation numbers can be left out if the element in question is commonly known to exist in only one oxidation state. That, for example, includes groups 1 and 2 but also elements such as zinc of which only one state is common.


Why does hand sanitizer leave your skin feeling cool? I noticed, and perhaps many others have too, that the application of hand sanitizer (mainly ethanol), leaves one's hands feeling rather chilly after application.

What is responsible for this phenomena? Is it the high heat of vaporization of hand sanitizer? However, explanation doesn't hold water; water has a standard heat of vaporization of 40.65 kJ/mol while ethanol has a heat of vaporization of 38.56 kJ/mol.

Could it be the low boiling point of ethanol? Hand sanitizer disappears (vaporizes) within seconds upon rubbing the hands together. Water, however, does not.

Additionally, how does one square a high heat of vaporization with a low boiling point? If it takes a lot of energy to vaporize something, then how can that something have a low boiling point?

As user137 noted, this is from evaporation of the alcohol. Evaporation of a liquid takes the enthalpy of vaporization away with it, making you cooler.  Just like how sweating helps keep you cool in the summer, and how swamp coolers (evaporative coolers) keep houses cool in the southwest.


Is the bond in HF ionic while it is covalent in HCl? Why would a hydrogen atom "donate" to fluorine in an ionic bond but not in $\ce{HCl}$?  Why would $\ce{H}$ and chlorine share instead of $\ce{Cl}$ just stripping it away like $\ce{F}$ does?

Bonds can be completely covalent as in $\ce{Cl-Cl}$, $\ce{H-H}$, etc. In these cases the electron density is shared equally between the two atoms.  Bonds can also be ionic as in $\ce{Na^{+} Cl^{-}}$ where much (~80%) of an electron has been transferred from the sodium atom to the chlorine atom.  Between these two extreme cases exist a continuum of bonds that are called polar, covalent bonds.  In such cases the electron density between the two atoms is not shared equally, but neither is the electron transferred extensively from one atom to the other.

Just as we can draw resonance structures for acetone and discuss how the real molecule is a weighted superposition of these two structures, we can draw resonance structures for inorganic compounds and, again, discuss how much the various structures count towards the "real" description of the molecule.  In the diagram below, two resonance structures are shown for sodium chloride.  The structure on the left represents a covalent bond and the structure on the right represents an ionic bond. Just as acetone is a mix of the two resonance structures, so is sodium chloride.  When the weighted mix of the resonance structures lies far to the left, we can say the compound or bond is covalent; when it lies far to the right we can say the compound or bond is ionic.  Everything in between represents a polar, covalent bond.

One can crudely estimate how polar a diatomic covalent bond is by using the following methodology.  The dipole moment of a molecule is given by
$$\ce{\mu~=~Q*d}$$
where the dipole moment is in Debye (D), Q is the partial charge on an atom measured in coulombs and d is the distance between the two atoms (the bond length) in meters.

In the case of $\ce{HF}$, the dipole moment is 1.86 D and the bond length is 91.7 x 10^(-12) m.  Noting that 1 D = 3.3356 × 10^(−30) C·m, solving for Q we find a partial charge on one of the atoms of
$$\mathrm{Q~=~\frac{[1.86 ~x~3.3356 ~x~ 10^{-30} ]}{[91.7~x~10^{-12}]}~=~6.77~x~10^{-20}~C}$$

Now since one electron equals 1.6022 × 10^(−19) C, by taking the ratio
$$\mathrm{\frac{[6.77~x~10^{-20}]}{[1.6022 × 10^{−19}]}~=~ 0.42}$$
we estimate that roughly 42% of an electron has been transferred from the hydrogen atom to the fluorine atom.  This gives us a rough idea of how ionic the polar, covalent bond is in $\ce{HF}$.  Repeating this calculation for $\ce{HCl}$ suggests that roughly only 18% of an electron has been transferred in this case.

Clearly the polar, covalent bond in $\ce{HF}$ has more ionic character than the analogous polar, covalent bond in $\ce{HCl}$.  The reason why the $\ce{HF}$ bond has more ionic character is the greater electronegativity of fluorine (3.98) compared to chlorine (3.16).


What happens to Aluminium in Autoclaved Aerated Concrete Aluminium is used in Autoclaved Aerated Concrete in order to produce the holes.

On Wikipedia I'm reading about the ingredients: sand, calcined gypsum, lime (mineral), cement, water and Aluminum powder.

Also it says "Aluminum powder reacts with calcium hydroxide and water to form hydrogen".

I'm wondering what Aluminium ends up as if anyone can help with detailed reactions.

$$\ce{2Al + 7Ca(OH)2 -> Ca[Al(OH)4]2 + 3H2 ^ + 6CaO}$$
or maybe
$$\ce{2Al + Ca(OH)2 + 6H2O -> Ca[Al(OH)4]2 + 3H2 ^}.$$
I think you'd get basically (ha) aluminium hydroxide.  This is effective because bases can dissolve the protective aluminium oxide layer that usually protects aluminium metal from corrosion.  The concrete is foamed by the hydrogen.


Proportion of gases filled in cylinder of sea-divers In my book it is given that earlier cylinder were filled with 2% oxygen and 98% helium. Nowadays it is filled with 11.7% helium, 56.2% nitrogen and 32.1% oxygen.

There are few questions which come up in my mind

Earlier 2% oxygen is changed to 32%. So was 2% oxygen enough for huaman?
98% helium is reduced to 11.7% why it is so?
I don't get reason why helium and nitrogen are filled?

Hint:
Try to use Henry's Law. According to which solubility of gases increases with increase in pressure. If we go deeper downwards, pressure increases and when we come to the surface, pressure decreases, so there might be that soem buubles are created in the blood which may be fatal. So air cylinders are made keeping this in mind.


Is this statement for the First Law of Thermodynamics correct? My Biochemistry book states the First law of Thermodynamics like this:
"The First Law of Thermodynamics states that in a closed system, energy is conserved".
My book defines Closed System as a system which can exchange energy with its surroundings but not matter.
My doubt is how can energy be remain conserved(constant) if the system can exchange energy with its surroundings.
Please clarify.

You can think of the first law of thermodynamics as an energy accounting method. It says that all inputs and outputs of energy should add up to the total change in the energy of a system. The exchanged energy is generally classified as work or heat. By "energy can neither be created nor destroyed," it is meant that nuclear reactions, in which the structure of nuclei changes with accompanying transformation of matter into energy and vice-versa, will not be considered. The reactions considered are ordinary chemical reactions involving making and breaking of chemical bonds resulting from interactions between electrons and nuclei.


