User talk:Smokeybrain

I normally use the second NFPA numerals, but I was just wondering whether I should use them or the first set. Thanks. --Cheminterest (talk) 01:17, 10 March 2010 (UTC)[reply]

That's the best info to communicate hazards to fire fighters and emergency workers.....I will use the same myself. Thanks for a very good thought provoking question.15:38, 10 March 2010 (UTC)

Your note

You're quite welcome, but what's the subject? I don't remember meeting you before, so I don't know what you're talking about :-) Nyttend (talk) 19:00, 11 March 2010 (UTC)[reply]

We have met Nyttend. You actually saved my talkpage from being deleted by someone who did not show any professional courtesy of offering explanation (an act of vandalism perhaps?). Therefore you have earned my note of thanks. Keep up the excellent work.Smokeybrain (talk) 02:06, 12 March 2010 (UTC)[reply]

Ah, okay; I'd forgotten. Actually, your userpage did qualify to be deleted: not because anything was wrong with it, but because it had been blanked by the only person who had edited it. The reason I didn't delete it was a bad reason: someone tagged it for deletion for a reason that didn't make sense; if it had been deleted, you would have been free to recreate it. Nyttend (talk) 03:04, 12 March 2010 (UTC)[reply]

Thank you so much for explaining. Now I understand. The blanking must have happened inadvertently. Your thoughtful action step is well appreciated by me. Thanks and have a great Pi Day today. Smokeybrain (talk) 22:43, 14 March 2010 (UTC)[reply]

Copper II hydroxide - copper II oxide equilibrium

I have a question about the equilibrium condition that exists between copper II hydroxide and copper II oxide. Here is the reaction: Cu(OH)₂ + heat ←→ CuO + H₂O When heat is added, it drives the equilibrium to the right, forming more copper II oxide. When it is exposed to excessive moisture, it absorbs water to form copper II hydroxide again. But why does copper II hydroxide turn into copper II oxide when it is moist? I think that that would contradict Le Chatalier's principle. Just a question. --Cheminterest (talk) 20:21, 16 March 2010 (UTC)[reply]

Sorry I could not get back to you soonber. You have asked a very good question, and I agree that it contradicts Le Chatalier's principle - if you consider the reaction that leads to water as the product, viz. Cu(OH)2 ----> Cu(II)O + H2O. Here are my two cents.....what if we consider the water as reagent, i.e. in the presence of excess water, is it possible that Cu(OH)2 will convert to Cu(H2O)4(OH)2 and ultimately to Cu(II)(H20)6 and 2 OH negative ions? This could be an equilibrium reaction that facilitates easy oxidation of free Cu ions (hexa aquo copper ions) thus leading to Cu(II)O at ambient temperature. e.g. as illustrated in the equation Cu(H2O)4(OH)2 + 2H2O <----> Cu(II)(H2O)6 + 2 OH

Chemiterest, kindly note that Smokeybrain is not an academic (teacher or professor) like some of the editors you will encounter in Wikipedia. Quite honestly, Smokey is as enthusiastic and fascinated with science (especially chemistry) as you are. Hopefully we may be able to get one of these expert editors to dispel this mystery by offering their constructive and insightful inputs. Who knows?? Smokeybrain (talk) 06:01, 19 March 2010 (UTC)[reply]

I thought of another possible explanation. The water molecules have a strong attraction for the water molecules in the copper II hydroxide, so they take it away when it is moist, converting it to copper II oxide. They do not readily leave if there is no water present. Anyway, thanks for an idea. --Cheminterest (talk) 20:28, 31 March 2010 (UTC)[reply]

Chemical garden

Have you had any experience making a chemical garden? If so, I would like to hear because I would like to try making one. You can post a {{talkback}} template on my page or just post your response on my page. Thanks. --Chemicalinterest (talk) 11:57, 17 May 2010 (UTC)[reply]