Jump to content


Senior Members
  • Posts

  • Joined

  • Last visited

Everything posted by ChemSiddiqui

  1. please see the answer to balancing the equation on the chemnet chem group page

  2. the idea about the Cp and Cv for solid has been established but for gases are you guys sayin that when a gas is compressed then we have changed the pressure so the volume is decreased and so does the Cp and vice versa? Sounds logically tough! And by the way thanks guys .
  3. Hi all, I read this a while ago and was wondering why the this was so; Cp>Cv for a gas while for solids its the same. I tried to do some research but got absolutely nothing. Any ideas. Help appreciated. thnx
  4. I know the peak can't be N2 because N2 doesn't have a dipole hence no absorption peak. The wavenumber matches the co as well as the peak shape so I think my suspicion that there were some traces of CO in the cell. Thanks for your interest and help though!
  5. No John, I tried the experiment my self and do get some absorption peak. I was quite confused. But now i am thinking that it maybe that the cell i used to put N2 into had been used to put some CO in before so maybe the N2 was mixed with some traces of CO when I put N2 in the cell and it showed some absorption peak
  6. Hi all, I was just wondering and was infact a bit curious about a question thats was bring put forward to me. In FTIR, N2 is used to take the background right and it is supposed to have no absorption peak becuase it is not a dipole but still the N2 show absorption peak....! I don't know how that can be. May be there is a reason that the in N2 molecule bond if not vibrate but does rotate that show the absorption peak. I have no clue....any help????!
  7. o i see. I am shit in maths and just don't want to pay attention to some trivial things which i realise can be a disaster but i must get my acts straight in maths.
  8. nomenclature follow a set of rule for organic molecules. Its always important to state where any subtituent will be at the any organic compounds.
  9. so it did work then(at least use to) .....yeh it is a shame it did look like a good program to use
  10. don't know what to suggest....I am sure where[country] you live there will be a british High commision office. your best source of information will be them. good luck
  11. no herm it didn't. I just said thank you as a gratitute of your intention to help. I did all the calculations myself and filled pages of answer (it wan't just oxalic acid whose fragementation pattern I was after!!!!!!).
  12. well actually i was going through it last night and I figured that we have to convert the wavenumber into frequency in nm so I used the formula V(bar) = 1/Lambda and made lambda the subject and got lambda = 1/V(bar). That was 1/16300 which equalled 6.134 x 10^-5 cm. To change it into the nm scale i simply divided it by the power of 10 to -15 and the final answer came as 6.134 x 10^-15. Now I don't know if its the right answer or not but my head said it is.
  13. A. The avogadro number/constant isn't 2.022 x 10^23 its 6.02 x 10^23 so please correct your facts. b. you just have to divide the number by the avogadro constant to get the value. c. And please don't use comma to denote a 'dot'. Its mathematically wrong. Good luck and Welcome to the world of chemistry.
  14. Hi I was just wondering if you can help me out on this. Note: this isn't a homework question ok! How you convert the wavenumber in cm^-1 into nm. I know if it said that you had to convert that wavenumber into nm^-1 it would be X 10^10. You people think that I might have to multiply it by 10^-10? Any help appreciated.
  15. i had a great time with him although not a blast as you put it ...ha ha. Good luck with your preparations for exams.

  16. thanks hermanntrude! Appreciate it!
  17. hi all, I just wanted to ask what the fragmentation pattern of oxalic acid will be. Its not a homework question so I think its appropriate for this section. I think that the m/z = 17 corresponding to OH will be lost leaving the fragment C2HO+ behind. I can't think of the other fragment for it. Any idea folks!
  18. when i use 12M HCl in the laboratory I find it very concentrated. Not a very good idea to use 12M for zinc becuase that way you will lose a lot of hydrogen and the acid will eat away the metal. Good to hear you have done the experiment though!
  19. delta(small triangle) sign tells you about the difference of anything. So when its delta E it means difference in energy that occured from forward to the backward reaction. The formula is delta E = bond energy of reactants - bond energy of products. Look out the bond energy data from a chemistry data table and work out delta E urself
  20. religion and science mix perfectly in my opinion and I don't think this question would simply put a lock to this thread just because it concerns religion...no way in my opinion it won't!
  21. hi, sorry to hear that you are having rough time but thats the part of life. I have my brother visiting me today from london and i am relaxing this weekend after working so hard this last week. today after such a long time i will get to surf and contribute in the forums. bye for now

  22. yeh finally sinking my teeths into it. What about you what re you doing these days. are you studying? I meant what are you studying! ok good luck with your studies got to go.

  23. hey Thanks, its alright about wishing late. I haven't had a good eid because I have started University and just don't get time to have any fun. I did my eid away from family you see so thats why.

    I won't log in too much from now but i have other commitments which are related to my studies but thanks anyway again.

  24. No problem, It would be good for you to come here and have a tour of the country no doubt. Well, good luck!

  25. you have to be more precise. Micro-biology is a big field and you must zero on some topics[about microbiology] to enable us to help you. And you can always google your query for answers!
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.