Can 100% covalent bonds exist? Every covalent bond has some ionic character and every ionic bond some covalent character. I can understand why a completely ionic bond is an ideal situation. But completely covalent bonds can exist(?).

Take the case of $\ce{H_2}$ or any other diatomic molecule between identical atoms. How can these have any ionic character?

According to Pauling's famous The Nature of the Chemical Bond , 3rd edition, at page 73:

  In the hydrogen molecule a quantum-mechanical treatment has shown that the two ionic structures $\ce{H+H-}$ and $\ce{H- H+}$ enter into resonance with the extreme covalent structure $\ce{H-H}$ only to a small extent, each ionic structure contributing only about 2 percent to the normal state of the molecule.

He goes on to state that other diatomics such as chlorine should have even less ionic character.

So all molecules, even homonuclear diatoms, have some ionic character.

For further information on diatomic hydrogen's ionic character see section 1.5 of the book and The Normal State of the Hydrogen Molecule Journal of Chemical Physics vol. 1, pages 593-596.

See also Overlaps of Trial Functions for the Hydrogen Molecule. II. Covalent and Ionic Character of H2 J. Chem. Phys. 23, 176


Why does the $V_{\mathrm{max}}$ of an enzyme not change with competitive inhibition? Shouldn't it decrease since there are fewer active sites? My textbook states that $V_{\mathrm{max}}$ doesn't change for competitive inhibition. This makes no sense to me, since decreasing the amount of active sites available should lower the rate of an enzyme catalyzed reaction.

My attempt at rationalizing this is that $V_{\mathrm{max}}$ is dependent on the rate at which the enzyme/substrate complex dissociates into free enzyme + product by the equation Kcat * [E]. So, since it only takes into account the enzyme/substrate complex, we basically don't consider any enzymes with inhibited active sites.

Is that reasonable?

You can think of $V_{max}$ as describing the nature of the active site - it's deftness at converting substrates to the product, the rest is just probability. Competitive inhibitor does not change properties of the active site - they just hang there for some amount of time until the E-I complex dissociates hence they don't affect the maximum theoretical conversion rate of that enzyme.
Competive inhibitors only decrease the chance of inhibitor binding to the enzyme. Thus you can always raise the concetration of your substrate to the state that probability (now the other way around) of inhibitor binding the enzyme will become negligible with regard to the substrate allowing it to work at his maximum rate.


Solubility of Calcium Fluoride in water $\ce{CaF2}$ is insoluble in water, but $\ce{CaCl2}$, $\ce{CaBr2}$, $\ce{CaI2}$ are soluble. Why is this so?

Many factors contribute to the solubility of a compound in water. In this case, the main difference between the compounds is the bonds’ strength.

Fluoride is very electronegative, and thus the bonding forces between $\ce{Ca^2+}$ and each fluoride anions are very strong, they are tightly held together in the crystal form. A few water molecules do not have enough strength to get in between and dissolve it.

Chloride is less electronegative, thus $\ce{CaCl2}$ has weaker bonds, then comes $\ce{CaBr2}$ and at last $\ce{CaI2}$, the weakest of them all. You may also expect increase in bond length depending on the crystal structure, which may also contribute to the solubility properties of the given compound. So weaker the bond strength, greater the solubility – at least that’s the expected.

In fact, $\ce{CaCl2}$ is more soluble than $\ce{CaF2}$, and $\ce{CaBr2}$ is even more – as expected. However, I did some research and the Wikipedia article on $\ce{CaI2}$ claims it’s less soluble than $\ce{CaBr2}$, which is not the expected, but it does not provide a source for that information. It might be true, though, since there are a variety of factors that may or may not contribute to it’s solubility in water, such as it’s crystal structure and the particular characteristics of iodide.

You can do the experiment yourself to find out which one is more soluble, since all solubility facts and numbers we know come from experiments. I couldn’t find anything on the solubility of $\ce{CaI2}$ in my CRC Handbook of Chemistry and Physics either, so I can’t provide you with a precise answer on why it’s less soluble than $\ce{CaBr2}$ (if in fact it is), but I hope you can now understand some of the factors that make $\ce{CaF2}$ the least soluble compound on that list.


How can the intense color of potassium permanganate be explained with molecular orbital theory? In the permanganate ion, manganese is in the $+7$ oxidation state, therefore it is a $\ce{d^0}$ ion. $\ce{d^0}$ and $\ce{d^1^0}$ ions don't absorb visible spectrum radiation because there are no electrons to jump to higher crystal field orbitals (or in the case of $\ce{d^{10}}$ ions, there are no free orbitals for electrons to jump to). I think that the explanation to this question has something to do with the ligand donating electrons into $\ce{Mn^{VII}}$'s empty d orbitals or molecular orbital theory or $\ce{d-d}$ transition or none of them.

Also, chromate is also colored wherein $\ce{Cr^{VI}}$ is a $\ce{d^0}$ ion, therefore its not absorbing radiation due to electrons jumping to higher orbitals in the crystal field. What is causing it to absorb visible radiation?

Let’s take a look at a qualitative MO scheme for a tetrahedric transition metal complex whose ligands have three p-type orbitals each. On the left of figure 1 you have the metal orbitals ($\mathrm{3d}$, $\mathrm{4s}$ and $\mathrm{4p}$) and on the right the twelve degenerate ligand p-orbitals (transform as $\mathrm{a_1 + e + t_1 + 2t_2}$). Only orbitals of the same symmetry will mix, so the $\mathrm{t_1}$ ones stay where they are originally. I have colour-coded both the $\mathrm{t_2}$ (red) and $\mathrm{e}$ (blue) orbitals.

Figure 1: Qualitative MO scheme of a tetrahedric complex with σ and π bonding between metal and ligands. Double vertical lines represent electron pairs.

Note that the image is qualitative only. The order of energy levels can be different; it is not possible to predict it a priori. The more electronegative a side is, the lower its orbital energies will be. Manganese(VII) is lacking seven electrons so the orbital energies will be very low, will oxygen profits from its high electronegativity. All in all, the energies will be somewhat similar.

We can consider the filled orbitals to be mostly ligand-centred and the empty ones to be mostly metal-centred, in full alignment with the picture of a $\ce{d^0}$ $\ce{Mn^{VII}}$ core and four $\ce{O^2-}$ ligands with electron octets.

