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hydrogen_sulphide
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Posted on 09-25-06 4:26
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dear all, if any one of you would like to help me get the 10 percent of overall grade in my chemistry just by solving these 3 problems i would be thankful .
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RajaHarischandra
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Posted on 09-25-06 4:44
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Hydrogen Sulphide, To ask for a help in Chemistry in such questions is a contrary to your name(H2S). While you register for the class, if you have noticed or not, there is a thing called "academic Honesty". I am sure that posting such thing asking for helo to enhance your grade is totally against the academic Honesty also note that professors are very serious about such sorts of matter. As being one of your best wishers, I suggest you to go through some basic general chemistry to find the solution rather than pleading for help in public forum like Sajha.
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bugtraq
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Posted on 09-25-06 4:47
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Prem Charo
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Posted on 09-25-06 4:59
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WTF!!! Dude, why did you can the question? You don't even want to put enuff effort to type it. Well! Type the question and I might consider helping you, aite.
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SBL
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Posted on 09-25-06 6:04
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VincentBodega
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Posted on 09-25-06 6:13
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Dont cheat and allow others to cheat. What if the same H2S is the doctor who will be treating you 10/15 yrs down the road? I heard that general chemistry is one of the prerequisites for pre-med. People should be discouraged to post their homeworks, let alone exams, on this website so that others can do the homework for them.
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passinthru
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Posted on 09-25-06 6:22
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ya, seriously wtf... get out of that mentality of yours...You know why US is where it is & what Nepal is where it is (about 200 years behind Overall), because of this mentality. Trying to reach somewhere without putting the legitimate hard work. It's all Karma, dude. If you work hard, you will till the land and get greater benefits than if you steal someone else's fruits. I am didactic but so what.(f u)
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NaakPore
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Posted on 09-25-06 9:36
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Solution for question no. 1 Molarity= Moles/L Density is given, so we can use it as a conversion factor,(* = Multiply) 1.22g/1mL * 1000mL/1L * 1mol/97.19g = 12.56 mol/L = Molarity Molality = moles of solute/ kg of solvent Lets us consider 100 g of solution. Then we have 40g of solute(KSCN) g of solvent = 100g - 40g = 60g of solvent = 0.06 kg of solvent so, 12.56mol/0.06kg= 209.33mol/kg= Molality Mole fraction of KSCN= Moles of KSCN/ Sum of moles of all component Try this yourself.
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hydrogen_sulphide
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Posted on 09-26-06 9:52
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thank you naakpore bro, tra number 2 and 3 question ko ley solve garidene ho kuniii, please ma aasirwaad dinthey , raamro dulaii paye bhanera...
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sly_evil
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Posted on 09-27-06 1:44
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haha rajaharischandra le hasayo yaaar hehee ..smokeeeeeeeeee
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SBL
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Posted on 09-27-06 4:16
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please help gara yaar ullai
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NaakPore
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Posted on 09-28-06 2:11
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Is yours problem due today?
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