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

My mathematics is not good dude. Please try to do your homework yourself.. Just kidding

kukur, aru lai sodhne ho aafno assignment pani? plagiarhythm bhanne kura thaha chha ki chhaina?

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.

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