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Chemistry 18 Online
OpenStudy (anonymous):

A substance of unknown mass absorbs 138 kilojoules of energy, going from 298 to 303 Kelvin. If the specific heat of the substance is 7.11 J/g·°C, how much of the substance was present? Answer 3.88 kg 32.3 g 3.88 g 32.3 kg I need help setting up this equation. q = mcΔT 138 = 5*(7.11)triangleT i have this but I feel like its wrong

OpenStudy (anonymous):

First you need to convert kilojoules to joules 138*1000=138000 J And you must convert kelvins to celsius 298-273=25 C 303-273=30 Now 30-25= 5 So your equation is now 138000=x * 7.11 * 5 138000=35.6x So x=3876g or x=3.88kg D

OpenStudy (anonymous):

Woops I mean A haha

OpenStudy (anonymous):

Thanks so much for your help!

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