What the mass of aluminum?
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The question is either incomlete or it doesn´t make sense. Are you looking for the atomic mass, density or somethig else.\nAs it is, the best answer to your question would be: "depends on how much aluminum".
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Pascal (Same as Newton per Meter sqaure)\nmm of Hg (Mercury)
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Formula mass is the mass of one mole of material molecule; while atomic mass is the mass of one mole of a single atom. e.g. the atomic mass of hydrogen (H) is 1, while its molecular mass is 2 since the molecule has two hydrogen atoms (H-H or H2)
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309grams.\nThe dry mass is 58grams.
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an ore sample weighs 15N (newton) in air. when the sample is suspended by a light cord and totally immersed in water, the tension in the cord is 10.80 N. find the volume and the total density of the sample.. \n\n\nApparent loss in weight=(15 -10.8 )N\nMass of water displaced=(1.50-1.08)kg\n                                        =0.42 kg\nVolume=0.42 l(since cm3= l)\n1 kg= 1 l (litre)\n\nTherefore,volume =0.42 litre\n                             =21/50000 m3\n\n\nsince density equals to mass over volume;\ndensity=0.42 (divide by) 21/50000\n           =100000 kg m-3\n\n\n\n\n\nThis is called Archimedes principle\n\nBelow is a table that will help you to change units from Wikipedia. \n\n\n\nName of unitSymbolDefinitionRelation to SI units\n\ncubic metre (SI unit)\nm³≡ 1 m × 1 m × 1 m\nlitre\nL≡ 1 dm³\nlambda\nλ≡ 1 mm³= 1 μL\ndrop (metric)\n ≡ 1/20 mL= 0.05 mL\nminim (Imperial)\nmin≡ 1/480 fl oz (Imp) = 1/60 fl dr (Imp)≈ 0.059 193 880 208 333 mL\nminim (U.S.)min≡ 1/480 US fl oz = 1/60 US fl dr= 0.061 611 519 921 875 mL\ndrop (U.S.) (alt)gtt≡ 1/456 US fl oz≈ 0.064 854 231 mL\ndrop (Imperial) (alt)gtt≡ 1/1824 gi (Imp)≈ 0.077 886 684 mL\ndrop (U.S.)gtt≡ 1/360 US fl oz= 0.082 148 693 229 1(6) mL\ndrop (medical) ≡ 1/12 mL= 0.08(3) mL\ndrop (Imperial)gtt≡ 1/288 fl oz (Imp)= 0.098 656 467 013 (8) mL\ndash (U.S.) ≡ 1/96 US fl oz = ½ US pinch= 0.308 057 599 609 375 mL\ndash (Imperial) ≡ 1/384 gi (Imp) = ½ pinch (Imp)= 0.369 961 751 302 08(3) mL\npinch (U.S.) ≡ 1/48 US fl oz = 1/8 US tsp= 0.616 115 199 218 75 mL\npinch (Imperial) ≡ 1/192 gi (Imp) = 1/8 tsp (Imp)= 0.739 923 502 604 1(6) mL\nfluid scruple (Imperial)\nfl s≡ 1/24 fl oz (Imp)= 1.183 877 604 1(6) mL\nfluid drachm (Imperial)\nfl dr≡ 1/8 fl oz (Imp)= 3.551 632 812 5 mL\nfluid dram (U.S.); U.S. fluidramfl dr≡ 1/8 US fl oz= 3.696 691 195 312 5 mL\nteaspoon (Canadian)\ntsp≡ 1/6 fl oz (Imp)≈ 4.735 510 416 667 mL\nteaspoon (U.S.)tsp≡ 1/6 US fl oz= 4.928 921 595 mL\nteaspoon (metric) ≡ 5 mL\nteaspoon (Imperial)tsp≡ 1/24 gi (Imp)= 5.919 388 020 8(3) mL\ndessertspoon (Imperial)\n ≡ 1/12 gi (Imp)= 11.838 776 041 (6) mL\ntablespoon (Canadian)\ntbsp≡ ½ fl oz (Imp)= 14.206 