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can anyone kindly help me to solve q#4 october november 2016 component 12?
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it's okay, got it !!! thanks anywaycan anyone kindly help me to solve q#4 october november 2016 component 12?
Look here's the thingAns :21/9
Sulfur can be oxidised in two ways.
S(s) + O2(g) → SO2(g) ∆H o = –296.5 kJ mol–1 2S(s) + 3O2(g) → 2SO3(g) ∆H o = –791.4 kJ mol–1
Sulfur trioxide can be made from sulfur dioxide and oxygen.
2SO2(g) + O2(g) → 2SO3(g)
What is the standard enthalpy change for this reaction? A –1384.4 kJ mol–1 B –989.8 kJ mol–1 C –494.9 kJ mol–1 D –198.4 kJ mol–1
please somebody explain me this twist in the hess's law and the point of doing it. i would be grateful.
The reactionSulfur can be oxidised in two ways.
S(s) + O2(g) → SO2(g) ∆H o = –296.5 kJ mol–1 2S(s) + 3O2(g) → 2SO3(g) ∆H o = –791.4 kJ mol–1
Sulfur trioxide can be made from sulfur dioxide and oxygen.
2SO2(g) + O2(g) → 2SO3(g)
What is the standard enthalpy change for this reaction? A –1384.4 kJ mol–1 B –989.8 kJ mol–1 C –494.9 kJ mol–1 D –198.4 kJ mol–1
please somebody explain me this twist in the hess's law and the point of doing it. i would be grateful.
Thank you so much for the reply. I wanted to ask why don't we multiply enthalpy change for sulfurtrioxide with 2 .First of all reverse the first reaction and multiply it by 2 as we do in writing equations for contact process this will also change the sign of enthalpy change required for this reaction i.e 2(296.5)=+593 make an overall equation of the both equations given to get 2SO2(g) + O2(g) → 2SO3(g)
then
-791.4+593= -198.4 thus D is correct option.
cuz the reaction is like 2SO2 + O" -> 2SO3Thank you so much for the reply. I wanted to ask why don't we multiply enthalpy change for sulfurtrioxide with 2 .
its okaya big big big thanks to you from my side. but isn't it a bit misleading to write -781.4 per mole. They could have just written enthalpy change = 781.4 . i hope i am not annoying you. this per mole thing is what is causing me the whole confusion anastasia grey113
You don't need to revise all of them.Just go with cobalt and copper.Hey Guys! Long time no see. I hope your studies are going great this year!
Could an A level student help me out here. Is there any easy way to memorize all the Transition metal complex reactions regarding Cobalt, Manganese, copper etc., and their products?
Thanks, but is there any easy way to memorize the products and the reactions for cobalt and Copper ? Im having a really hard time with Ligands this year!You don't need to revise all of them.Just go with cobalt and copper.
Not really. You just have to memorise them as they are. There's no concept.Thanks, but is there any easy way to memorize
Hey where have you been all the time. Seeing you almost after a year though Welcome back to the XPC!Hey Guys! Long time no see. I hope your studies are going great this year!
Could an A level student help me out here. Is there any easy way to memorize all the Transition metal complex reactions regarding Cobalt, Manganese, copper etc., and their products?
May june 2016/12, q#7 why the answer is D instead of C, any comments please?
Well first write down the volumes of each gas.T is an alcohol, CxHyO. A gaseous sample of T occupied a volume of 20cm3 at 120°C and 100kPa. The sample was completely burned in 200cm3 of oxygen (an excess). The final volume, measured under the same conditions as the gaseous sample, was 250cm3 . Under these conditions, all water present is vaporised. Removal of the water vapour from the gaseous mixture decreased the volume to 170cm3 . Treating the remaining gaseous mixture with concentrated alkali, to absorb carbon dioxide, decreased the volume to 110cm3 . The equation for the complete combustion of T can be represented as shown. CxHyO + zO2 xCO2 + y/ 2 H2O (i) Use the data given to calculate the value of x ..
the answer states that x= 3
How?????!!!!!!!!
Thx for help
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