woelen

Normally I don't like to give sources of chems, but giving a source for silver metal is not too hefty I think . You can get very pure silver (99.95+%) at http://www.emovendo.net at a price of $13 per 30 grams. Shipping worldwide is another $3 or so. If you combine orders from this company, then shipping almost can be neglected. This is a very nice source of very pure silver.

This article may help you. http://www.chem.uidaho.edu/faculty/ifcheng/Chem%20253/labs/Experiment%206.doc Whether silver ntirate is a primary standard or not is debated somewhat. I think it can be a primary standard, provided a really high grade sample is used. In the paper, I have given here, the silver nitrate must be standardized. You could do the other way around, standardizing your NaCl.

No, I'm a completely home-made chemist, but I've read a lot of books, Internet pages and did a lot of experiments. That is how I gained my knowledge. In my daily working-life I'm an ICT-consultant and software engineer.

No, I do not agree with you. If in one beaker, you have dissolved 1 mol each of NaNO3 and KCl and in the other you have dissolved 1 mol each of KNO3 and NaCl, then in both beakers you will have 1 mol of each of the following ions: Na(+), K(+), NO3(-), Cl(-). If then you slowly evaporate the solutions, in both cases the same salt first crystallizes, because both beakers contain the same ions in the same concentrations.

There is not more O2 in it. You have the best method of getting the O2. Really, this stuff is not that interesting. You have at most 10% by weight of free oxygen, probably much less, due to impurities in the stuff. You can explain that easily. Look at the formula of the stuff. It is something like Na2CO3.H2O2.xH2O. There is only one free oxygen atom in this entire compound and if you look at the weight percentage, then you only have 10% or so. If the compound also is impure (due to mixed in detergents, perfumes, surfactants, etc), then you can imagine that the percentage of free oxygen maybe is in the range of 5 to 7% by weight. So, 30 grams of this compound gives 1.5 to 2 grams of oxygen, which is roughly 1.5 liter of gas.

Oh yes... concentrated sulphuric acid is a children's toy compared to oleum. No, both in oleum/SO3 and in sulphuric acid, the oxidation number of sulphur equals +6. It has to do with the EXTREME dehydrating properties of oleum. Concentrated sulphuric acid already is strongly dehydrating, but this is nothing compared to the dehydrating powers of oleum.

I did another interesting experiment with the selenium. I now have CS2 and this opens up new possibilities. Black selenium does not dissolve in CS2 at all. I have some powdered black selenium, consisting of many minute crystals. It does not dissolve, nothing at all. I also have those little corpuscles, 99.999% Se, from http://www.emovendo.net, an element-collectors supplier. These pieces are grey and they dissolve in CS2 a little bit. The smooth silky-looking surface if these corpuscles becomes a little pitted and irregular, so apparently they are a mix of different Se-allotropes, probably made by melting the element. Again, I made red selenium with black Se, HNO3 and Na2SO3 and shook this solution with some CS2. The red stuff for a large part dissolves in the CS2, the solution in CS2 becomes yellow/green/brown. Next, I let the CS2 evaporate. Nice little crystals remain. Unfortunately, these crystals are not red, but very dark red/brown/grey. I'm not sure what allotrope this is. it may also be a mix of allotropes. But it is nice to see that there is a solvent for selenium. Btw, sulphur dissolves in CS2 REALLY well. If you take 1 ml of CS2 and you add a large spatula of flowers of sulphur to this, then the sulphur dissolves almost at once and a nice yellow solution is obtained. It is really surprising to see how well sulphur dissolves in CS2. Selenium dissolves with more difficulty.

Exactly, this is what I wanted you to find out yourself. You cannot balance the equation and the reaction is not possible. Indeed, Cu2O will dissolve in HCl, forming the complex CuCl2(-). If the HCl is absolutely free of oxygen and there is no contact with air, then you would get a colorless solution. In practice, however, dissolving Cu2O results in formation of a dark brown liquid, because O2 oxidizes part of the CuCl2(-) and then the mix of copper (I) and copper (II) is formed, in which the deep brown multi-valence complex is formed. I have the impression that you understand this quite well now. As you see, sometimes doing the math yourself may be very enlightening.

