I can't find it on the web anywhere. If someone can explain to me in words what it looks like and what the central atoms are and what is bonded to what, that would be great.
138094
It might help if you spelled "naphthalene" correctly. \nhttp://www.madsci.org/posts/archives/may98/894550073.Gb.r.html \nThis website has the structural pictures you can use to do your Lewis dot structure.  \nNaphthalene is an aromatic compound.  It's two aromatic rings sharing a side.
344666
1) Coordinate covalent bond is a covalent bond that both electrons are donated by the same atom. In the case of CO, C has 4 valence electrons and needs another 4 to reach 8. For O, it has 6 valence electrons and needs another 2. In order to make both atom happy (which both have to have 8 electrons), the only way is to have the following structure\n\n. .\nO\n:::\nC\n. .\n\nIn this case, both C and O has one lone pair. In between them, there are three bond pairs, however, one of this bond pair is made up using both electrons from O. You can easily count the number of electrons for C and O to see this.\n\n\n2) Water molecules can form hydrogen bond because every H atom can have a small attraction towards the O atom from another water molecule and vice versa. This H-bond in water increases the attraction between adjacent water molecules, and make it harder to break them, thus, leads to a higher water tension (think of it all the water molecules are sort of connecting between each other, and is not easy to break it). \n\nWhen water turns to ice, this H-bond force the water molecule to freeze at a position that every water molecule are connected to the adjacent molecule through H-bonding structure. If you draw out this V-shape water molecule and start connecting them with the another water molecule, you can see how the water molecules are bonded when they are in ice form. This structure is not a very good way to pack molecules, and leads to lots of open space in between molecules. As a result, for the same amount of water, when it turns to ice, it takes up more space. As a result, leads to a lower density, so ice floats on water.
503314
non polar  \nit's linear in shape ,so the magnet moment equals zero \nthe lone pairs around the two O atoms \nevery O atom has 2 lone pairs\nthe central atom is C atom
509065
It is very simple. Last band gives you that answer. You need to see G=G bond and it is follow by TAG. It is definatly eukaryotic.
449931
Zero.  Carbon has 4 valence electrons, so it likes to form 4 bonds to give it a stable octet.  C is the central atom in CBr4, and since it's hooked up with 4 things, it's happy, and got nothing left over.  CCl4, CF4, and CH4 are similar to it in that respect.
406412
Here is what I got\nhttp://img.photobucket.com/albums/v628/kerahna/lewis.jpg\n\nPretend the Xs are all dots. I just draw Xs to keep track of which electron belongs to which atom. They are all the same.\nI haven't checked the formal charges yet, I'm going to do that now.\n\nEDIT: The formal charges are good.
497170
C, Al(OH)3...\nWould you like an explanation? Okay, I'm supposed to be learning about this in Chemistry right now, so, I think this is how it works...\n\nIf you study the periodic table of elements, then you will find the number of valence (or outer) electrons each element has. Al has 3, O has 6, and H has 1. Before we start, it is important to know that a filled orbital has 8 valence electrons... okay? \nO and H combine, so they have a total of 7 valence electrons. You can imagine this by thinking of the letters OH surrounded by 7 dots in a circle, 2 pairs of dots on each side and 1 alone (a filled orbital with have four pairs opposite each other). The lone pair needs an electron, so it can borrow one from aluminum, which has 3 electrons. Aluminum loses its electron, so now it has 2, and the OH is now full, because it has 8 electrons. However, the aluminum has to get rid of its other two electrons, so another OH is brought in. The aluminum gives up an electron and the second OH is now full. Finally, the last OH pair is brought in and the aluminum gives away its last electron. So now, it is empty and the OH's are full. I guess they're balanced. Anyway, you are now left with 1 Al and 3 OH's, making Al(OH)3. Do you understand now? \n\nAl (3), and OH (7) (because O=6 & H=1, which equals 7)...\nAl - 1 = (2), so now OH = (8) (now it's full)\nAl - 1 = (1), so a 2nd OH (7) is brought in, making it OH (8)\nAl - 1 = (0), so a 3rd OH (7) is brought in, making it OH (8)\n\nIt took 1 Al and 3 OH's, so the new compound is Al(OH)3\nHope this helps!!!\nP.S. sorry for any confusion earlier when I tried to "draw" them. The computer changed the spacing so all you could see was a mess of stars and stuff. : )
500279
PCl3 has the same shape as NH3 (tetrahedral with P in the center, the 3 Cl at the 3 corners and a lone pair at the fourth corner) So the polar covalent bonds don't cancel out each other\n\nSO3 is planar trigonal and thus the dipole moments cancel out each other and it is non-polar\n\nCF4 has a tetrahedral shape and since all bonds are the same they cancel each others dipole moment;the molecule is nonpolar\n\nCH2F2 has the approximately the same shape as CF4, but since not all bonds are the same their angle is a bit different and the sum of the dipole moments is not zero; thus the molecule is polar\n\nSO2 is a bent molecule (because of the sulphur's lone pair), think of it in a way as water with S in the place of O and O in the place of H so it is polar.\n\nCO2 is a linear molecule (O=C=O) so the dipole moments of the two identical polar bonds cancel out each other; the molecule is nonpolar\n\nO=C=S has different polar bonds, so although they have opposite directions the result will be a slight dipole moment, so the molecule is polar.\n\nIn all your questions you have to compare a polar and a nonpolar molecule. Of course the one with the strongest dipole-dipole interactions in each case is the polar molecule.