The electronic transition that causes the intense purple colour corresponds to an excitation from $\mathrm{t_1}$ to $\mathrm{e}$ — $\mathrm{t_1}$ should always be the highest ligand orbital, while tetrahedric symmetry typically implies that $\mathrm{e}$ lies lower than $\mathrm{t_2}$ on the metal’s side. The higher the central metal’s oxidation state, the lower its orbitals and the closer it is to the ligands’ p-orbitals.

Since the transition is from a ligand-centred to a metal-centred orbital, it is called Ligand to Metal Charge Transfer (LMCT). Oversimplified, one could think of it as a one-electron reduction of manganese by an oxygen ligand to intermediately create $\ce{Mn^{VI}-O^{.}}$.

The transition is obviously spin-allowed (filled to empty orbitals) and the Laporte-rule is invalid in tetrahedric complexes as there is no centre of symmetry. Therefore, the observed colour is very intense.


When will acid-base reactions occur before substitution reaction For example, given the reaction below:

The substrate is secondary, the solvent is aprotic and it is to react with a strong bulky base. From this, I assumed that the reaction would be $\ce{E2}$. However, it instead begins with an acid-base reaction where the tert-butoxide removes the proton of hydroxide and then proceeds with a $\ce{Sn1}$ reaction where the oxygen attacks the carbon of the leaving group to form an epoxide.

How can you determine if an acid-base reaction will take place before proceeding with a substitution reaction?

Acid-base reactions (like the one shown, involving proton transfer between electonegative atoms) are almost always extremely fast - for all practical purposes instantaneous (diffusion-controlled), and therefore much faster than substitution reactions. In the example given, the bromocyclohexanol is of comparable acidity to tert-butanol (maybe even more acidic), so we would expect instant proton transfer, and the observed result.

The initial proton transfer might not be so favorable if you used a much weaker base, so you could get a different result - or no reaction.


How does pH level change the concentration of dissolved oxygen (DO) in sea water? I am planning to investigate the relationship between $\mathrm{pH}$ and dissolved oxygen.

I would like to have a reason without taking in account the plants, animals and decomposer roles.

I want to know if $\mathrm{pH}$ affects the concentration of dissolved oxygen. For example:

If I added $\ce{HCl}$ to water, would the concentration of DO change?
In same way, if I add $\ce{NaOH}$, would the concentration of DO change?

The equilibrium of dissolved-to-gas-phase oxygen is not affected by the aqueous hydronium ion concentration ($\ce{H3O+}$), and thus the pH of water does not affect dissolved oxygen concetrations of sea water.


Radical formation via tritium to helium decay Following on from this question about tritium water, I was wondering whether the decay of tritium to helium-3 has ever been used in organic synthesis. Such a transition would presumably result in no strong $\ce{C-He}$ interaction and thus radical or cation formation at $\ce{C}$.

Here are two reasons why this approach would be impractical for organic synthesis:

(1) The half life of tritium is 12.3 years, so the decay is too slow to be practical as part of a synthetic method

(2) While on a scale of nuclear reactions, tritium decay is relatively low in energy emitted, it still dwarfs a typical chemical reaction (18.6 keV or 429,000 kCal/mole) and may cause degradation of the molecule (or the bulk material) in unpredictable ways.

But it could be useful for other types of chemical studies. I will defer to others in the forum to elaborate.


Why is are weak acid-strong base and strong acid-weak base reactions irreversible? In the case of both are strong, I can see why it's irreversible, because the acid is very willing to give a proton and the base is very willing to take one.
Now here are my two main questions:

In the case of both are weak, we say the reaction is reversible. Does this mean the reaction happens then the products react again to give reactants, or that the reactants don't run out and the reaction doesn't happen at "full potential"?
In the case of a reaction between a strong base and a weak acid (or the other way around), why is it always irreversible? I mean since one of them is weak, shouldn't it give out an unstable product which is willing to react again, like

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

The $\ce{CH3COOH}$ is a weak acid. The $\ce{CH3COO-}$ isn't stable so shouldn't it react again?

Yes the following reaction is reversible.

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

There is an equilibrium between the species $\ce{CH3COOH, OH-}$ and $\ce{CH3COO-}$ in aqueous solution.

Thus the gist is that particular $\ce{CH3COOH}$ and $\ce{CH3COO^-}$ molecules aren't stable, but rather continuously react.


What kind of filter can remove zeolite dust from ethanol? I only really have high school level knowledge of chemistry, and also I don't really own any lab equipment.
I'm using 3A zeolite molecular sieves to dry ethanol (and sometimes isopropanol) and they seem to be working great.
However, I'm always left with a bunch of dust.
I can let the dust settle over night and then pour off the top in the morning, but it's still got an orange tint to it. I can keep repeating that process but it starts to take a really long time.
I tried running it through a coffee filter but the particles are too small for the coffee filter to catch.
What should I look for in a filter paper / what other material could I use to filter the remaining dust out of the ethanol?
I don't want to distill it, I don't have the setup for that.

Since this never got a proper answer I thought I'd come back with a follow-up.
Essentially the root cause was low quality sieves (I had just picked the cheapest I could find on the internet). There are probably other things that work but all of the following captured the dust when used alone:

Whatman #6 filters do a decent job. I have #597 too but I don't recall if they worked here. (Whatman filter grade chart).
A bed of sodium bentonite over a #597, which  I was using to filter other stuff, also incidentally captured the sieve dust.
I never actually tried celite; but it was recommended in the comments so that could be a solution, too.

And the following also helped with the issue:

Simply letting the solutions sit for a day then pouring off the top left a significant portion of the dust settled at the bottom. I'd do this anyways before running it through a filter, to lighten the filter load.
More importantly, they were just cheap sieves. The second time I used them after baking them there was significantly less dust, and by the third time the issue was basically solved.

The next time I get cheap sieves I think I might wash them in the solvent first just to get the initial dust load out, recharge them, and then use them for their intended purpose. Although I suspect that if I balance the time, cost, and energy of doing this with the extra cost of better sieves, it's probably a good idea just to pay for the better sieves.
Pretty obvious in retrospect but maybe a fellow non-chemist will find it useful some day.


Why does there need to be a planar intermediate in 2,2'-dibromobiphenyl when one of the phenyl groups rotates? I was in class and my professor brought up the example of 2,2'-dibromobiphenyl:

In her example, one of the bromines was actually pointing into the page and the two phenyl groups were not actually on the same plane of the page. Let's call the phenyl with the bromine that points into the page group 1 and the one that is on the page group 2. She then said that in order for the group 1 phenyl to point out from the page (a conformer) there will be a planar intermediate where the bromines are very close to each other during the rotation (just like in the picture), and this will require a large amount of energy (20 kcal/mol) which will again drop off once the group 1 rotates till it's pointing out of the plane. My question is, why doesn't it just rotate the other way where the bromines will not come into close contact and there will not be a planar intermediate? Am I overthinking this?