531 25 mL\ntablespoon (U.S.)tbsp≡ ½ US fl oz= 14.786 764 782 5 mL\ntablespoon (metric) ≡ 15 mL\ncubic inch\ncu in≡ 1 in³= 16.387 064 mL\ntablespoon (Imperial)tbsp≡ 5/8 fl oz (Imp)= 17.758 164 062 5 mL\npony ≡ 3/4 US fl oz= 22.180 147 171 875 mL\nfluid ounce (Imperial)\nfl oz (Imp)≡ 1/160 gal (Imp)= 28.413 062 5 mL\nfluid ounce (U.S.)fl oz (US)≡ 1/128 gal (US)= 29.573 529 562 5 mL\nshot ≡ 1 US fl oz= 29.573 529 562 5 mL\njigger ≡ 1 ½ US fl oz= 44.360 294 343 75 mL\ngill (U.S.)\ngi (US)≡ 4 US fl oz= 118.294 118 25 mL\ngill (Imperial); Noggingi (Imp); nog≡ 5 fl oz (Imp)= 142.065 312 5 mL\ncup (Canadian)\nc (CA)≡ 8 fl oz (Imp)= 227.3045 mL\ncup (U.S.)c (US)≡8 US fl oz ≡ 1/16 gal (US)= 236.588 236 5 mL\ncup (metric)c≡ 250 mL\nbreakfast cup ≡ 10 fl oz (Imp)= 284.130 625 mL\npint (U.S. fluid)\npt (US fl)≡ 1/8 gal (US)= 473.176 473 mL\npint (U.S. dry)pt (US dry)≡ 1/4 bu (US lvl)= 0.550 610 471 357 5 L\npint (Imperial)pt (Imp)≡ 1/8 gal (Imp)= 0.568 261 25 L\nfifth ≡ 1/5 US gal= 0.757 082 356 8 L\nquart (U.S. fluid)\nqt (US)≡ 1/4 gal (US fl)= 0.946 352 946 L\nquart (U.S. dry)qt (US)≡ 1/32 bu (US lvl) = ¼ gal (US dry)= 1.101 220 942 715 L\nquart (Imperial)qt (Imp)≡ ¼ gal (Imp)= 1.136 522 5 L\npottle; quartern ≡ ½ gal (Imp) = 80 fl oz (Imp)= 2.273 045 L\nboard-foot\nfbm≡ 144 cu in= 2.359 737 216 L\ngallon (U.S. fluid; Wine)\ngal (US)≡ 231 cu in= 3.785 411 784 L\ngallon (U.S. dry)gal (US)≡ 1/8 bu (US lvl)= 4.404 883 770 86 L\ngallon (Imperial)gal (Imp)≡= 4.546 09 L\nbeer gallonbeer gal≡ 282 cu in= 4.621 152 048 L\npeck (U.S. dry)\npk≡ ¼ US lvl bu= 8.809 767 541 72 L\npeck (Imperial)\npk≡ 2 gal (Imp)= 9.092 18 L\nbucket (Imperial)\nbkt≡ 4 gal (Imp)= 18.184 36 L\ntimber foot\n ≡ 1 cu ft= 28.316 846 592
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An ion differs from the original element in terms of the number of electrons.\n\nThe mass of an electron is about 1/1836 of a proton or a neutron.\n\nTherefore there would be a tiny difference in atomic mass, however in most cases it's fairly negligable.
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why do you need it?  how much do you want to spend to figure out what it is, or what alloy it is. \n\nyou can just do the density measurement\nmeasure the radius of the wire and length. \ndo the pi r squared length deal and then weigh it, and run it through the density equation. \n\nyou need a ruler and a calipers or micrometer that reads in inches. \n\nso lets say you have .125 inch wire.  divide that by 2 to get the radius. =  .0625"\nthen square it.  .0039" and multiply by 3.14. \n\nand multiply it by the length in inches  \n\nthen weigh it.  in lbs. \n\ndensity of al is between .096 lb/cu in. and .100 lb/cu in\n\nyou might want to buy wire that is certified as the type of aluminum you need.  the stuff is cheap.  you can get at the electical supply store or maybe a hardware store. \n\nwiring used for electical applications is 8000 series alloy for modern applications \n\nolder wire is 1350 alloy.
