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kukur
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Posted on 12-11-08 11:23
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propane is often stored and transported in its liquid form.how much energy is required to boil 428.37 g of propane at it boilin point(-42degrees)? delta H vaporization= 15.7 kj/mol of propane
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sajhabahadur
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Posted on 12-12-08 12:27
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My mathematics is not good dude. Please try to do your homework yourself.. Just kidding
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ctal
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Posted on 12-12-08 3:04
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kukur, aru lai sodhne ho aafno assignment pani? plagiarhythm bhanne kura thaha chha ki chhaina? 
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Poon-Hill
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Posted on 12-12-08 4:16
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Propane = 3 carbon and 8 Hydrogen. so (12*3 + 1*8) moles per gram = 44 moles/gm
428.37 gm = 428.37*44 moles = 18848.28 moles
1 mole = 15700 J
Therefore 18848.28 moles = 2,95,918 KJ
Its been ages...so i might be wrong if BP is also accounted in calculation.
Now Pay Up.
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