I think that your professor was just mentioning a rotational angle vs relative energy plot. It is really important to understand relative energy difference between different points in such plots. The situation when two bromines are in the same plane and close to each other will represent the highest energy spot, this actually means that the probability of the molecule to be in such spot is extremely low, but technically not zero, that is why some reaction that require such spot may happen. In the picture below you can see the plot for biphenyl, in case of the dibromobiphenyl one of the peaks will be much higher then the others. But you are actually right, molecule itself would prefer to rotate the other way.


How to explain that an increase in temperature favours an endothermic reaction? I'm trying to explain how a change in temperature favours an endothermic/exothermic reaction from a collision theory point of view, but I'm not really familiar with the concept just yet. For an increase in temperature, can I say that:

  An increase in temperature leads to a greater frequency of collisions between reactants and products, leading to an increased rate of reaction. Since the rate of the endothermic reaction increases more than the exothermic reaction, the equilibrium shifts towards the reactants/products.

If this is right, I'm struggling to see why the endothermic reaction rate will increase more with this logic. The exothermic reaction has a lower activation energy, so shouldn't its rate of reaction increase more than endothermic since it has a lower energy requirement to complete the reaction? Unless the increase in particles with enough energy for the exothermic reaction is comparatively lower to the increase in particles with enough energy for the endothermic reaction (aka proportion)?

I believe that collision theory in itself wouldn't be enough to explain this behavior. You can explain it by having Arrhenius equation and the activation energy in mind. What you got wrong was the effect of temperature on reaction rates depending on the activation energies. Actually, the speed constant of a reaction with higher activation energy will be more significantly affected by variation in temperature. This can be verified easily by manipulation of the Arrhenius equation. Apply ln to both sides to obtain:

$\ln{k}=\ln{A}-\frac{E_a}{RT}$

The graph of ln(k) as a function of (1/T) is a straight line with slope's magnitude directly proportional to activation energy.

Now, assuming that the endothermic reaction has higher activation energy than its inverse, it will be more affected by, say, an increase in temperature. While both reactions will become faster, the speed constant of the endothermic reaction will increase to a greater degree and that will shift the equilibrium favoring the products of the endothermic reaction.
If the temperature is decreased, the endothermic reaction will be slowed down to a greater degree, favoring the products of the exothermic reaction.


What makes C=O more stable that C(OH)₂ When there are $2$ hydroxy groups on same carbon, they tend to undergo dehydration ( of course with exceptions like chloral where $\ce H$ bonding prevails).

Why does this happen? We are breaking a sigma bond in order to get a pi bond between $\ce C$ and $\ce O$ ?

Bond stability is the answer.
The energy required to break a $\ce{C-O}$ bond is $\pu{85.5 kcal/m}$; the energy given off when a $\ce{C=O}$ bond is formed is around $\pu{178 kcal/m}$ (reference).  These bond energies suggest that a carbonyl will generally be a few $\pu{kcal/mol}$ more stable than the corresponding gem-diol $[(2 \cdot 85.5) - 178 = \pu{-7 kcal/mol}]$.
[Entropic effects also play a role (see @Martins comment), but while bond enthalpies will change within a series of compounds the entropic effects should be similar and roughly cancel out when we compare compounds.]
However, the $\pu{7 kcal/mol}$ difference noted above isn't large, so there is a delicate balance in the equilibrium between a carbonyl compound and its corresponding gem-diol. Since $\pu{1.4 kcal/mol}$ will shift an equilibrium by roughly a power of 10, small factors like dipole moments, inductive effects, etc., are enough to stabilize one side of the equilibrium more than the other and significantly shift the equilibrium from one side to the other.
If water is present, then an equilibrium will be established between a carbonyl compound and its corresponding gem-diol.  Below is an interesting comparison of acetone and formaldehyde.  A seemingly small change (changing $\ce{H}$ to $\ce{CH3}$) shifts the equilibrium from one side to the other (by 6 powers of 10!).

Here is a link to a Wikipedia article that has a few more interesting examples of carbonyl-diol equilibria.
In the case of chloral ($\ce{CCl3CHO}$) it is the strong electron withdrawing inductive effect of the $\ce{CCl3}$ group that destabilizes the already positively polarized carbonyl carbon driving the equilibrium to the diol side.
As an aside, if you're not familiar with the history of chloral (or as it is often called, chloral hydrate) google "Mickey Finn" or "knock out drops."

As I've mentioned above a change of just a few $\pu{kcal/mol}$ can markedly shift the equilibrium between a carbonyl and the corresponding gem-diol. Lest the reader think that gem-diols are a rarity, here are a number of examples of compounds that exist predominately in the gem-diol form:

ninhydrin (because of the H-bond)
$\ce{R-(CO)-(CO)-(CO)-R'}$ (middle ketone houses the gem-diol; see detailed explanation)
1,1-dihydroxycyclopropane ($\mathrm{sp^3}$ carbon relieves ring strain)
floral hydrate (similar to chloral hydrate)


CAS numbers for salts of molecules Often times Pharma molecules come as different salts where the Cation changes. Do these have typically the same or different CAS numbers?

e.g. Take Glucosamine Hydrochloride.
https://en.wikipedia.org/wiki/Glucosamine

It is available as as both the Na and K salt. But I cannot seem to find if they all have different CAS numbers.

What I have now is: 3416-24-8

Further complication seems that there is a Glucosamine Hydrochloride as well as a Sulfate. The wikipedia page describes both but I cannot see different CAS numbers. The sulfate itself seems available as either a Sodium or Potassium salt.

Any thoughts?

Every different compound has a different CAS number, even if they are enanthiomers when isolated and published (their racemic mixture has another CAS number). No exception! See following examples:

L-(−)-Glyceraldehyde: CAS Number [497-09-6].

D-(+)-Glyceraldehyde: CAS Number [453-17-8].

DL-(+/-)-Glyceraldehyde: CAS Number [56-82-6].

For your question, I found the CAS Number of D-(+)-glucosamine hydrochloride, which is [66-84-2]. The CAS Number [3416-24-8] is for glucosamine (free base).

The CAS Number of D-glucosamine sulfate is [14999-43-0]. Each molecule contains two molecules of glucosamines and one molecule of $\ce{H2SO4}$, the two acedic hydrogens on each of amine groups. Its molecular formula is $\ce{[C6H14NO5]2.SO4}$, giving the formula weight: $\pu{456.42 g/mol}$.

