Comments for Adrian Dingle's Chemistry Blog http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog Sat, 12 May 2012 13:04:27 +0000 hourly 1 http://wordpress.org/?v=3.3.2 Comment on Comments on, and answers to, the 2012 AP Chemistry FRQ’s by Adrian http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-161 Adrian Sat, 12 May 2012 13:04:27 +0000 http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-161 Oh, there are many simpler ways to answer that, BUT that was the way I 'saw' it! Oh, there are many simpler ways to answer that, BUT that was the way I ‘saw’ it!

]]> Comment on Comments on, and answers to, the 2012 AP Chemistry FRQ’s by sperri http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-160 sperri Fri, 11 May 2012 22:37:45 +0000 http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-160 Hi Adrian, on question 2, in order to get the oxygen pressure it seems much more straightforward to simply subtract the pressure of the hydrocarbon from the total pressure since no change in temperature or volume is reported Hi Adrian, on question 2, in order to get the oxygen pressure it seems much more straightforward to simply subtract the pressure of the hydrocarbon from the total pressure since no change in temperature or volume is reported

]]> Comment on Comments on, and answers to, the 2012 AP Chemistry FRQ’s by Adrian http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-159 Adrian Thu, 10 May 2012 18:20:35 +0000 http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-159 I wrote it backwards, sorry - it's been amended now. I wrote it backwards, sorry – it’s been amended now.

]]> Comment on Comments on, and answers to, the 2012 AP Chemistry FRQ’s by bmeighan http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-158 bmeighan Thu, 10 May 2012 16:21:47 +0000 http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-158 Adrian, I see that you had the right answer, just had the formula shown reversed and I made a math error in coming up with .209 min-1 Adrian, I see that you had the right answer, just had the formula shown reversed and I made a math error in coming up with .209 min-1

]]> Comment on Comments on, and answers to, the 2012 AP Chemistry FRQ’s by bmeighan http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-157 bmeighan Thu, 10 May 2012 15:35:30 +0000 http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-157 Adrian, thanks for posting the draft answers. I think you might have made a mistake on your draft on #3d. You used the first order integrated rate law to calculate k but you used ln[a] at time 0 - ln[a] at time t = -kt . The first order integrated rate law shows ln [a] at time t - ln[a] at time 0 = -kt . When I plugged into the equation as written on the formula sheet, I got -.209 min-1for k. Adrian, thanks for posting the draft answers. I think you might have made a mistake on your draft on #3d. You used the first order integrated rate law to calculate k but you used ln[a] at time 0 – ln[a] at time t = -kt . The first order integrated rate law shows ln [a] at time t – ln[a] at time 0 = -kt . When I plugged into the equation as written on the formula sheet, I got -.209 min-1for k.

]]> Comment on Comments on, and answers to, the 2012 AP Chemistry FRQ’s by Adrian http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-156 Adrian Thu, 10 May 2012 15:07:13 +0000 http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-156 Thanks for all the suggestions, typo corrections etc. I think I have fixed most things now, and you should be looking a version of the answers that has a time stamp of '5/10/12 10:48 AM' in the footer. I will pull these after another 24 hours. Thanks for all the suggestions, typo corrections etc.

I think I have fixed most things now, and you should be looking a version of the answers that has a time stamp of ’5/10/12 10:48 AM’ in the footer.

I will pull these after another 24 hours.

]]> Comment on Comments on, and answers to, the 2012 AP Chemistry FRQ’s by Adrian http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-155 Adrian Thu, 10 May 2012 14:10:25 +0000 http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-155 Breaking is endo (+ve), making is exo (-ve) and Delta H is the SUM of those numbers IF you use the signs in the manner that I did. I think my answer is correct. Breaking is endo (+ve), making is exo (-ve) and Delta H is the SUM of those numbers IF you use the signs in the manner that I did. I think my answer is correct.

]]> Comment on Comments on, and answers to, the 2012 AP Chemistry FRQ’s by elpeddi http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-154 elpeddi Thu, 10 May 2012 13:54:37 +0000 http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-154 On question 3, delta H is sum of the heat of formation of the products minus sum of the heat of formation of the reactants, so based on Hess's Law and bond enthalpy that makes the value negative not positve which means it would be exothermic in part C not endothermic (or am I totally off on my calculations here)? On question 3, delta H is sum of the heat of formation of the products minus sum of the heat of formation of the reactants, so based on Hess’s Law and bond enthalpy that makes the value negative not positve which means it would be exothermic in part C not endothermic (or am I totally off on my calculations here)?

]]> Comment on Comments on, and answers to, the 2012 AP Chemistry FRQ’s by Adrian http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-153 Adrian Thu, 10 May 2012 02:44:49 +0000 http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=785#comment-153 Thanks for the comments, tweaking as I go! Thanks for the comments, tweaking as I go!

]]> Comment on Hund’s Rule is the same as the urinal rule….. by 21stcenturychem http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=587#comment-152 21stcenturychem Sun, 09 Oct 2011 11:52:15 +0000 http://adriandingleschemistrypages.com/AdrianDinglesChemistryBlog/nfblog/?p=587#comment-152 Okay. The analogy worked perfectly. Okay. The analogy worked perfectly.

]]>