The activation energy for a reaction is quoted as 120 (卤8.4) kJ mol. What is the percentage error (uncertainty) in the result?
Please show your workings. Thanks.How to calculate the percentage error?
(120+8.4 - 120-8.4)/120 x 100How to calculate the percentage error?
(8.4 / 120) * 100 = 7%
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