The CAS Number of D-glucosamine sulfate potassium chloride is [216699-44-4]. Each molecule contains two molecules of glucosamines, one molecule of $\ce{H2SO4}$, and two molecules of $\ce{KCl}$. Its molecular formula is $\ce{[C6H14NO5]2.SO4.2KCl}$, giving the formula weight: $\pu{605.52 g/mol}$. However, NIH website listed [1296149-08-0] as the CAS Number of glucosamine sulfate Potassium chloride, which must be issued a fairly recent time. Spectrum Chemical Co. sells the compound as a dietary supplement under the same CAS number.


What is the reason that tertiary alcohol pKa is higher than primary alcohol pKa? A tertiary alcohol has a $\mathrm{p}K_\mathrm{a}$ of 18 while a primary alcohol has a $\mathrm{p}K_\mathrm{a}$ of 16.
Shouldn't the tertiary alcohol be more acidic since its conjugate base is more stabilized due to induction?

The short answer is that the steric hindrance of the tert butyl alcohol makes it hard to add an protonate the hydroxy group.

Based on this lecture from the vanderbilt university. The answer is due to steric hindrance:

The hydroxy group of the tert butyl alcohol is intramolecular stabilized by hydrogen bonds (2nd image), between the methyl hydrogens and the oxygen. In the other hand, i.e. the hydroxy group of methanol is intermolecular stabilized with another methanol molecules ("Behave more like a water") (3th image). Therefore, the equilibrium $\ce{ROH}\rightleftharpoons\ce{RO^-}+\ce{H^+}$ is more displaced to the right with the methanol than with the tert-butil alcohol.

The protonation of ter butyl alcohol is slower than methanol (Figure 4th and 5th), for the same reason: steric hindrance.


By how many pKa units will the pKa value of a ketone be minimized by addition of a Lewis acid I want to enolize a ketone, for instance acetone with a $\mathrm{p}K_\mathrm{a}$ value of about 20 (in $\ce{H2O}$).

I'd like to know if there exists a rule of thumb to predict how the $\mathrm{p}K_\mathrm{a}$ of the ketone will decrease if we add one of the following Lewis acids:

a) $\ce{(c-Hex)2BCl}$ (chlorine anion can be eliminated)

b) $\ce{Bu2BOTf}$ (triflate anion is a good leaving group)

c) $\ce{NaI}$ (sodium is bigger than lithium, and I expect the bond to oxygen is less strong meaning the decrease of the $\mathrm{p}K_\mathrm{a}$ will be less than in d))

d) $\ce{LiI}$ (iodine is bigger than lithium, therefore good hard-soft-mismatch; and oxygen is hard as well)

e) $\ce{HCl}$ (As @SYK suggested in his answer of my recent question
[1], the proton is the strongest Lewis acid as possible and will lower the $\mathrm{p}K_\mathrm{a}$ as much as possible, that even a very weak base, may be of $\mathrm{p}K_\mathrm{a}~9$ as $\ce{NH4Cl}$ will (or won't it?) form the cis-enolate)

I'm especially interested in the lowering values for a), b) and c). My questions:

Is my qualitative lowering of $\mathrm{p}K_\mathrm{a}$ values a < b < c < d < e correct?
Do you have somes guesses for decreasing values for a, b and c ?

I used to work in a group that specialised on the total synthesis of polyketides. For obvious reasons, syntheses of polyketides rely on aldol additions a lot. Yet, I never heard of any simple rule of thumb for the $\mathrm{p}K_\mathrm{a}$ value differences so I suspect there isn’t one.

Both boron reagents enolise ketones practically by themselves, $\ce{NEt3}$ is typically added to capture any excess protons. On the other hand, I have never heard of lithium or sodium enolising a ketone by themselves.

Finally, $\ce{HCl}$ (or any strong proton source for that matter) will protonate the ketone but not to the enolate. Instead, the protonated ketone with $\ce{C=\overset{+}{O}-H}$ will be obtained.


What's wrong with this gas law problem? I recently took a chemistry test in which the following problem was offered:

a 137 mL gas sample is collected over $\ce{H2O}$ at $753\ \mathrm{mmHg}$ and $22\ \mathrm{^\circ C}$. What is the volume of the dry gas at STP if the vapor pressure of $\ce{H2O}$ at $22\ \mathrm{^\circ C}$ is $22\ \mathrm{mmHg}$?
a) $106\ \mathrm{ml}$
b) $117\ \mathrm{ml}$
c) $123\ \mathrm{ml}$
d) $126\ \mathrm{ml}$

Here's my work:
To start with, I used Dalton's Law of partial pressures:
$$p_\text{total}=p_\text{gas}+p_{\ce{H2O}}$$
$$p_\text{gas}=731\ \mathrm{mmHg}\ \left(\text{at}\ 22\ \mathrm{^\circ C}\right)$$
Then I used the following proportion to find the volume of the gas:
$$\frac{p_\text{gas}}{p_\text{total}}=\frac{V_\text{gas}}{V_\text{total}}$$
$$V_\text{gas}=133\ \mathrm{ml}\ \left(\text{at}\ \mathrm{^\circ C}\right)$$
Which made sense to me because if the total number of moles of the gas mixture is not changing, and the temperature is not changing the ratio of the volumes should be the same as that of the pressures (shouldn't it?)
Finally I did a bit of dimensional analysis to change from $22\ \mathrm{^\circ C}$ to STP:
$$133\ \mathrm{ml}\ \cdot\ \frac{731\ \mathrm{mmHg}}{760\ \mathrm{mmHg}}\ \cdot\ \frac{273\ \mathrm K}{295\ \mathrm K}$$
$$V_\text{gas}=118\ \mathrm{ml}\ \left(\text{at STP}\right)$$
I feel like I'm missing/forgetting a key concept here because each individual step made sense in my head (and still makes sense now). The answer the prof gave was d. This is bugging me a lot and I would appreciate any help.

Let's derive the appropriate realtionship...

$$PV = nRT\tag{1}$$

so

$$\dfrac{PV}{T} = nR\tag{2}$$

but $nR$ is constant for the given sample of gas, thus

$$\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\tag{3}$$

For the sample the volume is 137 ml, the dry gas has a pressure of 753-22=731 mm Hg, and the temperature is 22 °C = 295 °K.

Now here is a bit of a wrinkle. What is STP? Until 1982 IUPAC defined STP as 760 mm of mercury at 0 °C = 273 °K. After 1982 IUPAC has defined STP as  750.06 mm of mercury at 0 °C = 273 °K.

