# woelen

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1. ## quick chem pH question

For sake of brevity' date=' I introduce the symbol H for [H'] and OH for [OH]. For water we have H*OH = 10^-14 The acid HCl is a strong acid. At such low concentration, it can be assumed totally split. Because for each OH(-) ion from water, we also have one H(+) ion, we now can say that H = OH + 10^-8, or OH = H - 10^8. Now plug into the equation for water: H * (H - 10^-8) = 10^-14. H*H - 10^8 * H - 10^-14 = 0. This is a quadratic equation in H. Solving this for H yields H = 1.05*10^-7. Taking the -log10() of this yields 6.98, so the pH indeed equals 6.98. The critical part of this computation is that at such very low concentrations, the contribution of the autoprotolysis of water cannot be neglected anymore.
2. ## Chlorine

If you have NaCl in solution, then this is split in ions Na(+) and Cl(-). The negative ions are attracted towards the anode, where they give off electrons and combine to Cl2. The positive Na(+) ions move towards the cathode. At the cathode, however, no metallic Na is formed, but hydrogen. The positive ions only take care of conduction of electricity, but it is the water, which is reduced. If you want to produce Cl2 without too much O2 by means of electrolysis, then you need to use a graphite anode (copper dissolves) and a voltage, which is not too high.
3. ## Chlorine

Positive ion: cation Negative ion: anion Positive electrode: anode Negative electrode: cathode

5. ## Carbonyls/Carbonyl Clathrates of Iridium

For simple ions: yes. Examples: oxide, chloride, bromide, sulfide, etc. For more complex ions: no. For these situations you have to use some 'rules' for determining the oxidation state of all elements in such ions. Another poster has given some of these 'rules', together with exceptions. Oxidation state is a nice concept, but it must be used with care and you always have to be aware for pitfalls, when using this concept. For budullewraagh: Metals can sometimes be in really weird oxidation states. The nitroso-complex of iron (produced in the brown ring test for nitrate, but also made by adding a nitrite to an acidic solution of a ferrous salt) has iron in its +1 oxidation state. The ion is [Fe(NO)](2+). The nitroso-ligand has formal charge +1, so the oxidation state of the iron must be +1. What to think of rhenium, which can be brought in the -1 oxidation state, by adding zinc to an aqueous solution of a rhenium salt?

7. ## Chlorine

I think you exchanged the role of anode and cathode. Cl2 is formed at the anode and H2 is formed at the cathode. Another point is that you WILL get gaseous contaminants. At the anode you also will have formation of oxygen. Formation of chlorine gas and formation of oxygen are two competing reactions. The conditions must be carefully controlled in order to minimize the production of oxygen at the anode.
8. ## Acid Concentration

Do not let the speed of the reaction mislead you. Many metals, which react with acids, according to the electropositive series, do so very slowly. I have tried with iron, chromium, titanium, tin, etc. All react very slowly, even in 30% HCl or 30% H2SO4. Even zinc is not that fast. The only real fast one I found is magnesium and with some patience, aluminium also can be made to react fast in concentrated acids. Of course the alkali metals also are really fast.
9. ## oxidation states

Oxidation state is just a tool for bookkeeping of electrons. Bookkeeping is done, such that it gives a good impression of how far an element is oxidized (or reduced). Some elements gain electrons easily (they act as oxidizers and they are reduced). The most common is oxygen. The oxo-group (or oxide ion) is said to have oxidation state -2. Hydrogen usually has oxidation state +1 in its compounds (except in metal hydrides, where it has oxidation state -1). Using these simple rules, one can easily compute the oxidation state of other elements. Example: MnO4(-). Total charge: -1. Charge per oxo-group: -2. So, you need +7 for manganese in order to get a total charge, equal to -1. Keep in mind though, that in MnO4(-), the manganese does not really have a charge equal to +7. In many real life compounds, the charge is not distributed as extremely as oxidation numbers would suggest. So, in reality, the charge on the manganese will be just a little over 1 and the charge on each of the oxo-groups is a little below 0, such that the total charge still remains -1. The usefulness of oxidation numbers is in that they easily tell you whether an element is strongly oxidized or not. In my experiment, you have Fe(OH)3, with Fe in the +3 oxidation state. In the ferrate ion, the iron has formal oxidation state +6. So, from this playing with numbers, you immediately see, that the ferrate is an iron-species, which is oxidized further than ferric hydroxide. Sometimes the concept of oxidation number is flawed. Consider the deep blue compound CrO5. This does not contain chromium in the +10 oxidation state, but it contains chromium in the +6 oxidation state, with an oxo-group attached to it and two peroxo-groups attached to it. For one oxygen atom we have oxidation state -2 and for the other four oxygen atoms we have oxidation state -1. A better formula is CrO(O2)2. So, with oxidation numbers you have to be careful and you have to know your compounds. Just another nice one: try to determine the oxidation state of S, C and N in the thiocyanate ion SCN(-) and in the thiosulfate ion S2O3(2-). Here you'll also see that the concept of oxidation number sometimes is not that easy.

