How can you figure out the notation (sp3/sp2/sp) for each Carbon in a molecule?
823065
Well, whether the bond is a single, double, or triple bond also determines the hybridization of the sigma bond.\n\nFor example, if you have two pi bonds in addition to the sigma bond, the sigma bond is an sp hybridized bond because two p orbitals on each atom are consumed making the two extra pi bonds.\n\nIf you have one pi bond in addition to the sigma bond, the sigma bond is an sp2 hybridized bond because only one p orbital on each atom is consumed to make the single pi bond.\n\nIf no p orbitals are used to make pi bonds all p orbitals are available to make the sp3 sigma bonds.\n\nJust count the number of bonds.  If one the sigma bond is sp3.  If two the sigma bond is sp2.  If three bonds, the sigma bond is sp.
36077
The person who has previously answered this question is very narrow-minded.  Yes, CO2 is a molecule, but it's made up of atoms!  \n\nHere's what you do to covert: \n\nFirst, find the molar mass of CO2.  You do that by adding up the molar mass of carbon and two times the molar mass of oxygen; both are read off of the periodic table. \n(carbon)+2*(oxygen)=(12.0 g/mol)+2*(16.0 g/mol)=44.0 g/mol  \n\nNext, convert the .015 g CO2 to moles using the figure calculated above:\n(.015 g CO2)*(1 mol/44.0 g)=.66 mol CO2\nNote how the g units will cancel each other out (g/g) to leave mol.\n\nNext, convert the moles of CO2 to moles of atoms.  Each molecule of CO2 has 3 atoms (1 atom C, 2 atoms O), so each mole of CO2 has 3 moles of atoms.  \n(.66 mol CO2)*(3 moles atoms/1 mol CO2)=2.0 moles atoms\n\nFinally, convert the moles of atoms to number of atoms using Avogadro's number, 6.022*10^23.  \n2.0 moles atoms*(6.022*10^23 units/mole)=1.2×10^24 atoms
795424
First, use the ideal gas law PV=nRT to solve for n, the number of moles.  You are given P,V, and T.  R is the gas constant 62.364 mmHg*L/(mol*K).  Now you have the number of moles of the unknown and you were given its mass.  Thus, you can find the molar mass of your unknown.  This will help to determine your molecular formula when you compare this value to that of the empirical formula.\n\nTo determine the empirical formula: Assume a 100g sample.  Then you'll have 65.45 g C, 5.49 g H, and 29.06 g O.  Determine the molar ratios of each compound and divide by the least common multiple.  In this case, you get a ratio of C:H:O of roughly 3:3:1.  So your empirical formula is C3H3O.  Compare the molar mass of the empirical formula (55 g/mol) with the molar mass calculated above.  Multiply the components of the empirical formula by a scalar as necessary to match the molar mass of the unknown.
51155
bond energies: CH 413, CC 346, CO single bond 358, OH 463, CO triple bond 1072. CO double bond 732. Add up all the bond energies on the left side of the arrow, then add up all the bond energies on the right and subtract one from the other.  Be sure to multiply the bond energy of each type with how many of each type you have.
428107
first count molecules of carbon in CH3Co2H, in this case its one, 12.01 grams of carbon in each mole of CH3Co2H, divide 12.01 by the total weight mass of CH3Co2H which is 133.91 and you get .0897 or 8.97%
776335
2C8H18 + 2502 ---> 16CO2 + 18H20\nit sometimes helps to first solve H then C then O in that order.
179528
1 mole of a substance means 6.023 x 10^23 molecules..so thr will be 6.023 x 10^23 no of carbon disulphide molecules. now each CS2 molecule has one carbon and 2 sulphur atoms  .so there will be 6.023 x 10^23 atoms of carbon and 2 x 6.023 x 10^23 sulphur atoms
912712
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 4.793 g x (12 / 44) = 1.307 g\nhydrogen: 2.452 g x (2 / 18) = 0.272 g\noxygen: 2.018 - 1.307 - .272 = 0.439 g\n\n2) convert to moles.\ncarbon: 1.569 / 12 = 0.109 mol\nhydrogen: 0.272 / 1 = 0.272 mol\noxygen: 0.439 / 16 = 0.0274 mol\n\n3) divide each molar amount by the lowest value to find small integers:\ncarbon: 0.109 / 0.0274 = 4\nhydrogen: 0.272  / 0.0274 = 10\noxygen: 0.0274 / 0.0274 = 1\n\nThe empirical formula of the substance is (C4H10O)n\nThe atomic masses add up to 74, which is also the molar mass of the compound. Therefore n=1 and we find C4H10O
240972
If you have a system which has, say, A, B, C and D mols of substances 1, 2, 3, and 4, respectively, then:\nmol fraction of substance 1 is A/(A+B+C+D) and so on for the other 3 substances, e.g. for substance 2 it's B/(A+B+C+D)
461580
C                    1s2 2s2 2p2\nSi                   1s2 2s2 2p6 3s2 3p2\nGe                   (Ar) 4s2 3d10 4p2\nSn                   (Kr) 5s2 4d10 5p2\nPb                   (Xe) 6s2 4f14 5d10 6p2\n\nConsult a good chemistry book.
839940
1. 2C6H6 + 15O2 --> 12CO2 + 6HOH\n\n2. C12H22O11 + 12O2 --> 12CO2 + 11HOH\n\n3. CH3COCH3 + 4O2 --> 3CO2 + 3HOH\n\n4. C2H5OC2H5 + 6O2 --> 4CO2 + 5HOH\n\n5. C4H9OH + 6O2 --> 4CO2 + 5HOH
1032656
C4H10\n\nCH3-CH2-CH2-CH3
859471
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 9.630 g x (12 / 44) = 2.626 g\nhydrogen: 4.928 g x (2 / 18) = 0.548 g\noxygen: 4.055 - 2.626 - .548 = 0.881 g\n\n2) convert to moles.\ncarbon: 2.626 / 12 = 0.219 mol\nhydrogen: 0.548 / 1 = 0.548 mol\noxygen: 0.881 / 16 = 0.0551 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.219 / 0.0551 = 4\nhydrogen: 0.548  / 0.0551 = 10\noxygen: 0.0551 / 0.0551 = 1\n\nThe empirical formula of the substance is (C4H10O)n\nThe atomic masses add up to 74, which is also the molar mass of the compound. Therefore n=1 and we find C4H10O
501994
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/posfe.html#c3\n\nThe link above may be of some help. Input the values you want or need.
842808
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 9.847 g x (12 / 44) = 2.686 g\nhydrogen: 5.038 g x (2 / 18) = 0.560 g\noxygen: 4.146 - 2.686 - .560 = 0.900 g\n\n2) convert to moles.\ncarbon: 2.686 / 12 = 0.224 mol\nhydrogen: 0.560 / 1 = 0.560 mol\noxygen: 0.900 / 16 = 0.056 mol\n\n3) divide each molar amount by the lowest value\ncarbon: 0.224 / 0.056 = 4\nhydrogen: 0.560 / 0.056 = 10\noxygen: 0.056 / 0.056 = 1\n\nThe empirical formula of the substance is (C4H10O)n\nThe atomic masses add up to 74, which is also the molar mass of the compound. Therefore n=1 and we find C4H10O
834060
def C6H14. i'm positive. hope this helps!!!
171003
in chemistry, u calculate coefficients based upon the subscripts of each element in the individual compounds which are involved in a chemical reaction