This is a difficult one. In order to understand this, I need to introduce the concept of hydrolysis. This cannot be solved by using the simple Ksp-story as explained before. All metal ions, when dissolved in water, form aqua-complexes. When iron (III) sulfate is dissolved, then in water, you get Fe(3+) ions. However, these ions do not exist as bare ions, in fact, water molecules form a complex with metal ions. The Fe(3+) ion happens to form hexaaqua complexes (many metals form hexaaqua complexes, but some form tetraaqua complexes or even other numbers). So, a solution of ferric sulfate contains ions [Fe(H2O)6](3+), nice formulation: [math][Fe(H_2O)_6]^{3+}[/math] These aqua complexes have a central metal core, the oxygen atoms of the water-ligands are pointing towards the metal core and the two hydrogen atoms are pointing outwards. The charge of the ion as a whole is not simply located on the metal core, but it is distributed over the whole hexaaqua-metal ion, that is why I write [Fe(H2O)6](3+), with the (3+) charge distributed over all atoms between []. For many metals, the story ends here. The water molecules are called ligands of a complex. Complexes can have other ligands, e.g. NH3, CN(-), Cl(-), NO(+), but here the ligand is H2O. For iron (III), another effect comes into play. The charge of the total ion is distributed, such that a relatively large part is on the H-atoms, pointing outwards. This makes it fairly easy to loose a H(+) atom. Hence, the hexaaqua-iron (III) ion is an acid! [math][Fe(H_2O)_6]^{3+} <-> [Fe(H_2O)_5(OH)]^{2+} + H^+[/math] The ion can even loose a second and a third proton! If three protons are lost, hydrous ferric hydroxide is formed: [math][Fe(H_2O)_3(OH)_3][/math] This phenomenon that a metal salt solution is acidic is called hydrolysis. It is an extremely important concept and one of the key-concepts for understanding reactions in inorganic chemistry. So, a water-ligand can change into a hydroxo-ligand by splitting off a proton. This is what you'll observe when sodium bicarbonate is added. The H(+) from the acidic hexaaqua iron (III) ion is taken up by the bicarbonate ion, CO2 bubbles out of sulution and water is formed. Finally iron (III) hydroxide is precipitated. -------------------------------------------------------------- Many metals hydrolyse in solution. Ferric ions do to quite some extent. Real hexaaqua iron (III) ions are almost colorless. The yellow/brown color of ferric solutions are due to the pentaaqua-hydroxo iron (III) ion. A metal like bismuth hydrolyses much more. Even in 1 mol/l acid, it still hydrolyses so much, that bismuth hydroxide is precipitated or in the presence of nitrate, basic bismuth nitrate (the Ksp of Bi(3+)/OH(-)/NO3(-) is quite low). If you want to see what color real iron (III) has, without hydroxo-ligands, look here at the section for oxidation state +3: http://woelen.scheikunde.net/science/chem/solutions/fe.html Here some misconceptions about the color of ferric salts and their solutions are covered.

Yes, you could say it like this. Easy to remember. The Ksp-things, however, also give you some quantitive powers. There are large tables with Ksp values for many salts. With such a table nearby you can predict which combinations give a reaction and which not. An example of such a table is this: http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm

If you have a solution of ionic compounds, then you really have separate ions in solution. Suppose you take 1 mol of KNO3 and 1 mol of NaCl and you dissolve both solids in 1 liter of water and label this beaker 'A'. Suppose you take 1 mol of NaNO3 and 1 mol of KCl and you dissolve both solids in 1 liter of water and label this beaker 'B'. Now, you give both beakers, labeled 'A' and 'B' to a super-chemistry-expert and you ask him/her which beaker contains KNO3+NaCl and which contains NaNO3+KCl, then that expert will have NOT A SINGLE WAY to determine which was put in 'A' and which was put in 'B'. The expert only can observe that the beakers contain Na(+), K(+), NO3(-) and Cl(-) ions. So, solutions of ionic compounds really have separate ions in solution and not entities like NaNO3 and so on. Many ionic compounds have limited solubility, which is specified by means of a number Ksp (solubility product), which for an ionic compound [math]C_nA_m[/math] (C=cation, e.g. Na(+), A is anion, e.g. NO3(-)) equals [math]Ksp = max(\[A^{m+}\]^n\[b^{n-}\]^m)[/math] As soon as the product of the ions in solution exceeds Ksp for the given compound, then it precipitates from solution, such that the concentration becomes equal to Ksp. So, Ksp tells in some sense how much of a compound can be in solution. When two compounds with large Ksp (e.g. NaCl and AgNO3) are mixed in a single solution, then because of the fact that all ions exist separately (see first part of this lengthy post), the following four Ksp's come into play: Ksp(NaCl) Ksp(AgCl) Ksp(NaNO3) Ksp(AgNO3) Ksp(AgCl) is much smaller than all other Ksp's and you can expect AgCl to precipitate out of solution. So, a reaction occurs, when the product of ionic concentrations of one of the possible combinations of ions exceeds the value for the Ksp for that compound. So, when CuSO4 and NaOH are mixed, then the following Ksp's come into play: Ksp(CuSO4) = max([Cu(2+)]*[sO4(2-)]) Ksp(Cu(OH)2) = max([Cu(2+)]*[OH(-)]*[OH(-)]) Ksp(Na2SO4) = max([Na(+)]*[Na(+)]*[sO4(2-)]) Ksp(NaOH) = max([Na(+)]*[OH(-)]) Here, the value of [Cu(2+)]*[OH(-)]*[OH(-)] easily exceeds the maximum allowed value, because Ksp(Cu(OH)2) is very small. So, a precipitate of copper hydroxide is formed. As Xeluc states, when two solutions of ionic compounds are mixed and none of the Ksp's is exceeded for any of the four combinations, then no reaction occurs. The four different ions simply coexist in the solution.

OK, let's go carefully through this post... Now you came to rescue me . Yes, this is right, with the added notion, that both the Cu(2+) and Cu(+) are complexed, so: [math]Cu(s) + Cu^{2+}(compl) -> 2Cu^+(compl)[/math] With (s) I mean solid and with (compl) I mean complexed in some way. Under these conditions there are no free copper (II) and copper (I) ions. No, this is not correct. O2 does act as oxidizer, but there definitely is no intermediate Cu2O. HCl is not oxidizing at all in this place, although it plays an important role. What really happens is that the complexed Cu(+) ions are oxidized by O2 with the help of H(+) ions from the acid: [math]4Cu^+(compl) + O_2 + 4H^+ -> 4Cu^{2+}(compl) + 2H_2O[/math] The H(+) ions are not oxidizing, but they help the oxidation to be carried out. Many oxidizers need H(+) for their oxidizing properties, e.g. permanganate, dichromate. What I did was introduce the notation (compl). I did this to tell you that the reaction between copper (II) and copper metal indeed occurs, but not as simple ions. In reality, it is the complex ion [math]CuCl_4^{2-}[/math], which acts as oxidizer. The resulting ion is [math]CuCl_2^-[/math]. This latter ion contains copper in the +1 oxidation state. When both ions are present, then even more complex things are formed, the deep brown mix-valency complexes, probably ClCu(μ-Cl)CuCl, which contains both copper in the +1 and copper in the +2 oxidation state (or two copper ions in the +1.5 oxidation state). There is another thing I want to point out. You state that Cu2O is oxidized by HCl to CuCl2 and H2O. Just as an exercise, try to balance the following equation: Cu2O + HCl ---> CuCl2 + H2O Please do this exercise and give feedback here. You'll learn a LOT of this exercise! After your feedback I'll come back on this topic. Good that you found out yourself. No peroxide can be formed here. You'll get something between (1) and (2). A little copper will be oxidized and you get a complicated mix of copper (I) and copper (II) complexes, together with some basic stuff, but the reaction will come to an end really soon. The acid is really needed.