11. ## another question

I assume that this picture is 1,2-cyclohexane-diol (1,2-dihydroxy cyclohexane). In this molecule, the bonds between the C-atoms are single. This means that the two items, connected at the C-atoms are pointing downwards and pointing upwards. So, you can have that both OH-groups are pointing upwards (relative to the plane, in which the 6-membered-C-ring is) or both of them are pointing downwards, or they point in opposite directions. Because the C-atoms are in a ring, there is no free rotational motion, so, a OH-group pointing downwards will remain so forever (assuming that the molecule as a whole is kept in the place and not rotated around an axis in the plane). In this way, just from plain 3D-geometry, you can easily see that a molecule is meso or not. In the picture you give, both OH-groups are pointing upwards.
12. ## another question

Indeed, they are the same, they are not enantiomers. The drawing of molecules like these in 2D is somewhat confusing. Imagine the right picture, and rotate the molecule, such that the H is pointing upwards. Then you have CH3 coming out of the plane, towards you, Br in the plane and NH2 going out of the plane, pointed away from you. This is the same situation as in the left picture.
13. ## Yet another nice experiment - selenium allotropes

Today, I did another nice experiment, now with selenium. I transformed the black allotrope into the red allotrope. Less risky than the chromyl chloride experiment, but fun also . This may also be interesting for people who collect elements. Black selenium is affordable. It can be obtained for just over \$10 per 30 grams, including shipping worldwide from http://www.emovendo.net . This is sufficient for all the experiments with selenium you can ever dream of . For my experiment you only need approximately 10 mg, so you have some left for other experiments as well :D. For the experiment, see my site again: http://81.207.88.128/science/chem/exps/selenium/index.html Have fun and stay green.
14. ## organic experiments

Oxalate forms beautiful complexes with iron and chromium. Mix a solution of oxalic acid with a solution of ferric chloride or ferric sulfate and then add some sodium hydroxide. You'll get a nice green complex, trisoxalato ferrate (III), [Fe(C2O4)3](3-). The color of the complex is shown here in the form of its ammonium salt: http://81.207.88.128/science/chem/compounds/ferric_amm_oxalate.html With your acid and a ferric salt, you can make this complex yourself. You can even isolate the potassium salt of this, because that is only sparingly soluble. If you do the same with a ferrous salt, then you get a deep yellow complex, bisoxalato ferrate (II), [Fe(C2O4)2](2-). A similar experiment you can do with chromium. Add some oxalic acid to a solution of potassium dichromate and also add a small amount of dilute hydrochloric acid or sulphuric acid. Then slightly heat. The liquid will turn deep purple. This is an oxalato complex of Cr(3+). I do not know its precise formula though.
15. ## A colored gas, containing chromium

Buy a good old book on chemistry. Pre-war books are best. These books describe all kinds of compounds, which can be made with fairly common chemicals. Especially books in the German language are very nice. Apparently before WW II, the German chemists were focussing more on compounds and processes. I have books with 1000+ pages from 1920 and 1915 and these describe all kinds of compounds. With these books at hand and a little reasoning you can come up with many cool experiments. Modern books focus much more on theory, which on its own is OK, but sometimes they tend to go too far with this. Just knowing the basic properties of common compounds makes chemistry much more fun.

17. ## A colored gas, containing chromium

Yes, you can do nice things with chromyl chloride. You can dissolve this stuff in relatively inert organic solvents, such as ligroin (Dutch: 'laagkokende wasbenzine' / 'petroleumether'). The solution in ligroin then can be used for other experiments, such as oxidation of organic compounds in non-aqueous solvents. The stuff, however, should not be stored, as it is quite unstable and with some organics, it forms explosive mixtures or may cause fire. It also eats almost everything. It is even worse than bromine on storage. I kept it in the little bottle for just a few hours and in that short time, the plastic screw-cap already was severely corroded. You can make a solution in ligroin, simply by adding some ligroin to the mixture. With a glass pipette, you can carefully take away the ligroin layer, with some of the CrO2Cl2 dissolved in it. The ligroin remains above the sulphuric acid layer.
18. ## A colored gas, containing chromium

I want to share with you a nice experiment with a colored gas, which contains chromium. It looks like bromine, but in reality it is a very volatile metal-compound! http://81.207.88.128/science/chem/exps/volatile_chromium/index.html Have fun. If you repeat the experiment, please be very careful. The gas is really toxic!

20. ## n-spheres.

