What is the general formula for an alkene?
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CnHnx2\n\nFor instance, the simplest alkene is ethene, or C2H4.
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Add a second carbon atom.  The molecular formula for methans is CH4.  The molecular formula for ethene is C2H4.
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a variety of hydrocarbon that has linear bond of 6 carbons and 2 nitrogen. The rest of atom is hydrogen. Its chemical formula is C6H16N2 and the structure is   \nCH3-CH2-NH-CH2-CH2-NH-CH2-CH3
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An alkane follows the formula CnH2n+2 (n=any real number), meaning for every carbon, there are 2 hydrogens.  Of the choices, C7H16 is an alkane.
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1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 7.849 g x (12 / 44) = 2.141 g\nhydrogen: 3.213 g x (2 / 18) = 0.357 g\noxygen: 3.215 - 2.141 - .357 = 0.717 g\n\n2) convert to moles.\ncarbon: 2.141 / 12 = 0.178 mol\nhydrogen: 0.357 / 1 = 0.357 mol\noxygen: 0.717 / 16 = 0.0448 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.178 / 0.0448 = 4\nhydrogen: 0.357  / 0.0448 = 8\noxygen: 0.0448 / 0.0448 = 1\n\nThe empirical formula of the substance is (C4H8O)n\nThe atomic masses add up to 72, which is also the molar mass of the compound. Therefore n=1 and we find C4H8O
841680
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 5.752 g x (12 / 44) = 1.569 g\nhydrogen: 2.354 g x (2 / 18) = 0.262 g\noxygen: 2.356 - 1.569 - .262 = 0.525 g\n\n2) convert to moles.\ncarbon: 1.569  / 12  = 0.131 mol\nhydrogen: 0.262  / 1  = 0.262 mol\noxygen: 0.525 / 16 = 0.0328 mol\n\n3) divide each molar amount by the lowest value\ncarbon: 0.131  / 0.0328  = 4\nhydrogen: 0.262 l / 0.0328  = 8\noxygen: 0.0328 / 0.0328 = 1\n\nWe have now arrived at the answer: the empirical formula of the substance is C4H8O
959805
HBr, -1\nHOBr, +1\nBr2, 0\nHBrO4,  +7\nBrF3  +3\n\nThere is nothing to explain.  There are rules, and you need to learn them and follow them.\n\nHere is a site with a good set of rules\nhttp://www.science.uwaterloo.ca/~cchieh/cact/c123/oxidstat.html
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Ethylene (or IUPAC name ethene) is the simplest alkene hydrocarbon, consisting of four hydrogen atoms and two carbon atoms connected by a double bond. Because it contains a double bond, ethylene is called an unsaturated hydrocarbon or an olefin.\n\nThe molecule cannot twist around the double bond at room temperature, and all six atoms lie in the same plane. The angle made by two carbon-hydrogen bonds in the molecule is 117°, very close to the 120° that would be predicted from ideal sp2 hybridization.
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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
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Chemisty, A Formula For Answers
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1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 7.432 g x (12 / 44) = 2.027 g\nhydrogen: 4.057 g x (2 / 18) = 0.451 g\noxygen: 3.383 - 2.027 - .451 = 0.905 g\n\n2) convert to moles.\ncarbon: 2.027 / 12 = 0.169 mol\nhydrogen: 0.451 / 1 = 0.451 mol\noxygen: 0.905 / 16 = 0.0566 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.169 / 0.0566 = 3\nhydrogen: 0.451  / 0.0566 = 8\noxygen: 0.0566 / 0.0566 = 1\n\nThe empirical formula of the substance is (C3H8O)n\nThe atomic masses add up to 60, which is also the molar mass of the compound. Therefore n=1 and we find C3H8O
215604
C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
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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
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(1) Find the empirical formula.  Assuming a 100 gram sample,\n\n40.00 g C x (1 mol C/12.011 g C) = 3.33 mol C\n6.71 g H x (1 mol H/1.0079 g H) = 6.66 mol H\n53.28 g O x (1 mol O/16.00 g O) = 3.33 mol O\n\nSo the ratio of C:H:O is 1:2:1, and the Empirical Formula is C1H2O1.\n\n(2) Now the molecular formula of something is some multiple of the empirical formula, or (C1H2O1) x z.  We have to figure out what this z is.\n\nIf it's a gas, and the data given are: \nP = 2931.6 mmHg x (1 atm/760 mmHg) = 3.86 atm\nV = 189.0 mL = 0.1890 L\nT = 23.0C = 296 K\nR = 0.0821 atm-L/mol-K\nso using the Ideal Gas Law, PV=nRT, n  = 0.0300 mol\n\nThe molar mass of the compound is mass/moles = m/n = 1.80 g/0.0300 mol = 60.0 g/mol.\n\nThe Empirical Formula is C1H2O1; one empirical formula unit has a mass of (1 x 12.011) + (2 x 1.0079) + (1 x 16.00) = 30.03; if the molecule weighs 60.0 g/mol, then the z is 2 (in other words, it requires two empirical formula units to make up the molecule) nad the molecular formula is C2H4O2.
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http://en.wikipedia.org/wiki/Alkenes\nhttp://en.wikipedia.org/wiki/Alkynes\nhttp://en.wikipedia.org/wiki/Benzene
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Are you sure you did not make a typo?\nA Hydrocarbon with the formula C7H17 would be very hard to find.  Usually, the number of Hydrogens in a ["normal"] Hydrocarbon can be determined by doubling the number of carbons and adding two.\nIn the case we have something that looks like Heptane (7 carbons), so I would expect there to be 16 Hydrogens (7*2 + 2).  If you align all the carbons in a row and single bond them to eachother to allow for the maxmium number of C-H bonds, where would you place the 17th Hydrogen atom?\n\nIf you did mean C7H17, then the balanced reaction would be,\n4 C7H17 + 45 O2  =  28 CO2 + 34 H2O\n\nHowever, if you did indeed mean Heptane (C7H16), the balanced reaction would be,\nC7H16 + 11 O2  =  7 CO2 + 8 H2O
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The level of unsaturation\n\nAlkanes = all C-C single bonds\nAlkenes = at least one C=C double bond\nAlkynes = at least one C-C triple bond.