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195.09 g\n\nhttp://www.noblemind.com/search.exe?keyword=Platinum+Molar+Mass&var=2
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The mass of protons and neutrons are nearly identical, the mass of an electron is about 1/1840th the mass of a proton, thus it is negligible, so you can't really tell its there.
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There are 32 empty aluminum cans to a pound. Each can weighs 0.5 ounce.
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The mass of the earth has been calculated as 6 x 10^24 kilograms. (Mass is a little different than weight, but let's not get into that.) Here's a link on how this calculation was done.
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the answer is\n2Al + N2 ----> 2AlN
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All right, you know what M means, right?  It's short for moles per liter.  You want to have 0.250 moles of Al2(SO4)3 in each liter of solution.  You're preparing 0.4 L of solution, so you need (0.4 * 0.250) = 0.1 moles of Al(SO4)3.\n\nNow the molar mass of each element is listed on most periodic tables.  That's how much one mole of that atom weighs.  A molecule weighs as much as all the atoms in it put together, so a mole of Al2(SO4)3 weighs as much as two moles of Al, 3 moles of S and 12 moles of O.  That's 2(26.982) + 3(32.06) + 12(15.999), according to my periodic table.  So if you add it all up, a mole of Al2(SO4)3 weighs 342.132 grams.  You only need a tenth of a mole, so you use 34.2132 grams.
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Aluminum is 2.9 times as dense as iron. On the other hand, iron has roughly half the specific heat as aluminum.\n\nSpecific heat is measured in terms of how many calories of energy it takes to raise the temperature of one gram one degree centigrade. \n\nGiven two equal masses of aluminum and iron, the aluminum with the higher specific heat will take more energy to heat and give of more energy cooling. As a result, it will take longer to heat and cool. \n\nHowever, given two equal volumes of aluminum and iron, the aluminum will have only 1/3 the mass and therefore will need throw off only 2/3s as much heat to cool as the iron.
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You don't have to 'get' equivalent weights of substances, its very simple to calculate them.\n\nFirst of all, you need to calculate the molecular wt or know the atomic wt of the compound or element you're dealing with.\n\nThe general formula for the equivalent wt of a compound is \n\nEquivalent weight = Atomic or molecular wt/ x\n\nx has different definitions in different contexts:\n\nIf for an element, x is the valency\n\nEg- E.W. of magnesium = 24/2 = 12\n\nIf for an ion, x is the number of units of charge\n\nEg- E.W. of Al+3 ion = 27/3 = 9\n\nIf for a compund, x is the total amount of formal charge present on the cation/anion, i.e. the number of ions multiplied by the charge on each radical.\n\nEg- For NaCl, the value of x would be 1\n    For MgSO4, the value of x would be 2\n    For AlCl3, the value of x would be 3\n    For Ba3(PO4)2, it would be 6,...\n\nIf for an acid, x is the basicity(no. of ionizable H+ ions)\n\nEg - 1 for HCl, 2 for H2SO4,..\n\nThe same applies for a base, x is the no. of ionizable OH- ions.\n\nNow for KMnO4 (acidified or not),\n\nThe molecular wt is 39+55+64 = 158\n\nThe value of x is 1\nSo the E.W. is 158/1 = 158\n\nHowever, if you are interested in the E.W. of Mn in KMnO4, which is more relevant in redox reactions(using acidified KMnO4), calculate the oxidation state of Mn first. It comes out to be +7.\n\nSo the E.W. is 55/7
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The mass is a property of an object, a fundamental measure of the amount of matter in the object. The usual symbol for mass is m and its SI unit is the kilogram.\n\nThe weight of an object is the force of gravity on the object and may be defined as the mass times the acceleration of gravity, w = mg. Since the weight is a force, its SI unit is the newton.
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Atomic mass is based on the average amu (atomic mass units) of C 12 therefore 35.5 amu is the average atomic weight for Cl 17