I need to know the atomic size for the element carbon?
412415
You can't get the absolute number(size).\nBecause different books,websites say different values.\nAny way IUPAC value is internationally accepted.\n\nMany IUPAC publications are available over the Internet. For example, the Green Book ("Quantities, Units and Symbols in Physical Chemistry, 2nd edition, 1993") can be downloaded in its entirety from http://www.iupac.org/publications/books/gbook/green_book_2ed.pdf. The 2005 version (a work-in-progress) can also be downloaded.\n\nAn important IUPAC supplement ("Recommendations for nomenclature and tables in biochemical thermodynamics, 1994") is available at http://www.chem.qmul.ac.uk/iubmb/thermod/.
697475
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
335552
a) The difference in weight of Carbon-12 and Carbon-13 is the weight of one neutron. So 1-neutron has the mass:\n(2.158*10^-23)-(1.993*10^-23)\n\nc) The relative atomic mass of the carbon in the sample is:\n[(1.993*10^-23)*0.989] + [(2.158*10^-23)*0.011]\nor just the average weight based on the isotopes' abundance. \n\nb) I don't get the "relative atomic mass of carbon-13", because for a pure sample of 100% carbon-13, the relative atomic mass equals the absolute atomic mass of carbon-13. So it should just be 2.158*10^-23.
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
752142
The element with the largest atomic radius is cesium.  Its electronic configuration is [Xe] 6s1.\n\nFrancium could be larger but I do not believe there is data on its atomic radius yet.
40150
The atomic # of the element is the # of protons and (some times electrons) in the element
21316
I didnot find atomic radii data but font covalent radii data here in the link below. But i think atomic radii must not be available because of the very short life of the nucleide. Its half life is measured in nanoseconds(25 or so/0
874456
If they are linear, then it's quite easy. \n\n1. Number the carbon atoms, trying to assign the lower number possible to double and triple bonds.\n2. Check for double or triple bonds an their numbers.\n3. Write a preffix that expresses the number of carbon atoms:\n\nif its 1 , "meth"\nif its 2 , "eth"\nif its 3 , "prop"\nif its 4 , "but"\nif its 5 , "pent"\nif its 6 , "hex" and now you can carry on.\n\n4. Write the number of the double or triple bonds present.\n5. Write a suffix for the type of compund (no double bonds: "ane", double bonds, "ene", triple bonds "yne")\n\nExamples are:\n\nCH3-CH2-CH=CH2     ____   1-butene (or butene)\n\nCH3-CH2-C≡C-CH3     ____    2- butyne\n\nCH3-CH2-CH2-CH=C=CH-C≡CH   ____    octa-3,4-diene-1-yne\n\nIt can be weird for some cases. So practice.
905163
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 65.45% x 4.13 g = 2.703 g\nhydrogen: 5.49% x 4.13 g = 0.227 g\noxygen: 29.06% x 4.13 g = 1.200 g\n\n2) convert to moles.\ncarbon: 2.703 / 12 = 0.225 mol\nhydrogen: 0.27 / 1 = 0.227 mol\noxygen: 1.200 / 16 = 0.0750 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.225 / 0.075 = 3\nhydrogen: 0.227 / 0.075 = 3\noxygen: 0.075 / 0.075 = 1\n\nThe empirical formula of the substance is (C3H3O)k\nThe atomic masses add up to 55. To find k we need the molar mass of the compound. Use PV=nRT and plug in the data given *in the correct SI units*. Solve for n the number of moles in the sample. You problably find 5.51/55 or 5.51/110 so that k=1 or 2
29071
Well, look in the periodic table of the elements and find the molecular weight. The molecular weight tells you how many grams are in one mole of that element. Once you know the amount of grams that correspond to 1 mole, you do the math.\nGood luck!
149295
carbon, hydrogen and oxygen.
43823
Carbon, with 4 electrons in its outer shell is the most versatile element. It can form single, double, and triple bonds with a wide variety of elements, creating just about every living thing we see. Carbon is known to form at least 10 million different compounds. Elements, such as Nitrogen, with 5 electrons in the outer shell, and Boron with 3, are much more limited.
916738
becoz carbon have very special property\nthe speciality of carbon is its 4 valency electrons,,,, which can be form its octet to achieve stablity by loosing, gaining the 4 electrons,,,,,, or it will share electrons to form octet \n\neven carbon shows catenation property i.e. carbon have tendency to form long chains relating to other atoms which is existing in periodic table\n\ncarbon is basic element for life which is existing on earth
926891
just count how many of each element you have, and make sure they add up\n\n\nC2H6 + 5O2 ---> 2CO2 + 3H2O\n2 carbon 6 hydrogen 2 oxygen----> 1 carbon 2 oxygen 2 hydrogen 1 oxygen\n\n\nso to even it out it would be...\n\nC2H6 + 7O2--->2CO2 + 3H2O\n2 carbon 6 hydrogen + 7oxygen--->2 carbon 4 oxygen + 6 hydrogen 3 oxygen\n\niff you add up each element they are the same on each side
471015
google it and type 'periodic table, first twenty elements'\nn u'll get sumthing, far quicker than lettin sum1 write them up 4 u.
969783
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 6.406 g x (12 / 44) = 1.747 g\nhydrogen: 5.244 g x (2 / 18) = 0.583 g\noxygen: 4.664 - 1.747 - .583 = 2.334 g\n\n2) convert to moles.\ncarbon: 1.747 / 12 = 0.146 mol\nhydrogen: 0.583 / 1 = 0.583 mol\noxygen: 2.334 / 16 = 0.146 mol\n\n3) divide each molar amount by the lowest value\ncarbon: 0.146 / 0.146 = 1\nhydrogen: 0.583 / 0.146 = 4\noxygen: 0.146 / 0.146 = 1\n\nWe have now arrived at the answer: the empirical formula of the substance is CH4O.\nThe atomic masses add up to 32, which is also the molar mass of the compound.
1113568
The chemical element with the highest melting point is carbon, (4300–4700 K) (4027–4427 °C,7280–8000 °F).