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8 Apr 2009 18:28:39 IST
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40. A yellow turbidity , sometimes appears on passing H2S gas even in the absence of second group radicals. This happens because a) sulphur is present in the mixture as an impurity b)fourth group radicals are precipiated as sulphides c)the H2S is oxidised by some acid radical present in the solution d)the third group radicals are precipitated 
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Only change is eternal.
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8 Apr 2009 21:04:04 IST
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i think option b) is correct.
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8 Apr 2009 21:09:30 IST
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i can't get it JPS ....plzzzzzzz help
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Don't trouble the trouble until the trouble troubles you
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8 Apr 2009 21:14:43 IST
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it is bcoz both both II and IV group radicals are precipitated as sulphides by H2S.
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8 Apr 2009 21:34:20 IST
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ANS IS b ..
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8 Apr 2009 21:43:24 IST
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aye.....how can u tell from the oxidation no whether a substance wioll or will not be oxidised by O3... plz do reply
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8 Apr 2009 21:48:18 IST
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Substance which are in their highest oxidation state are generally notoxidised by O3 since substance in highest oxidation state act as oxidising agent and O3 is also oxidising agent..Hence most likelyno reaction is there between a two Oxidising agent.
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8 Apr 2009 22:34:31 IST
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Nitrate test question is d) Got it off the web: If a bromide or iodide is in the unknown solution, a ring due either to free bromine or to free iodine forms and the iron (II) sulfate is not part of this reaction. However, if bromide or iodide is already known to be in the unknown solution, add silver sulfate solution to precipitate the bromide or iodide as a silver salt and then test the filtrate for the nitrate ion.If a nitrite is in the unknown solution, a diffuse brown ring forms. To eliminate nitrite, add a concentrated solution of urea, then dilute sulfuric acid and warm until effervescence of nitrogen stops. Then test for nitrate.....cheers!!
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8 Apr 2009 22:37:45 IST
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Nitrate test question is d) Got it off the web: If a bromide or iodide is in the unknown solution, a ring due either to free bromine or to free iodine forms and the iron (II) sulfate is not part of this reaction. However, if bromide or iodide is already known to be in the unknown solution, add silver sulfate solution to precipitate the bromide or iodide as a silver salt and then test the filtrate for the nitrate ion.If a nitrite is in the unknown solution, a diffuse brown ring forms. To eliminate nitrite, add a concentrated solution of urea, then dilute sulfuric acid and warm until effervescence of nitrogen stops. Then test for nitrate.....cheers!!
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8 Apr 2009 23:02:30 IST
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Ans 40) its c Ans 39) its d
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9 Apr 2009 11:49:48 IST
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Easy Mirka Q - 
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-na- |
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9 Apr 2009 11:52:17 IST
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(A) and (B)
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9 Apr 2009 11:52:59 IST
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i think C
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9 Apr 2009 11:54:52 IST
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it is A. SO3 exists in trimer form in solid state as S3O9. but the structure you have given has an extra S in place of O. it is O
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9 Apr 2009 11:55:51 IST
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but structure is not correct...S has more than 8 electrons..
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9 Apr 2009 11:57:19 IST
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so what, In H2SO4 too, the total no. of e of S is more than 8
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9 Apr 2009 11:58:07 IST
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SO3 exists as linear chain as well as cyclic structure in solid phase ~ in gas state its is as shown in (c) ~`
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9 Apr 2009 11:59:01 IST
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there S can expand its octet as it is in gaseous form....Q is mentioned abt solid form..
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9 Apr 2009 12:00:14 IST
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i think b is correct option......
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9 Apr 2009 12:00:31 IST
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well I know that A is correct. If B is correct too, thanks for telling me. I didn't know that SO3 can exist as linear chain.
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