tag:blogger.com,1999:blog-4415570722844642835.post1902664274390884321..comments2024-03-05T22:36:18.660-08:00Comments on Christina's Chemistry Blog: Final Exam Review #27crbowmanhttp://www.blogger.com/profile/04278646988556852649noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-4415570722844642835.post-15709680109443152932011-05-16T17:10:03.379-07:002011-05-16T17:10:03.379-07:00Thanks Sahil! You're right about the [OH-] val...Thanks Sahil! You're right about the [OH-] value. I just reentered it into the calculator and 1.7 X 10^-11 is the correct answer. I will be sure to fix that!crbowmanhttps://www.blogger.com/profile/04278646988556852649noreply@blogger.comtag:blogger.com,1999:blog-4415570722844642835.post-39996122704162138592011-05-16T16:34:37.836-07:002011-05-16T16:34:37.836-07:00I disagree with your value for [OH-] on part B. I ...I disagree with your value for [OH-] on part B. I think it should be 1.7x 10^-11, which I found using Kw. The answer you provided does not check, but you method is correct in my opinion. I think you may have mistyped the numbers into the calculator.<br /><br />Other than that, great job. Your explanations were clear and thorough.Sahil Vanjanihttps://www.blogger.com/profile/16485325517035425140noreply@blogger.comtag:blogger.com,1999:blog-4415570722844642835.post-27719786142381144492011-05-16T12:43:44.838-07:002011-05-16T12:43:44.838-07:00Christina, you did a great job explaining your tho...Christina, you did a great job explaining your thought process and guiding me through the steps in your problem. As far as I can see all your problems and sig figs are done correctly.<br />Thanks!Madisonhttps://www.blogger.com/profile/15814701593069064268noreply@blogger.com