i need help determing lewis structure for CH2Br2?
361732
Lewis structure of CH2Br2\n\nIts identical to the lewis structure of methane but just replace two hydrogen atoms with bromine atoms (remember to add the three lone pairs around both bromine atoms).\n\nNow about polarity:\n\nThe C-Br bond is then considered to be polar. Having bromine slightly negative and carbon slightly positive. Over all the molecule is polar.\n\nSymmetry:\n\nAnd yes dibromomethane (CH2Br2) has two planes of symmetry, one horizontal and one vertical. In order to verify this, you have to look at the molecule in three dimension and not two dimentional (as drawn on a paper)
824584
I'm only helping you because you seem to have done the rest of the questions!\n\n1. Butene is C4H8\n\n2. N and O are very close in the Periodic Table\n\n3. NaBr contains a metal and a non-metal\n\n4. Your diagrams have come out as unintelligible. Look for a molecule with a C=C double bond and no oxygen.\n\n5. CO2 contains non-metals only.\n\n6. Ethane is C2H6, and ethene is C2H4.  So they both contain 2 C atoms.\n\n17. See 3.
613727
try:\nhttp://lb.chemie.uni-hamburg.de/static/data/50_zzk46n8n.html
801483
b. Tertiary bromide. Br2 dissociates into Br. + Br. The first Br. abstracts a proton preferentially from CH3CH2-C-H to give the most stable free radical (most stable because most hindrered and so least likely to revert to hydrocarbon again). That free radical abstracts Br from a new Br2, giving a tertiary bromide and generating a new Br. to continue the chain reaction.
960668
http://www.stolaf.edu/depts/chemistry/courses/toolkits/123/js/lewis/\n\nI agree it is too hard for us to draw lewis structures in a forum like this. What helped me learn lewis structures is by practicing them on sites like these. If you follow each step you can do lewis structures for anything.
805169
alright...\nconsider the R.H.S.\nh2o has two o-h bonds which account for 2x467 and since thre are 2 mol. u get 2x2x467--------(1)\nco2 has two c=o bonds which account for 2x799 and since u hav 4 mol. u get\n4x2x799-------------(2)\naddin 1 and 2 u get 1868+6392=8260\nconsider the L.H.S.\nu hav in c2h2, two c-h bonds which account for 2x413--------(3)\nand one triple bond accounting for 1x839---------(4)\nnow u hav 2x(3+4)=2x(826+839)=3330------(5)\nand u hav 5x(o=o) which acc for 5x(495)=2475----(6)\naddin 5 and 6 u get 5805\ntherefore ur ans is 5805-8260=-2455\nand since u want for one mol of acetylene u get -1227.5 kj/mole
763539
In the Lewis structure, dots left alone want to pair up with those of another atom.  If there are 5-6-7 dots all around, the atom will accept electrons to fill it last layer.  If there are 1-2-3 dots, the atom will want to give those electrons away.\n\nSo you look at your Lewis structures, and try to pair up alone dots from one giving atom to a receiving atom. \n\nFor your peroxyde, you would get H.   and :.O.: twice so one link between and H and a O, another link between the other O and the other H, and then the two O's link together.\n\nFor you chlorine, there are 7 dots all around each atom, so they will share their un-matched electron in a covalent bond.  \n\nFor the CH4O, their are 4 dots all around the carbon, so you would think that one H will bond to each (that's 4 H then) but to add the oxygen, you have to insert it between a H and the C (that the hydroxyde radical).  It is written CH3H usually to show the -OH polyatomic ion.\n\nFor the C2H4, the carbon has 4 dots all around it, and they will bond to each other.  Each will link to two hydrogen (to make up the four in the molecule), leaving two dots on each C, so this means that there will be one double bond between the two carbon atoms.\n\nFinally, the CH2O os trickyer.  That carbon, with its four dots, will have one H bond to each one dot, so there are two free electrons left.  That oxygen has six dots, two pairs and two free.  This means that each free electron of the O will link up to one of the free electrons on the C, which means that there is one double bond between the carbon and the oxygen.  You'll have H-C-H for the first part and C=O for the second part.\n\nIt's kinda hard to type these :-)  Hope this helps anyway
