Capa Posted January 24, 2009 Report Share Posted January 24, 2009 (edited) HiI am doing a design experiment and I have so far decided to investigate the enthalpy change of combustion for ketonesHowever, to be able to do this I need to know the theoretical value and I have no idea how to find it. Does anyone have an idea of what I could do and does the experiment look appropriate?I would be very thankful for any answersCapa... Edited January 29, 2009 by Capa Reply Link to post Share on other sites More sharing options...
Max Posted January 24, 2009 Report Share Posted January 24, 2009 However, to be able to do this I need to know the theoretical value and I have no idea how to find it.Check the data booklet. It has values for some carboxylic acids. Reply Link to post Share on other sites More sharing options...
IBdoc Posted February 9, 2009 Report Share Posted February 9, 2009 Quick question: In a chemistry lab, do you propagate the uncertainty and do the normal calculations in one single step (e.g: (5 +/-0.1) + (4 +/0.1) = 9 +/-0.2) ? Or is it better to separate and propagate the uncertainty in another step? Reply Link to post Share on other sites More sharing options...
heartworthy Posted February 9, 2009 Report Share Posted February 9, 2009 Quick question: In a chemistry lab, do you propagate the uncertainty and do the normal calculations in one single step (e.g: (5 +/-0.1) + (4 +/0.1) = 9 +/-0.2) ? Or is it better to separate and propagate the uncertainty in another step?When adding (or subtracting) numbers, you add the uncertainties. So 5±0.1 + 4±0.1 = 9±0.2To find the percent uncertainty, you would divide the uncertainty by the value and multiply by 100. So (0.2/9)*100Hope this helps Reply Link to post Share on other sites More sharing options...
IBdoc Posted February 9, 2009 Report Share Posted February 9, 2009 I meant, do we have to explain how we calculated the uncertainty or can we just plug it in without explaining in details?For example, do we present our work like this: 5±0.1 + 4±0.1 = 9±0.2 OR like this: 5+4=9 and 0.1+0.1= 0.2, therefore the answer is 9±0.2 (this is just a simple example it gets more complicated when we multiply or divide)I hope u got what I mean. Thanks Reply Link to post Share on other sites More sharing options...
Tilia Posted February 9, 2009 Report Share Posted February 9, 2009 I meant, do we have to explain how we calculated the uncertainty or can we just plug it in without explaining in details?For example, do we present our work like this: 5±0.1 + 4±0.1 = 9±0.2 OR like this: 5+4=9 and 0.1+0.1= 0.2, therefore the answer is 9±0.2 (this is just a simple example it gets more complicated when we multiply or divide)I hope u got what I mean. Thanks I do the first way, and this far my teacher hasn't complained. Reply Link to post Share on other sites More sharing options...
Mahuta ♥ Posted February 9, 2009 Report Share Posted February 9, 2009 I meant, do we have to explain how we calculated the uncertainty or can we just plug it in without explaining in details?For example, do we present our work like this: 5±0.1 + 4±0.1 = 9±0.2 OR like this: 5+4=9 and 0.1+0.1= 0.2, therefore the answer is 9±0.2 (this is just a simple example it gets more complicated when we multiply or divide)I hope u got what I mean. ThanksPersonally, I would not allow a chance for the examinor to mark me down if he/she happens to see mine. It doesnt take that long anyways. Reply Link to post Share on other sites More sharing options...
IBdoc Posted February 9, 2009 Report Share Posted February 9, 2009 I did mine the first way but I'm going to ask my teacher tomorrow just in case! Anw, thanks for the answers! Reply Link to post Share on other sites More sharing options...
ezex Posted February 11, 2009 Report Share Posted February 11, 2009 HiI am doing a design experiment and I have so far decided to investigate the enthalpy change of combustion for ketonesHowever, to be able to do this I need to know the theoretical value and I have no idea how to find it. Does anyone have an idea of what I could do and does the experiment look appropriate?I would be very thankful for any answersCapa...I love it how the discussion went completely off topic...Capa, here's something that might be helpful: Combustion of Ketones Reply Link to post Share on other sites More sharing options...
Capa Posted February 24, 2009 Author Report Share Posted February 24, 2009 Haha, tahnk you very much ezex.... and yeah, I haven't replied before this cause I stopped looking at this thread went it went faaaaar off Reply Link to post Share on other sites More sharing options...
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