exchemist

Any wave carries energy. A wave is just a travelling disturbance in some medium. It takes energy to displace the medium away from its equilibrium state, and that is the energy carried by the wave. A wave can't be said to have any particular mass. A photon is a travelling disturbance in the electric and magnetic fields. Mass does not need to come into it.

Starch is a term for polysaccharide polymers, assembled from glucose monomer units. However there can be differences in molecular weight and in degrees of chain branching, not to mention grain size and possibly other physical characteristics, depending on the source.

My sentiments exactly, Doctor. I hope to God @toucana is right about Trump being behind bars before the election. If the US gets taken over by an authoritarian personality cult, we get worryingly close to the power blocs of Orwell’s 1984, turbocharged by the surveillance state, internet disinformation and AI.

Well, it starts to get interesting. I did not find any instructions for demonstrating how to do this using lycopodium spores but instead came across this amateur site: http://www.microscopy-uk.org.uk/mag/indexmag.html?http://www.microscopy-uk.org.uk/mag/artoct20/ms-brownian.html which seems to claim the grains are too big to show Brownian motion themselves and it is only when they burst and release their contents that there are particles small enough to demonstrate the effect. This site also says the hydrophobic properties of lycopodium spores are due to air trapped by hairs on their surface. If so, when they burst, this would not apply to their contents - which might then be able to disperse in the body of the liquid. So maybe we are getting somewhere.

This does not seem to address my point. I suppose one advantage of the particles clinging to the glass surface would be that they all stay in one plane and thus can be easily kept in focus when examined with the microscope. I would have thought their motion might be inhibited somewhat by contact with the glass, but apparently it is not enough to prevent the demonstration from working. I suppose if they are hydrophobic they will probably not be strongly stuck to the glass either, since that too is a polar medium. Perhaps what I need to do is find some detailed instructions for how to set this demonstration up, and see what they say about it.

Yeah but I'm not questioning whether or not it is used to show Brownian motion. Evidently it is. It is just that my previous understanding of the demonstration, that the particles were in the body of the liquid and being buffeted in all directions by water molecules, can't be correct. They must be clinging to the glass and sliding around in 2 dimensions only.

I've given you the example of the towels my wife bought, which do exactly this. They came from a mid-price fabric and furnishing store in France (Bouchara, in Vannes, I think), so were respectable quality but not top notch. So yes it can happen. This is why you are always advised to wash dark coloured garments together, separately from light ones. You can wash them with jeans, dark socks and underwear, dark blue or dark green shirts and trousers, etc. and there will be no effect. So just do that, like everybody else, and get on with life.

I don't know anything about clothing brands, but all dyes can run if not fixed well. Black, being so dark, will be particularly obvious when it happens. I have some maroon coloured bath towels (a slightly mad choice of colour by my late dear wife, hem hem) which still run a bit after 15 years, so I have to wash them only with other reddish items. But I also have a (Barbour) shirt with dark blue, green and cream colours in it, that is absolutely fine. Clearly, the manufacturer had the wit to realise that putting these colours together required the dyes to be well fixed, to avoid the shirt being wrecked the first time the customer washes it. Barbour is a good - and expensive - brand however. I suspect your items may not be the best quality if the black runs. However the good-ish news is that black, unlike say blue or red, will just make other items in your wash a bit grey, rather than changing their colour in an obvious and potentially more damaging way. It will probably be best to wash them with other dark clothes and not at a high temperature. There's a bit of explanation here from a detergent manufacturer, which may shed some light on the issue: https://www.persil.com/uk/laundry/laundry-tips/washing-tips/knew-one-colour-runs-wash-another.html

Thanks - something I did not know. But that raises a question in my mind about its use to demonstrate Brownian motion. My understanding is this is done in a water droplet on a microscope slide. But if the grains are hydrophobic, then presumably they will not be in the bulk liquid but clinging to the glass of the slide and being nudged this way and that, along the surface of the glass. Is that your understanding? Or is a surfactant used to to get them to disperse into the water?