What you say is correct. These microscopic hairs on their feet are so thin and numerous, that they provide sufficient vanderwaals force to keep the animal sticking to the surface. Currently research is done with nanotubes on a substrate, which could be put in shoes and gloves, allowing trained humans to 'walk' on vertical walls, glass, etc. As I stated in my previous post, in theory, when the surface of two smooth panes of glass is really really smooth, it would be possible to stick them together. But even the smoothest glass has nano-scale wrinkles, irregularities and so on, such that this prevents to two panes from sticking to each other. The gecko's feet, however, have hairs, which perfectly adapt to all these irregularities and that is why these animals can walk on such surfaces, even upside down. So, the gecko's do not grip the imperfections of the surface, but their littles hairs snugly fit into these imperfections. Just, because of these little hairs fitting well in all kinds of imperfections, allows these animals to walk on virtually every surface, beit smooth or rough.

Xeluc, I must admit that I made some big mistake with this copper riddle . You were totally right. You are right with your conclusion that it could be copper chloride, anhydrous. I was fooled by the observation of the light yellow/brown stuff I made by boiling a solution of CuCl2 in HCl to dryness. The stuff I had made before is not pure anhydrous copper chloride, but it most likely was something with HCl incorporated in the solid. Now, I've heated some of my reagent grade CuCl2.2H2O very slowly and carefully, taking care not to overheat it in order to avoid the making of basic copper chloride. The product I obtained is exactly the same as on the picture, but now as a dry powder. A nice dark brown and dry powder, which dissolves in water, giving a green solution, without any residue. With this observation, I regard this riddle as solved . The webpage is updated (not removed).

Many covalent solids and liquids are kept together by so-called vanderwaals forces. These are attracting forces between molecules and atoms. So, if you tear apart a piece of paper, or a solid covalent crystalline compound like sugar, then you are not breaking molecular bonds, but vanderwaals forces between molecules. These forces only are relevant at very low distances (at the molecular level) and they are the result of a delicate second order balance of electrostatic forces. The reason for the existence of vanderwaals forces is that, although molecules as a whole are neutral, the charge density is non-zero (positive in nuclei, negative in the electron clouds). Just as an exercise, think of the following system: P--N.......................................P--N Here, P and N are charges of the same magnitude, but with opposite sign. Now suppose P--N is a rigid stick, with P and N having a distance d. Now, also suppose that the distance with ............ equals r. You'll see that if you compute the force between the two sticks (purely electrostatically), that this is non-zero, if d is not equal to 0, although the total charge of each stick equals 0. This is a very simple model, but it does demonstrate the basics of vanderwaals forces. You'll also see, that for r much larger than d, the vanderwaals forces are amazingly small. If you do the math, then you'll see that they fall off with r^4. The reason why paper and so on do not stick together again is that when you put the pieces next to each other, only at very few points you manage to bring the molecules sufficiently close to each other that there are relevant vanderwaals forces. In theory, when you have two very smooth surfaces, and you bring together these two, they would stick together. Certain animals use this effect, such as certain gekko's. They can walk upside down, even on smooth metal and glass surfaces. Flies also can do that, but they use vacuum between their legs and the surface they are walking on. Gekko's, however, are too heavy, relative to their contact-area and they use the stronger effect of vanderwaals forces.

Is that true? Of course, you need to be careful with any chlorate mixture, but why would a sulfATE mixture be more sensitive. In fact, sulfate is rather inert. I know about the really dangerous mixes, with sulfIDE mixed in (e.g. barium carbonate from a pottery supplier, which also contains some barium sulfide). The danger of sulfide, mixed with chlorate is understandable, because sulfide is the most reduced form of sulphur. Even with sulfITE, I can imagine, but with sulfATE, no. Another dangerous thing is mixing of copper salts with chlorate. So, mixing copper sulfate with pot. chlorate may be quite dangerous, because the copper ions act as a catalyst on certain decomposition reactions. If ammonium ion is present, then also the dangerous TACC can be formed.