598319
http://www.sbuniv.edu/~ggray/CHE3345/chp14.html\nhttp://web.odu.edu/sci/tisenhou/chem321/chap12/index.htm\n\nbest luck
543388
Hmm, kinda hard to help you here, since we can't see all your options for the first two.  The ellipses (. . .) indicate that you have more in your message than is being displayed.  Perhaps if you retyped them, it would be easier for us to help.\n\nHere's what I *can* help with:\n\nHalogens:Reactive as Noble Gas:STABLE\n(the halogens are very reactive gases, while the noble gases are extremely stable, almost to the point of being inert)\n\nOxygen:molecule as chloride:ION\nOxygen gas is a molecule, while chloride is the ionic form of chlorine\n\nRatio:subscript as oxidation number:CHARGE\nthe ratio of components in a mixture is denoted by the subscripts, as the oxidation number of an atom is denoted by its ionic charge
92358
Just count the (sigma) bonds and the lone electron pairs and memorize the cooresponding geometric shapes.  Sorry, but it's the only way.
1049588
I think that you are trying to find the net ionic equation. \nWhich in this case would be \nAg+ (aq) + Br- arrow AgBr (solid)\nif you are trying to find a single replacement reaction?\nthen you need to put Br- instead of NO3- and cancel out the NO3 - on the product side? \nTried my Best.
985077
chlorine, Cl2, bromine, Br2, or iodine, I2 or fluorine
441718
Key points to look for include:\n\nInfrared\n-NO2 bonds: 1515-1560, 1345-1385\nAlkenes: C-H: 3020-3080 (moderate), 675-1000\n         C=C: 1640-1680(variable) \nIf wet: Large OH hump in the 3500 area... should obliterate most other peaks in the area.\n\nI am not as up to scratch with NMR.\n\nThis is a common compound and an IR and NMR spectra should be easily found on the internet. I recommend chaecking them carefully as jus tlooking for the peaks above will probably not be highly useful.\n\nEven better than the internet is a resource book, any laboratory or university library should have such resource books.\n\nPersonally I recommend that if you are doing much organic chemistry to get 'Organic CHemistry' by Morrison R.T. and Boyd R.N., published by Prentice Hall.\nThis book has an excellent guide for spectra.
512329
Strictly speaking, the Lewis theory is not applicable to Xenon compounds (& if XeCl3 exists at all). The proper way goes by the Valence Bond theory. In that structure, Xe forms 3 single-sigma bonds with 3 Cl. That leaves out 5 valence electrons which is 2 lone pairs & 1 unpaired. This should (if possible) thus dimerise as Xe2Cl6, each segment being an octrahedron with 2 lone pairs. there should be 2 such octahedra (parallel planes) sharing 1 bond.
644051
See if this article has a diagram\n\nhttp://jchemed.chem.wisc.edu/Journal/Issues/1999/Apr/abs542.html
166454
I can supply you said chemicals.\n(C2H5NH)(CH3)C6H3OH \nPlease contact me by email: halsliu@gmail.com
799473
1\na) H2C = CH - CH2 - CH2 - CH = CH2\n ( 1,6-hexadiene )\n\nb) H2FC - CHF - CFH2\n ( 1,2,3-Trifluoropropane )\n\nc) H3C - CH2 - CH2 - CH2OH\n  ( 1-Butanol )\n\nd) H3C - CH2 - CH2 - CHO\n  ( Butanal )\n\n2.\na) Usually FeBr3 is used in addition to Br2 to brominate benzene, but for this:\n\nProduct: Bromobenzene\n\nb) \nProduct: Ethyl butanoate\n\nc) Dunno this one.\n\nd) \nProduct: 1,2-Dichloropropane