Oh great!

Why do you say lycopodium powder grains are hydrophobic?

Sure. What I had in mind though is that at each interaction there is one value made concrete out of a range of probabilities and this value goes on to set the conditions for the thing interacted with, which at its next interaction in turn makes concrete one value out of a range, etc etc. So at the end of the chain of interactions the outcomes are not determined exactly by the starting conditions.

Well both are indeed false. 1) There is no such thing as a "hydrophobic force". For a non-polar molecule to dissolve into water, it has to get between the water molecules. There will be attractive London forces between water molecules and the non-polar molecule, but its presence between water molecules will reduce their mutual hydrogen bonding and thus raise the energy of the solution, so it is energetically unfavourable - and will only happen to a slight extent. (Dissolution is still favoured entropically, so the free energy change won't be determined purely by the enthalpy change. The influence of entropy will be greater as the temperature goes up: ΔG = ΔH-TΔS. ) So there is no "repulsion" of any kind: it's just a reduction in net attractive force. 2) I had actually forgotten this 🙂, but D and L stereoisomers are a naming convention, relating chiral molecules to the enantiomers of glyceraldehyde, i.e. to the way those rotate the plane of polarised light. It does not mean and given D or L stereoisomer will rotate polarised light in a particular way. More here: https://chem.libretexts.org/Courses/Purdue/Purdue%3A_Chem_26200%3A_Organic_Chemistry_II_(Wenthold)/Chapter_22._Carbohydrates/22.03%3A_The_D_and_L_Notation

In the Apollo programme, the capsule was only pressurised to 5psi and the astronauts breathed pure oxygen: https://history.nasa.gov/SP-350/ch-4-4.html So the air pressure they experienced was only 1/3 of that at sea level, but the partial pressure of oxygen was actually above that at sea level.

I had always thought quantum theory knocked determinism on the head, a century ago. Which is why Einstein found it so disturbing and unsatisfactory.

Hoho. Grotesquely, I see there was even some festival in Italy in which a version of this supposed relic was paraded annually, on the feast of the Circumcision of Our Lord (on 1st January, the Octave of Christmas, the feast day having been renamed by Pope John XXIII in honour of Mary). It seems there was a determined attempt to stamp out this particular superstition in 1900 or so, under threat of excommunication. But the business of relics in general continues of course, tolerated, if not officially approved, by the church. Facetious jokes aside, theologically, the circumcision of Christ has been considered symbolically important because it inducted him as a Jew, subject to the Law of Moses, it was the day he formally received his name, Jesus, and it was the first occasion on which his blood was spilled (seen as foreshadowing his eventual sacrifice on the cross).

Are there relics of wooden objects supposed made by Jesus?

This sounds like homework. Before we respond, what are your thoughts about the answers?

OK thanks for coming back to clarify. If you dissolve 2 salts and all the combinations of ions are soluble compounds, then you won't get any displacement reaction as such, you will just get a mixed solution with all the ions dissolved. For instance, NaCl(aq) + CuSO₄(aq) will just give a solution with separate Na⁺, Cu²⁺, Cl⁻ and SO₄²⁻ ions, all happily solvated and swimming around. That's because there is no combination that has markedly lower solubility than the others. (If you were to concentrate the solution enough you you would eventually exceed the solubility limit of the least soluble combination and you would start to get that one precipitating out. I don't know without looking it up which one that would be.) In your example. CaCO₃ (chalk, limestone) is well known to have limited solubility so that would precipitate, leaving you with NaCL(aq). KCl has similar solubility to NaCl. But if you are asking if there is a rule to predict which salts have high solubility and which ones have low solubility that gets involved and is quite hard. There are slightly handwavy explanations e.g. that when cations and anions are of similar size they pack more efficiently in the crystal structure, with small gaps, and thus tend to have a high lattice energy. This makes them reluctant to trade that stability for the attraction of a cage of polar water molecules. This is one explanation for why BaSO₄ is insoluble, for instance. Both are big ions, Ba because it is in the 6th row of the Periodic table and sulphate because it is compound ion in the shape of a nice tetrahedron so it can pack more or less like a sphere. Both also have a double charge on them which increases electrostatic attraction in the crystal and also makes it a bit harder for water to stabilise them fully when solvated. Carbonate does not pack like sulphate as it is a planar triangle, and you need to know a bit of crystallography to understand how efficiently it can form a structure with Ca. But Ca also has a double +ve charge so is a bit more challenging for water to stabilise in solution. So, to he honest, this kind of argument strikes me as being better at rationalising after the fact than really predicting with confidence. If this is a school level question I think you will just need to know which of the common salts are insoluble and which are soluble. You won't get asked about obscure combos. There is a rule of thumb in the link here which may help a bit: https://chem.libretexts.org/Courses/College_of_the_Canyons/Chem_201%3A_General_Chemistry_I_OER/04%3A_Introduction_to_Solutions/4.05%3A_Solubility_of_Ionic_Compounds