107345
R-O-CH3 \n\n R is any carbon based group. I wish I had chem draw on this computer, but the structure is actually bent due to the symmetry (this involves group theory - a linear algebra based idea in quantum mechanics) of the orbitals involved in bonding. Its bent for the same reasons that water is bent, now your teachers may say that water is bent due to electronic repulsions but if thats true how come they draw the lone pairs so close to eachother??? Its because its not true, and its part of VSEPR theory which is just a practicle method of teaching structure to people who dont know too much physics, math, and chemistry.
512747
a) Tetrahedral shape. bond angle : 109.5\nNo lone pair on central carbon\n\nb) Linear shape. Bond angle: 180\nNo lone pair on central Be\n\nc) Trigonal pyramdial shape. Bond angle less than 109.5\na lone pair on Cl. 2 oxgen double-bonded and 1 oxygen single-bonded to Cl. 0 formal charge on Cl and the O=Cl. Formal charge on Cl-O is -1.
87111
www.visionlearning.com/library/module_viewer.php?mid=55 - 66k \n\nwww.biology.arizona.edu/biochemistry/ tutorials/chemistry/page2.html - 7k - \n\nhttp://en.wikipedia.org/wiki/Chemical_bond
44366
Information for all of the elements is given at the Wikipedia. Linked below is the article for carbon, with everything you asked for.\n\nTo see similar information for a different element, change the word "Carbon" in the URL to another element.
113848
Write down what you know about the complex:\n\n1) it's tetrahedral (sq planar)  --> ∆t  (t2 and e levels)\n2) CN   --> low spin\n3) d electrons of Ni(2+)  --> [Ar],d8\n\nBefore drawing the CFSE, state that ∆ is large for a square planar complex.  Since ∆ is large, electrons will always pair up, rather than occupying the dx²-y² level.  Since there will be no unpaired electrons, the complex will always be diamagnetic.  I cant draw the diagram here, but show the z²,x²-y² and xz,yz,xy levels, fill the diagram with 8electrons and pair them.
139808
I'm going to explain it really simplified, without the theory behind it, but:\n\nIf an atom has two bonds attached to it, it's an sp.\nIf an atom has three bonds attached to it, it's an sp2.\nFour bonds, sp3.\nFive bonds, sp3d.\nSix bonds, sp3d2.\n\nMake sure you count double and triple bonds toward the bond number.\n\nJust remember that on the test. The reasoning behind it has to do with the number of orbitals that are mixed together. An sp3 is a mix of one s orbital and 3 p orbitals, forming 4 orbitals that are between s and p in energy. An sp3d2 is a mix of one s, 3 p's, and 2 d orbitals, forming 6 orbitals, all of the same energy.\n\n\nHope this helped.
374397
You left out the "f" orbitals.\n\nFollow the increasing atomic numbers as you move across the periodic table, and you'll see that you have to go through the lanthanides before you go to the "d" block.\n\nmercury is  [Xe] 6s2 4f14 5d10
382850
no too sure since its been a while since iv done this, but its a simple enough molecule, C2Br4  2Br to each carbon and a double bond between the carbons, its a linear moleule, not sure about the electronegativity of the elements as my notes r all packed away but the Br atoms should sit slightly of veritcal from each other toward the outside of the molecule. and im sure that the two halves sit at 90 degrees to each other, not sure if u call it this but if u do then 2 Br attached to one carbon will sit along the x- axis of the molecule and the two Br attached to the other carbon will sit along the Z-axis, ask ur tutor or teacher if this is correct as im not 100% sure\n\ntetrahedral only applies to CBr4, thats because the 4 Br are comin off the same carbon, because there are 2 carbons it changes it all together\n   (ignore the full stops, needed them to position the atoms decently)\n\nBr&#92; ........ /Br\n.......C=C\nBr/ ........ &#92;Br\n\nThis is not perfect but imagine the top Br on the left aiming toward you at 45 degrees and the top one on the right aiming away at 45 degrees, hope this helps
444341
You need to look at the central atom and determine the number of lone pairs and bonding pairs of electrons (multiple bonds in this case will not count, as the orbitals used in those bonds will be orthogonal to the bonding axis and thus won't be hybridized).  2 is sp, 3 is sp2, 4 is sp3, 5 is sp3d, 6 is sp3d2.