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Everything posted by woelen

  1. This does not work with acid of concentration 3.75 mol/liter. Then the acid first needs to be concentrated by boiling away the water, until faint white fumes are produced. That, however, is quite a dangerous thing to do without the proper equipment (boiling the H2SO4 in a pan in the kitchen does not seem to be a good idea to me ).
  2. Mix with sodium chloride, just as 'etacude' proposes. Try to dissolve as much as possible NaCl by heating somewhat (no need to boil, that would drive off HCl). Let liquid cool down, remove the crystals (these are mainly Na2SO4 and NaHSO4) and then add some solid NaClO3. KClO3 also works but slower. You'll see a deep intensely colored yellow gas, ClO2. This experiment works great with acid of the concentration you mention. This experiment also works with concentrated HCl, but then you won't get rid of the H2SO4 . This is not a way to get rid of lots of H2SO4 though, because you should do the experiment only on a test tube scale! ClO2 is a very sensitive compound, which easily explodes (if you drip in some organic, such as ethanol or acetone, then it probably explodes, resulting in scattered glass and acid sprayed around!). The gas, however, is cool to see and when you keep the test tube horizontally, then you can actually 'pour' out the gas. If you do this close to the surface of some water, then you get really cool effects at the border of the glas cloud (kind of nebula on the water surface, which spreads out slowly). This experiment of course should be done outside.
  3. It is used as drain cleaner. Go to the shelf holding drain cleaners and select the one which contains 98-99% NaOH (caustic soda). There are also drain cleaners based on sulphuric acid and possibly also on other stuff, so take care to select the correct one.
  4. Punctuation? Capital letters? (broken shift key? ) Quotes (ima -> I'm a)? Please use normal language. Let's keep the level of this forum acceptable. If you want some respect' date=' then [i']show [/i] some respect by formulating your sentences correctly and being not too lazy to press the shift key on your keyboard! Sorry for you weldermanx that you happen to be the victim of my griefs . This message is directed to every member of this forum who does not use acceptable language.
  5. I missed that . In my browser it was in the second last list of posts and I clicked the last list .
  6. Yes, you're right, but that is something different than the net energy released or needed by a reaction. In my hill/marble analogue, the activation energy could be the energy needed to tap the marble or in the case of vegetation, to lift the marble over the vegetation, but still the total energy converted from potential to kinetic energy remains the same. This is exactly the same in chemistry. Any energy, initially put in the system for passing the activation barrier is regained, so it does not contribute to the net energy released (or taken) by the reaction.
  7. It might be, but only with considerable input of energy. As I explained in my lengthy post, a catalyst does not affect the energy released (or needed) for a reaction.
  8. As I stated earlier already in another post, the potential energy, stored inside a molecule cannot simply be described by means of a simple formula, such as given for the marble/hill metaphore. You need quantum mechanics to perform such computations. What you see in practice is that for many reactions the amount of energy is given, e.g. 2H2O2 --> 2H2O + O2 + xx KJ/mol (I don't know xx, many textbooks contains tables for many common reaction or half-reactions). What is important, it that xx does not depend on the path of the reaction (e.g. use of a catalyst or not). The hill/marble metaphore also shows this. Whether the marble moves in a straight line from top to floor or through a shaky and lengthy path, in both cases it looses the same height h and hence it looses the same amount of potential energy. What do you mean with high thermodynamics? Assuming that you mean high energy level, then I would say 'moderate'. Acid/base reactions can be fairly energetic (e.g. add solid NaOH to concentrated HCl), but in general redox reactions are much more energetic. NaOH+HCl can lead to boiling and splattering, but many redox reactions release so much energy that fire-like phenomena can be observed (e.g. mix of KClO3 and S, or mix of KMnO4 and glycerin). If I would classify (roughly) reaction classes according to energy release, then I would come up with the following (ordered from high to low): redox acid/base coordination precipitation But beware, this is just a rough list. There certainly are redox reactions, which are less energetic than many acid/base reactions, it is just the overall/on average behaviour. Wilco
  9. A really good way to remove iron oxide/rust stains is the use of oxalic acid, which is a very strong complex-former with iron (III) ions, while otherwise being _relatively_ mild. First try with citric acid though, because oxalic acid is quite toxic. If that does not work, then try oxalic acid.
  10. Energy, available in molecules and so on has no real macroscopic equivalent, I think it is OK to call this 'bond energy'. The best comparison with macroscopic energy is potential energy. This potential energy can be released in chemical reactions, producing heat (and possibly light). A real nice example is white phosphorus, where the four P atoms are arranged in such a way, that there is considerable bond strain. Compare this with a mechanical spring, which is about to break apart in two pieces due to over-stressing. This is what makes white phosphorus so reactive. When the bond (spring) breaks apart, then a lot of potential energy is converted to another form of energy (fast motion macroscopically, which is heat at the atomic/molecular level). When the spring snaps, then it may hurt you, when the P4-bonds break apart then it may really hurt you . I hope that this example makes things a little more clear to you .
  11. There is no safe way to mix these two substances. If you just want to do this for a fun experiment, then take mg quantities in the open air (e.g. 100 mg of KClO3 and 35 mg of Al, giving slight excess of KClO3). If you want to do more serious work making fireworks, then I would say, don't do it with Al+KClO3. There are safer and equally powerful mixtures.
