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hess' law and enthlapies of reaction
I understand the questions i have been doing but on the last few it said 'calculate the standard change when each of the following is burned completely under standard conditions of 298k and 100kpa'
i) 1.00kg hydrogen
ii) 1.00kg ethanol
iii) 1.00kg octane
does this affect the calculation or are they just trying to put you off.
Thankyou xxxxx
i) 1.00kg hydrogen
ii) 1.00kg ethanol
iii) 1.00kg octane
does this affect the calculation or are they just trying to put you off.
Thankyou xxxxx
indie_couture
I understand the questions i have been doing but on the last few it said 'calculate the standard change when each of the following is burned completely under standard conditions of 298k and 100kpa'
i) 1.00kg hydrogen
ii) 1.00kg ethanol
iii) 1.00kg octane
does this affect the calculation or are they just trying to put you off.
Thankyou xxxxx
i) 1.00kg hydrogen
ii) 1.00kg ethanol
iii) 1.00kg octane
does this affect the calculation or are they just trying to put you off.
Thankyou xxxxx
I think its to do with the number of moles.
Work out enthalpy of combustion normally as kJ mol-1, then work out the number of moles being burnt and multiply by kJ mol-1 to give the actual enthalpy change.
I think, although 'standard change' throws me off slightly.
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