For old IUPAC STP

$$V_2 = \dfrac{P_1V_1}{T_1}\times\dfrac{T_2}{P_2} = \dfrac{731\times137}{295}\times\dfrac{273}{760} = 121.95 = 122\text{ ml}$$

For current IUPAC STP

$$V_2 = \dfrac{P_1V_1}{T_1}\times\dfrac{T_2}{P_2} = \dfrac{731\times137}{295}\times\dfrac{273}{750} = 123.57 = 124 \text{ ml}$$

In looking at some possible variations, it seems the prof is using the wrong initial pressure, and the old IUPAC definition.

$$V_2 = \dfrac{P_1V_1}{T_1}\times\dfrac{T_2}{P_2} = \dfrac{753\times137}{295}\times\dfrac{273}{760} = 125.62 = 126\text{ ml}$$

NOTE

When collecting a gas over water you must subtract the partial pressure of water at the collection temperature from the atmospheric pressure, but you don't adjust the volume too.

Also this assumes that the atmospheric pressure given, 753 mm Hg, is for dry air.
If you were measuring the pressure with a laboratory barometer then you'd have to adjust the measured laboratory pressure for the relative humidity (partial pressure of water).


Comparing acidities of substituted and aromatic carboxylic acids When comparing the acidities of carboxylic acids, we primarily see the electropositivity of the carboxylic acid carbons, i.e. we see how effectively the negative charge on the carboxylate ion is dispersed upon ionization. Electron withdrawing groups increase the acidity by stabilising the anion via electron withdrawal.

Now consider the following two comparisons:-

  Formic acid ($\ce{HCOOH}$) and benzoic acid($\ce{C6H5COOH}$)
  Benzoic acid and 3-chloropropanoic acid ($\ce{ClCH2CH2COOH}$)

$$\begin{array}{c|c}
\hline
\text{Species} & \mathrm{p}K_\mathrm{a} \\
\hline
\ce{HCOOH} & 3.75\mathrm{-}3.77 \\
\ce{PhCOOH} & 4.19 \\
\ce{ClCH2CH2COOH} & 3.98 \\
\hline
\end{array}$$

(source)

My questions:

  Why is 3-chloropropanoic acid less acidic than formic acid. The negative inductive effect of chlorine helps in dispersing the negative charge on the carboxylate ion. It is not conjugated and too far to exert positive mesomeric effect. Therefore, the negative inductive effect should increase the acidity by improving the stability of the conjugate base.
  Why is benzoic acid weaker than formic acid? Unless I am mistaken, in several other places, the phenylic moiety is treated as slightly electron withdrawing, as in case of alcohol acidity, acidity of hydrogens in base catalysed reactions etc. I am unable to understand the resonance here or any other factor, that might cause this anomaly.

In the case of 3-chloropropanoic acid vs. formic acid, I suspect the disparity in expected acidity can mostly be explained by solvation effects. The 3-chloropropanoic acid is much bulkier than formic acid, and hence interacts with solvent molecules differently than formic acid does. There are two essential components to the thermodynamics of solvation:

Due to steric hindrance, it may be the case that the negative charge is less accessible to solvent molecules, and thus more weakly stabilized. I suspect the difference in enthalpy here is actually negligible, but I'm not able to find a reference at this time.
Perhaps more importantly, there is an entropic effect, in that larger molecules tend to require a larger number of solvent molecules to solvate them. These solvent molecules form highly ordered, cage-like structures, which is entropically unfavorable. I believe similar underlying mechanisms are at work in, e.g., the hydrophobic effect, micellization, protein folding, etc.

As Klaus Warzecha points out, if the chlorine were not as far-removed from the carboxylate moiety (since induction rapidly diminishes with growing distance), the strength of the inductive effect would trump the solvent effects.

In the case of benzoic acid, there are probably similar solvent effects occurring. Additionally, the benzene ring is actually electron-donating within the $\pi$-electron system of the molecule, which can only be destabilizing to the carboxylate conjugate base. Unlike in, e.g., phenol or aniline, the excess negative charge cannot actually be delocalized into the ring. Experimentally, the electron-donating effect of the ring (equivalently, the electron-withdrawing effect of the carboxylate) is confirmed by the fact that benzoic acid is severely deactivated in electrophilic aromatic substitution reactions by comparison to ordinary benzene. The benzene ring's electron-withdrawing effect within the $\sigma$-framework (rationalized, according to valence bond theory, by the greater s-character of $\mathrm{sp^2}$ carbons) of the molecule is probably quite weak. There may also be Coulombic destabilization between the strong $\delta^+$ of the carboxylate carbon atom and the $\mathrm{sp^2}$ carbons of the aromatic ring, again due to greater s-character, although that's speculative.


Bond angle in water vs in oxygen difluoride My teacher taught that bond angle $\ce{H-O-H}$ is larger than $\ce{F-O-F}$ due to less repulsion in the latter's bond pairs.
I know that bond angle $\ce{H-O-H}$ is $104.5^\circ$, but I'm confused with that of $\ce{F-O-F}$ because variable angles are given in articles on websites and in books. Some are $109.47^\circ$, $101.55^\circ$, $103.1^\circ$.

I want to know which bond angle is greater and why?

Water has a bond angle of 105º and oxygen diflouride has a bond angle of 103.8º, which cannot be explained by repulsion since fluorine is larger than hydrogen$[1]$. This is phenomenon is, however, explained by Bent's rule:

  Bent's rule is that in a molecule, a central atom bonded to multiple groups will hybridise so that orbitals with more s character are directed towards electropositive groups, while orbitals with more p character will be directed towards groups that are more electronegative.

p orbitals are oriented 90º from each other, so orbitals with more p character have smaller angles between neighboring atoms. The effects of this phenomenon can also demonstrated by comparing bond lengths.

$[1]$ Bent, H. A. Chemical Reviews 1961, 61 (3), 288.


How to tell between carbon powder or graphite powder? I have a black powder that I cannot reliably tell just by looking or touching whether it is graphite powder or charcoal powder.

Are there any simple tests that I can perform to tell whether it is one or the other?

What I have tried:

Adding it to water, it sinks to the bottom and leaves a slight oily looking film on the top. Upon stirring, it darkens the water and then gradually sinks back to the bottom after some time.
I tried writing some thick lines on a piece of paper with a graphite pencil and testing for conductivity (high resistance).
Same high resistance with the powder (granted this is very unrepeatable)
Rubbing it into paper it has a much blacker colour than the graphite (which to me looks like a really dark gray). However, I don't know if that is a reliable indicator.