  12. I was tempted to give this answer also at the original question, but I decided not to do so. From a pure theoretical/academic point of view you are right, but from a practical point of view I still would say that not all reactions are reversible. Have you ever seen appear wood from ashes and smoke ? If you want to explain why certain reactions are reversible and why others are not, then the concept of equilibrium does come into play again. In general, one can say that the closer the energy levels of the reactants and the products are, the easier the reversible reaction can occur, but this is not a guarantee that the reversible reaction also does occur always. Another factor, which determines whether a reaction is reversible or not is whether the reaction products remain available in the system or not. E.g. if one of the products is a gas and is completely insoluble, then the reaction is driven completely to one direction. Another way, in which this can occur is that one of the reaction products is taken away. An example of this is the equilibrium reaction of dichromate and chromate: (1) Cr2O7(2-) + H2O <---> 2HCrO4(-) (2) HCrO4(-) <---> CrO4(2-) + H(+) Reaction (1) is almost completely at the left, a solution of a e.g. potassium dichromate is only very faintly acidic. Now, if some excess lead nitrate is added, then the incredibly insoluble PbCrO4 is formed, which settles as a solid. This drives reaction (2) to the right and that in turn also drives reaction (1) to the right. The liquid becomes almost colorless and all dichromate is converted to chromate in the form of lead chromate. The liquid also becomes quite acidic. An other example is addition of a bicarbonate to an acidic liquid. The equlibrium HCO3(-) + H(+) <---> H2O + CO2 is driven to the right, because most of the CO2 escapes from the system as bubbles. If the system were kept under high pressure and no space for formation of gas bubbles, then the reaction would really be an equilibrium. Yet another factor is whether there are mechanistic pathways for reactions to occur. Even an energetically very favourable reaction does not need to occur if there is no suitable mechanism for the reaction to occur (this is why perchlorates are so inert in aqueous solutions). So, if there is a reaction with roughly equal energy levels at both sides of the arrow and there is mechanistic pathway for one direction, while there is not one for the other direction, then on a macroscopic level we would observe the reaction as being non-reversible.
  13. Wow, this is interesting! I have some older books and they all state that H2CO3 does not exist in the free state. So, the isolation must have been very recent. Somewhere in the year 2000 or just before? I unfortunately have no access to the article you mention, but from the weblink I can grasp the idea. Do you know something about the physical properties of H2CO3 (is it a gas, a liquid?), or was the isolation only just a few molecules? It is nice to read about such compounds, which seem so familiar, while in reality they have eluded chemists up to very recently. Now its time to wait till the first auctions on eBay appear for ultrapure carbonic acid, totally free of water :D. Maybe an interesting exercise for you... probably even harder than storing fluorine gas if only a few molucules of water cause an autocatalytic destruction of the compound
  14. No' date=' not all reactions are reversible. E.g. burnt wood and the gaseous reaction products cannot revert to wood and oxygen. No, the acid does not exist as such. The acid H2CO3 cannot be isolated. When CO2 is dissolved in water, then indeed the liquid becomes a little acidic, but the real reaction mechanism from CO2 and water to the acid is not really simple. The net equilibrium will be something like, without the H2CO3 species present in solution: CO2 + H2O <---> HCO3(-) + H(+) Another example of such a 'troublesome' acid is sulphurous acid (H2SO3). This acid also does not exist in the free state, nor in solution. No one has ever observed the species H2SO3. In water, when SO2 is dissolved, then the species SO3(2-), HSO3(-), SO2, S2O5(2-) and HS2O5(-) can be observed. So, observing that a liquid is somewhat acidic does not always mean that the corresponding acid exists! No, this is exactly what makes the computations of chemical reactions and structures so difficult. When a molecule or set of molecules need to be modelled, then the number of states for even very simple systems can be incredibly large. The same energy in a system can represent many many different configurations. In practice, one also frequently observes, that reactions are not 'clean'. Especially, when the different energy levels for differnet compounds are very close to each other, then a reaction can produce many different compounds (there are many competing reactions). An example of this is reduction of HNO3 by zinc. This reaction can yield NO, NO2, NH4(+), NH3OH(+), N2O, all in one 'dirty' mix. What is formed precisely, depends on temperature, pressure, concentration of reactants, etc.
  15. Yes, one can compute the energy contained in a certain molecule. This energy can be regarded as potential energy (that is the best macroscopic equivalent I can think of). The computations, however, are extremely complex. They require quantum mechanics. With lots of simplifications one can reason about how stable molecules will be and this reasoning frequently provides nice rules of thumb, but if one really wants good estimates of the free energy in molecules (and relatied to this, the stability) then detailed QM computations need to be done. There are software packages for doing this kind of computations, but these are not suitable for the beginning chemist and for more complex molecules than e.g. CH4, NH3, etc. quite some computational power is needed.
  16. Sorry for the misunderstaning. MY post was off-topic, not yours
  17. As jdurg already explained to you, making sodium metal at a small/home scale is incredibly difficult. Forget about it. The average hobbyist will NOT be able to make sodium metal at home. Attempting to do so probably will result in severe injury and fire if you do not know exactly what you are doing. Obtaining sodium is not easy at all for the general public and there are good reasons for that .
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