Yes, it does. A countable set of points is said to have dimension 0. An n-sphere indeed is a (curved) n-dimensional space of finite n-volume. For 1-dimensional spaces (a circle), the 1-volume is the length of the circle, for 2-dimensional spaces (an ordinary sphere, the surface of a ball), the 2-volume is the surface of the sphere, etc. The n-volume of an n-sphere is proportional to r^n, with r being the radius of the n-sphere, e.g. the 1-volume of a circle is 2*pi*r, the surface of a sphere equals 4*pi*r^2. For a 0-sphere, the 0-volume does not depend on the radius, it can be written as k*r^0, being a constant k. A three-sphere has nothing to do with time. It simply is a geometrical object. A nice way to visualize a 3-sphere is the following: Suppose you are on the sphere floating around in space (this is imaginable, because you are 3D as well I hope ) and start moving forward. If you keep on moving forward in a "straight line" in the same direction, then you'll eventually reach your original position again. If you cannot imagine this, then think of the analogon of a 2-sphere with you being on the surface (of e.g. a planet). If you walk to a certain direction and you keep on moving, then you'll end up at your initial position again. I place the words "straight line" between quotes, because of the fact, that such thing not really exists on a sphere. A sphere is not an euclidian space, although locally it approaches an n-dimensional euclidian space. In curved spaces, the concept of straight line must be replaced by the more general concept of "geodesic". Google is your friend on this subject.
21. ## organic experiments

What organic compounds are you referring at. Beware, organic chemistry is not the easiest to deal with. Organic reactions frequently are slow, incomplete and frequently not spectacular to see (input is colorless and output still is colorless). If you want to appreciate organic chemistry, then you really have to know what you are doing, otherwise you'll be disappointed quickly! Besides what I mentioned above, organics frequenly involve volatile solvents and reactants. This introduces a severe additional risk for your health. If I were you, stick to inorganics for the time being. This is easier, the chemicals can be obtained more easily and the risks for your health are less (there are some exceptions, but the average home chemist will not come across these usually).
22. ## How are radical valencies solved?

You cannot say that a certain atom (e.g. the halogen) has a radical, a compound or ion as a whole is a radical (or not). For the halogens, the compound ClO2 is regarded a radical, but Cl2O is not. Try to draw a Lewis structure of ClO2 and you'll see that you have trouble making such a model, while it can be done easily for Cl2O. The same difficulty exists for NO2 and NO, but not for N2O5, N2O3 and NO3(-).
23. ## Location of zeros of truncated taylor series of exp(x)

Hello, Suppose we have a truncated taylor series for exp(x), truncated at the term, just beyond the N-th power. Let's call these truncated series trunc(N, x). Some examples: trunc(0, x) = 1 trunc(1, x) = 1 + x trunc(2, x) = 1 + x + x^2/2 ... trunc(N, x) = 1 + x + x^2/2! + .... x^N/N! Here N! means the factorial of the integer N. Here, trunc(N, x) is an N-th degree polynomial over the complex numbers with variable x. The zeros of trunc(N, x) are on a very regularly shaped curve. The shape of the curve hardly depends on N. As an example I'll show the zeros for trunc(400, x) as a gif image, computed by means of my polynomial solving routine (which uses adaptive multiple precision arithmetic). http://www.woelen.nl/zeros-trunc400-exp.gif The image shows the distribution of the zeros of trunc(400, x) in the complex plane. Each red dot represents a single zero. As the image shows, the zeros are nicely distributed along a curve. My question is, is there a closed analytic expression for the curve, as function of N, or is there a closed analytic expression for the limiting curve, for N going towards infinity? When a plot is made for other values of N, then the curve looks very similar. For increasing N, the curve tends to blow up, but its basic shape hardly changes. If some of you has any idea about an expression for the curve, then I would be pleased to read about that. Thanks, Wilco PS: Source code of the polynomial equation solver, together with some test programs is available at http://woelen.scheikunde.net/science/software/mpsolve.html
24. ## How are radical valencies solved?

What is the valence shell? This concept works quite OK for e.g. sodium, chlorine, etc. For transition metals it does not work anymore. Many transition metals use electrons from their outer shell, but also from deeper shells in their chemical reactions. For the lantanides things get even more complex. With these, the second deepest shell also is involved in the chemistry of the elements, so one cannot simply speak of THE valence shell. As I stated already, for these elements, the only way to understand how it is is by performing mathematical simulations of quantum mechanical models.
25. ## How are radical valencies solved?

From this, I understand that you relate valency to reactiveness. You mean that the higher the valency, the higher the reactiveness? The fact is that valency is not related to reactiveness. The element fluorine has valency 1 and has no other possibilities. Yet, it is the most reactive element there is and also among the compounds (which is a much broader class) it is among the most reactive ones. On the other hand, an element as sulphur frequently has valency 2, but it is not really reactive. The concept of fractional valency I have never seen. Remember, the concept of valency is introduced to make reasoning about chemical compounds and bonds somewhat easier, but with many real-life molecules this concept gets flawed and more advanced concepts like molecular orbitals or even low-level quantum mechanics is needed in order to understand the structure of the compound.
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