I've also added a picture, if it helps:

EDIT:

Ok, I've made a paste using isopropyl alcohol and the carbon powder and made a small "stick" or rod of the stuff. I have now been able to achieve solid continuity at around 2 centimeters in distance and the resistance dropped from the high 40 - 50 kOhms for the loose powder, to the low 10s (ie 40 Ohms). If this helps anyone.

Short answer: You can't reliably tell the "difference" (by any simple means)

Longer version:

First of all, the graphite you were taught about at school is a pedagogical construct. Any macroscopic sample of "real" graphite would contain both the layered honeycomb (crystalline) bits as well as amorphous bits of carbon (in addition to God-knows-what bunch of other substances that managed to wiggle their way into your sample).

What about charcoal then? It's amorphous carbon with significant quantities of several kinds of organic compounds (you can't expect them all to decompose to carbon...), depending on the source.

Trying to class your sample (which clearly has other stuff in it... I can see white/gray specks on it in the first picture ;P) as "graphite" or "charcoal" is an effort in futility; neither "real" graphite nor charcoal have a sharply defined constitution.

Simple tests of the physical nature of the powder would not be able to accurately distinguish it from being "mostly graphite" or "mostly charcoal", mainly, because it's a powder. Had you had the sample before it was pulverized you could (easily) tell the difference; the qualities that most people attribute to graphite (luster, slippery-ness) aren't easily discernible when you turn it into a powder and then compare it with powdered charcoal.

About the tests you conducted:

Your attempt to subject the sample to various (physical) tests is laudable (more users ought to do that before posting "Identify this thingy"-type questions here... it may give us something to work with). However those tests, apart from sorely lacking in some sort of mechanism/procedure to cut down on sources of error, don't really provide any meaningful data that you or me or anyone could use to determine what your sample is (largely) composed of.

Let's break down what you did

  Adding it to water, it sinks to the bottom and leaves a slight oily looking film on the top. Upon stirring, it darkens the water and then gradually sinks back to the bottom after some time.

The film isn't really a giveaway. Even particles denser than water are known to "float" in these circumstances... well, they aren't technically "floating", it's the surface tension of the water that's holding it up. This is easily disrupted by stirring/agitation, and as you've noticed, it'll sink.

Graphites typically have have a much higher density (slight variation depending on purity and crystal defects) than water, so pieces of "graphite" would sink.

Blocks of charcoal typically have lower densities (varies depending on type/source of charcoal) than water and hence floats, but that changes when you pulverize them (and destroy most of the air cavities in the process).

Conclusion: No useful data.

  I tried writing some thick lines on a piece of paper with a graphite pencil and testing for conductivity (high resistance).
  Same high resistance with the powder.

That was a lot of effort... for nothing (you have my deepest sympathies).

Okay, I think I see your reasoning here:

"Graphite's pretty conductive (low-ish resistance) and charcoal is a poor conductor (very high resistance), so doesn't this test tell me something?"

Nope.

Thing about pure graphite (of the "Holy cow! A non-metal that conduct electricity!!!" fame) is that it only conducts (really well) along a particular plane; try another plane perpendicular to this and you'll find it to be a thousand times less conductive ( see answer to this question).

Even if your powder is (largely) graphite, surely the individual powder particles aren't going to align themselves into an extended plane between your terminals. They'll be oriented randomly, so expect the high resistance.

You first drew a line on a piece of paper using a pencil and measured its resistance to create a baseline to work up from. Smart. Just two things though:

1) Pencil lead isn't graphite (neither is it "lead" for that matter ;P); it's a mixture of graphite and clay. So unless you know for sure that the impurities in your sample are clay, this doesn't really serve as a good baseline. Besides, the graphite particles aren't all aligned in that plane I was talking about.
2) Your makeshift resistor ("thick line on a piece of paper") is a poor standard for another reason: dimensions. Your line has no uniform dimensions (length, breadth or even depth). If you need resistance from resistivity of a given material, or the other way around, you need the sample's dimensions. Neither your line nor your powder have fixed dimensions in this test.

Trying to figure out what your sample is by comparing its resistivity with that of another substance is a really neat idea, so don't get me wrong; the issue here is (1) the design of the experiment, (2) of the two substances you're trying to compare (graphite and charcoal), the latter's electrical properties are altered because it's been powdered and (3) your "control-test/sample" (the line-resistor) is neither graphite nor charcoal, making it a rather pointless "control sample".

Conclusion: No useful data.

  Rubbing it into paper it has a much blacker colour than the graphite (which to me looks like a really dark gray). However, I don't know if that is a reliable indicator.

I don't need to tell you how poor a test this is (you already seem to doubt the validity of the test yourself)

Conslusion: No useful data.

  EDIT: Ok, I've made a paste using Isopropyl Alcohol and the carbon powder and made a small "stick" or rod of the stuff. I have now been able to achieve solid continuity at around 2 centimeters in distance and the resistance dropped from the high 40 - 50 kOhms for the loose powder, to the low 10s (ie 40 Ohms). If this helps anyone.

Same argument as before: This test is pointless because your sample doesn't have fixed, uniform dimensions. Nor do the particles (if they're graphite; doesn't matter if it's charcoal) align themselves in extended planes.

As for the drop in resistance, I reckon it's because the sample isn't completely dry. The "isopropanol" you get at the drugstore is at least one-quarter water (not pure isopropanol). After you mixed it with your sample, a ton of electrolytes would've dissolved in the water (pretty sure your sample has loads of stuff that isn't carbon/ carbon-based). When you measured the resistance across this stick of yours, all that water hadn't evaporated by then. This is probably the (main) reason for the drop.

Conclusion: No useful data.

Interesting question by the way, you have my upvote ;)


Grignard reagent will act as a base or a nucleophile R- from grignard reagent (RMgBr) and gilmann's reagent (R2CuLi) although being a strong base reduces the carbonyl group and does not act as a base.

Why can't it abstract the alpha hydrogen forming an enolate ion?

As most of us chemists have understood and experienced, the Grignard reagent have undergone the many reactions with suitable substrates. Yet, the most familiar and probably the most used reaction is the addition to a carbonyl group to give $2^\circ$- or $3^\circ$-alcohol. The Grignard reagents' affinity for a corbonyl group in the presence of an acidic $\alpha$-$\ce{H}$ has been documented here. However, if at least $\ce{OH}$ group is present in the molecule, acid base reaction would be take over. For example, the usual Grignard reaction would not happen if your glassware is wet or not properly dried.