Yes we are I think. I'm rusty on all this so you may be just as likely to be right as me these days, but I think b is off-target as we don't know whether the reaction has been allowed to reach equilibrium. If you start with just left hand side reactants, say, it will take some time to generate enough right hand side products before equilibrium is reached. These thermodynamic quantities tell about what the equilibrium state will look like, not whether or not equilibrium has been reached. I think d is off target because a "spontaneous" reaction is generally said to be one in which ΔG is -ve.

It is unlikely domestic wooden artifacts of that type would have survived for 2000 years, even if Jesus did practice carpentry at some point, which seems unclear.

This is chemistry. If you insist on starting from first principles of physics every time when addressing a chemistry question, you will never reach an answer for anything but the most trivially simple systems. Physics can't accurately model anything in chemistry more complex than the hydrogen molecule ion. In chemistry the systems are far too complex for that type of approach. You don't seem to have understood the question that was asked. Nobody asked about energy release. You introduced that yourself and it's not relevant. I only included some comments about it in my reply because that was the aspect that seemed to interest you. Anyway, let's see if @jeff einstein responds to this thread. Then we may see if any of this has been useful.

I think the point abut c being wrong may be that the expression for K involves the product of concentrations of product species, raised to the appropriate power, divided by the product of concentrations of reactant species raised to the appropriate power. So K is not simply one concentration divided by another. (Also, strictly speaking K should involve activities rather than concentrations, though this may be too subtle for the test question you are describing.)

I don't think this is very helpful. It's obvious the reaction won't proceed unless ΔG for the process is -ve, but the poster will not have tables of Gibbs free energy to hand and should not need them. E=mc² is neither here nor there, and nor is activation energy in this context. Double displacement reactions generally involve ionic compounds in solution, exchanging partners, i.e. AB + CD -> AD + CB, A and C being cations and B and D being anions. What drives the reaction to the right is the removal of one of the two product species from solution, usually because it is insoluble and precipitates. All our poster needs to do is apply this idea to the examples he or she has been given. But of course it does require considering the charge on the various cations and anions involved, in order to come up with the right formulae for the reaction products, and knowing - or guessing - which of these salts are insoluble. An example would be BaCl₂(aq) + Na₂SO₄(aq) -> BaSO₄↓ + 2NaCl(aq). Both ionic compounds on the left are water soluble but barium sulphate is insoluble and precipitates, leaving only sodium chloride in solution. The driving force is the affinity for Ba²⁺ for SO₄²⁻, which prefer to bind to each other rather than to water molecules. (If you like to express it thermodynamically, ΔG is -ve for the process Ba²⁺(aq) + SO₄²⁻(aq) -> BaSO₄(s), or, conversely that barium sulphate has a +ve solvation energy in water, so it does not dissolve. However this is not really germane to the poster's question.)

But what does it mean to say the system “needs” to shed angular momentum? That suggests it has intent to do so, in order to keep itself together, which is not science. OK, I get that if it did not, it would not hold together, but why should these jets so conveniently arise?