what is the %weight of each element in the alloy below? (Units are mole fraction) Carbon0.04 Chromium 0.12 and
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take the mole fraction and multiply by the atomic weight of each which you can find on the periodic table. For carbon, it would be 12 x .04 = .48 Once you have all three, sum them and divide each by this sum. So, say you find Ch and I for the other two. Then the percentage by weight of Carbon would be the number you got for Carbon (.48) divided by (.48 + Ch + I).\n\nTo double check, your weight percentages should all add up to 100%.
24464
Group 6A contains oxygen, sulfur, selenium and tellurium.\n\na) Any of these form ions with a double negative charge.\n\nb) consequently, the formular can be expected to be Al2X3\n\nc) 18.55% percent Aluminium by mass means that 81.45% of the mass is taken up by X.\n\nThat's 9.275% per Aluminium ion and 27.15% per X ion.\n\nTo determine the atomic mass of X, you take the atomic mass of Aluminium (27 is exact enough for our  purposes), divide it by the mass percentage per Al ion and multiply by the mass percentage per X ion:\n\n27 / 9.275 * 27.15 = 79\n\nA quick look into the periodic table reveals that this is the atomic mass of Selenium.
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Carbon:  12.01 g\nNeon:  20.18 g\nIron:  55.85 g\nUranium:  238.03 g
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It seems that you have problems with moles and Avagadro's number. You will need to refer to your periodic table to get the molecular weights that your teacher wants you to use.\n\nSo get the molecular weights of the elements you are working with from your periodic table and substitute in the formula.\n\nI am using round numbers to show the substitution you should know how many decimal points you need to work to.\n\nHG atomic weight 200\nO atmic weight 16\n\nSubstitute in the formula to get the masses involved\n\n2 HgO(s) - 2 Hg(l) + O2(g) Hrxn = 181.6 kJ\n 2 (200g + 16g)   =  2(200g) + 2(16g) = 181.6 kj  \n\n432g oxide decomposes releasing 181.6Kj\n\nQuestion a)\nStraightforward ratio. X = the no of KJ required\n\n572g/432g = xkj/181.6kj\n\nYou'll need to calculate the molecular mass from you periodic table substitute it for the 432 and then refine the equation.\n\nQuestion b) Straightforward ratio. X = the no of KJ required\n\nYou'll need to calculate the molecular mass from you periodic table substitute it for the 432 and then refine the equation.\n\n289/181.6 = xg/432\n\nA mole is the amount of pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of carbon-12 (i.e., 6.023 X 1023). This involves the acceptance of two dictates -- the scale of atomic masses and the magnitude of the gram. Both have been established by international agreement. Formerly, the connotation of "mole" was "gram molecular weight." Current usage tends to apply the term "mole" to an amount containing Avogadro's number of whatever units are being considered. Thus, it is possible to have a mole of atoms, ions, radicals, electrons, or quanta. This usage makes unnecessary such terms as "gram-atom," "gram-formula weight," etc.
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Calcium phosphate: 13 atoms (3 Ca, 2 P, 8 O)\n4 Aluminum Carbonate molecules: 56 atoms: (36 O, 12 C, 8 Al)\nLead Nitrate: 9 atoms (1 Pb, 2 N, 6 O)\nThe Or threw me off on on this one.  I'm assuming it's Cr.  If it is it would be-\n2 Ammonium Dichromate molecules: 38 atom (4 N, 16 H, 4 Cr, 14 O).\nIf you are adding all the atoms for all the molecules together you get 116 atoms.
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First, it seems to have only 2 isotopes.  \n\nYou need to take each mass and multiply by its percentage.  Then add the two resulting numbers together.  You then have the weighted average mass for the element, or average atomic mass.  This weighted average atomic mass is the one reported on the periodic table of elements.\n\n35 x 40%\n+\n37 x 60%
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Of those three are elements: Gold, Platinum, and Aluminium\n\nOne of them is a molecular compound: Sugar (each sugar molecule is the same, but is composed of many elements)\n\nOne of them is an alloy: Brass (composed of a melt of several metalic elements)\n\nReally depends of what your definition of "pure substance" is.
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In the 76th edition it is Chapter 10 of the CRC - "Line Spectra of the Elements".  You may need to convert the numbers given to the units you desire.
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Cr has 6 electrons\nEach Benzene contributes 6, so that's \n\n6 + 6(2) = 18. CrPh2 does\n\nFor Co(CO)4-\n\nCo provides 9 (uncharged)\nNegative charge provides 1\nCO (carbonyl) provides 2 each, so that's 8\n\n9 + 1 + 8 = 18 electrons\n\nBoth organometallics obey the 18 electron rule.
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First find the mass weight of Strontium: 87.62 g/gmol\nNow find the mass weight of SrCO3: 87.62 + 12 + 3*16 = 147.62\n\nNow find the fraction of Sr in SrCO3: 87.62/147.62 = 0.594\n\nSo in order to get 100 ton of Sr which is only 59.4% of what you need SrCO3.  The the tonnes of SrCO3 you need is\n\n100 ton Sr --- 0.594\nx ton SrCO3 --- 1\nx = 100 / 0.594 = 168.4 ton SrCO3.\n\nNow, again, this is 2% of what is in the ore, so the amount of ore you need is\n\n168.4/2% = 168.4 * 50 = 8417 ton ore\n\nCheck\n8417 * 0.02 * 0.594 = 100  OK.
384517
(12/12+16)*100=42.85\n12---------atomic mass of carbon\n16---------atomic mass of oxygen
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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
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carbon is element # 6 and its molecular weight is 12.0
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d)Argon: 1s2 2s2 2p6 3s2 3p6\ne)Could be carbon, silicon, germanium, tin, or lead.  It's not specific enough.  And there is no such thing as an "elemen." Please treat chemistry with enough respect just type a "t" at the end of a word.  \nCarbon: 1s2 2s2 2p2\nSilicon: 1s2 2s2 2p6 3s2 3p2\nGermanium:  1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2\nTin: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4d6 5s2 4d10 5p2 \nLead: \n1s2 2s2 2p6 3s2 3p6 4s2 3d10 4d6 5s2 4d10 5p6 6s2 4f14 5d10 6p2
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Steel = 12 x 10-6/oC = 6.5 x 10-6/oF\n\nSee the site (the formula does not type through correctly)
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You answer this question by formula weight ratios.  One mole of Na2CO3.10H2O has a mass of 286.14 g.  It also contains 13 moles of oxygen (3 in the carbonate + 10x1 each water), which is 13x15.9994 = 207.99 g.\n\nThe mass percent oxygen is then (amt in question / amt total)x100% = (207.99/286.14)x100% = 72.688%, or 72.7% to the nearest tenth.
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1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 7.793 g x (12 / 44) = 2.125 g\nhydrogen: 3.190 g x (2 / 18) = 0.354 g\noxygen: 3.192 - 2.125 - .354 = 0.713 g\n\n2) convert to moles.\ncarbon: 2.125 / 12 = 0.177 mol\nhydrogen: 0.354 / 1 = 0.354 mol\noxygen: 0.713 / 16 = 0.0446 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.177 / 0.0446 = 4\nhydrogen: 0.354  / 0.0446 = 8\noxygen: 0.0446 / 0.0446 = 1\n\nThe empirical formula of the substance is (C4H8O)n\nThe atomic masses add up to 72.1 g, which is also the molar mass of the compound. Therefore n=1 and we find C4H8O