Albeit they were rarely published, it is well known fact that some side reactions with available acidic $\alpha$-$\ce{H}$ and some reduction reactions have been taken place during original Grignard reactions. These side reactions are frequently complicated the intended reactions, which may replace some or all of the intended addition reaction, reducing expected percent yield (Ref.1). Most importantly, the conversion of an alkyl halide to a deuterium labelled hydrocarbon has been achieved via Grignard reagent preparation (Figure A):

Keep in mind that these labelling are selective based on the alkyl halide. Recently, there was a method developed to selective labeling by using organozinc reagents (Ref.2). Although these are not conventional Grignard reagents, organozincs are close relatives (Figure B). Another example for Grignard reagents acting as a base is a method to prepare terminal alkynyl-Grignard reagents (Figure C).

Some times during Grignard preparation, you might experience no Grignard was evidently formed because you have loosen your alkyl halide reagent while substrate ketone was recovered unchanged. That is because, if your ketone is bulky like diisopropyl ketone (2,4-dimethylpent-3-one), following reaction may happen (look under an example for Grignard acting as a base):

If your Grignard reagent have an active $\alpha$-$\ce{H}$ to give, it can also acts as an reducing agent as in above case of Grignard acting as a reducing agent. This is the case of reaction between benzophenone and specific Grignard reagent (isobutyl magnesium bromde or 3-methylbutylmagnesium bromide) illustrated in Ref.1. The main products from that reaction were dibenzylcarbinol and isobutene.

References:

G. E. Dunn, John Warkentin, "An Isotopic Study of the Reducing Action of the Grignard Reagent," Canadian Journal of Chemistry 1956, 34(1), 13841–13857 (https://doi.org/10.1139/v56-007).
Aiyou Xia, Xin Xie, Xiaoping Hu, Wei Xu, Yuanhong Liu, "Dehalogenative Deuteration of Unactivated Alkyl Halides Using $\ce{D2O}$ as the Deuterium Source," J. Org. Chem. 2019, 84(21), 13841–13857 (https://doi.org/10.1021/acs.joc.9b02026).


Adding cyanide vs nitrite for nucleophilic substitution and why does it change when we use silver salts for it On adding $\ce{NaCN}$ to $\ce{RX}$, we get

$$\ce{NaCN + RX -> RCN + NaX}$$

And not $\ce{RNC}$. Clayden gives the explanation as follows:

  Although linear cyanide (which is isoelectronic with $\ce{N2}$) has a lone pair on nitrogen and a lone pair on carbon, the nucleophilic atom is usually anionic carbon rather than neutral nitrogen as the $\mathrm{sp}$ orbital on carbon is of higher energy than that on the more electronegative nitrogen, and therefore constitutes the HOMO.

So according to Clayden, Carbon being less electronegative, has a higher HOMO than Nitrogen, thus it's lone pair attack the LUMO, to form a bond as the less the difference in energy between HOMO and LUMO, the stronger is the bond

But when we consider the same case with $\ce{NaNO2}$, so according to Clayden, nitrogen being less electronegative than oxygen, Nitrogen should have a higher energy HOMO and thus in a nucleophilic substitution reaction it is from the nitrogen side, that  the reaction should take place but ...
$$\ce{NaNO2 + RX -> RONO + NaX}$$

The reaction takes place from the oxygen side. Why?

Also when we have the silver salts of them, e.g., $\ce{AgCN}$, the reaction becomes:
 $$\ce{AgCN + RX -> RNC + AgX}$$
Similarly,
 $$\ce{AgNO2 + RX -> RNO2 + AgX}$$

Why does it change?

Potassium cyanide ($\ce{KCN}$) is a predominantly ionic compound ($\ce{K+}$ is Hard Acid while $\ce{CN-}$ is Soft base) and would dissociate completely to give cyanide ions in solution. Cyanide ion ($\ce{CN-}$) is considered an ambient nucleophile, which means that the negative charge can associate with either carbon or nitrogen. However, the negative charge of cyanide ion concentrated on carbon than nitrogen. Thus, reaction with haloalkanes yields cyanides (nitriles) since $\ce{C-C}$ bond is thermodynamically more stable than $\ce{C-N}$ bond (simplest explanation).

Silver cyanide ($\ce{AgCN}$) is largely covalent and it does not dissociate easily to give $\ce{CN-}$ due to soft-soft interaction ($\ce{Ag+}$ is Soft Acid and $\ce{CN-}$ is Soft base). Thus, negative charge of $\ce{CN-}$ ion is mostly available at nitrogen, and hence the attacking centre of the nucleophile will be $\ce{N}$-atom. As a result, you get nucleophiluc substitution of haloalkanes to give isocyanides (isonitrile).

Same argument can be given for nitrite formation with $\ce{KNO2}$ and nitro formation with $\ce{AgNO2}$. In this case, $\ce{KNO2}$ is a predominantly ionic compound ($\ce{K+^-O-N=O}$) where one of oxygen atoms of nitrite ion has negative charge. Therefore, the nucleophilic attack of $\ce{NO2-}$ on haloalkane would result alkyl nitrite ($\ce{R-O-N=O}$). Silver nitrite ($\ce{AgNO2}$) is, on the other hand, largely covalent and would not dissociate easily to give free $\ce{NO2-}$ ions. As a results, the lone pair on nitrogen acts as an attacking site, forming nitroalkanes (high electronegativity of oxygen compared to nitrogen prevented it from donating its lone pair, as explainnation given by Clayden).


Why do we take the compressibility factor equal to one when graph is plotted? I saw the graph of compressibility factor against pressure I was confused to see that all the gases are started by 1 unit of conpressibility factor when pressure is zero. So why do we take the compressibility factor equal to one when the graph is being plotted?

I think that at $p = 0$ the compressibility factor ($Z = pV_m/RT$) itself is not particularly well-defined since $p = 0$, $V_m = \infty$ and $T = 0$. Zero pressure is not a state that we can reach.

However, if you use any equation of state and take the limit as $p \to 0$ (or $V_m \to \infty$, which is equivalent), you will find that $Z \to 1$. Try it with the virial equation of state:

$$pV_m = RT\left( 1 + \frac{B}{V_m} + \frac{C}{V_m^2} + \cdots \right)$$

This happens with all gases because the individual gases are identified by their virial coefficients $B, C, \ldots$ and the terms involving these constants vanish as $V_m \to \infty$.

A simple qualitative explanation is that in the limit of $p = 0$ all gases behave ideally; for an ideal gas, $Z$ is of course equal to $1$ because of the ideal gas law $pV_m = RT$.