I have the idea that you misinterpreted this sentence. The compound is endothermic, so the decomposition is exothermic. EDIT woelen: Sorry, jdurg already pointed out this reading-error.

When you add some oleum 65% to water, then it explodes, spraying around concentrated sulphuric acid and boiling water. If you get a drop of this on your skin, then instantly, you have a deep black charred wound. If you open up a bottle of 65% oleum, then you'll see an incredibly thick white plume of smoke. Open such a bottle in a normal living room. Within two or three minutes, you'll not be able to see the wall at the opposite side of the room. Besides that, the fumes are extremely corrosive and eat away your wall-paper, metal objects, your lungs, etc. You understand, why this stuff has no domestic applications? It only can be used in carefully designed reaction chambers (such as in industry) or in well-designed and very powerful fume hoods.

The 100% acids can be formed, but not by boiling a more dilute solution of the acid. E.g. 100% H2SO4 can be made from 96% H2SO4 by adding some SO3 to it. That SO3 reacts with the water and the acid: SO3 + H2O --> H2SO4 Pure HNO3 can be made by distilling 68% (azeotropic HNO3) with conc. H2SO4. 90+ % HNO3 can be made in this way, leaving the water behind in the sulphuric acid, due to its dehydrating properties. Making 100% HNO3 can be done by adding N2O5 to 90% HNO3: H2O + N2O5 ---> 2HNO3 N2O5 in turn is made by adding a large excess of P4O10 to 90% HNO3. The P4O10 takes up all water and also dehydrates some of the HNO3 (reverse of reaction shown above). As you can see, making the 100% acids is not easy at all, but technically speaking, it is possible. In fact, they are available as lab reagents, although at a very high price. You could even speak of over 100% acids, e.g. sulphuric acid, with excess SO3 dissolved in it, or HNO3, with excess N2O5 dissolved in it. The first one is available commercially as oleum with up to 65% of SO3 by weight, but not for the general public.

When you add hydrated copper sulfate to conc. H2SO4, then it remains blue. I tried that before, so the chloride is part of the reaction. I can try adding some anhydrous CuCl2 to sulphuric acid. I'll see if I can make some by heating CuCl2.2H2O. I must be absolutely sure that no remains of HCl are left in the solid and I must also be absolutely sure that no basic oxychloride is formed, so that's why I did not try it yet. Many hydrated salts decompose when heated, not by simply loosing water, but also loosing HCl. E.g. CuCl2.2H2O is prone to loosing HCl when heated: CuCl2.2H2O ---> Cu(OH)Cl + H2O + HCl 2Cu(OH)Cl ---> CuO.CuCl2 + H2O Here we have a basic copper chloride, also known as copper oxychloride. This is not a mix of the two, but a crystalline solid, with copper, chlorine and oxygen in a single crystal lattice. With this kind of compounds, the experimental outcome will be quite different when it is added to H2SO4. However, I certainly will try your suggestion and once I'm sure that I have pure anhydrous CuCl2 I'll do. One big problem with investigating this compound is that it is so hard to isolate. Remember, it is in concentrated H2SO4. So, simply boiling off the liquid is out of the question (H2SO4 boils at appr. 300 degrees C and the fumes are insanely corrosive and lethal). Hot H2SO4 at 300 C eats everything in seconds. Filtering also is not an option with the viscous and very corrosive liquid. Conc. H2SO4 really is nasty stuff. If you have some of your CuCl2, could you please compare the colors/darkness and let me know? The pictures I have made are fairly close to the real color. The compound in the acid really is almost black as you can see on my site.

Can you give me a reference? I hardly can imagine that this reaction is endothermic, but of course you may be right. I never studied the NI3/NH3 system in detail. The only reason, where an endothermic reaction may be dangerous is where solid compounds are converted to gaseous compounds at high speed, but I do not know an example of an endothermic reaction with that property.