just a bit confusled (1 Viewer)

fashionista

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hiiiiya!! these are sum questions from a practise half yearly we were given...i was just wondering if i got the first two questions right. (i think the first one is wrong)
 

Ragerunner

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You are correct.

Unless I read it wrong. the picture is too small!
 

fashionista

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im sorry..but the site wudnt let me put it any bigger....does that mean i got both of em rite?? wackadoooooooo thanku!!!
 

Ragerunner

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I'm pretty sure you are because the other answers sound wrong.
 

04er

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Well, the next question does tell you that its an exothermic reaction, so I guess 1(b) is correct. I just don't understand how this can be concluded from the graph :confused:
 
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Constip8edSkunk

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yep they r both right . when CO2 is dissolved the reaction is exothermic, so if extra heat is applied, the equilibrium shifts 2 the sde wif less heat, ie the side of the reactants. hence the solubility of the CO2 decreases, this is shown by the graph
 

fashionista

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WOOHOOOOO!!!! thanx evereebodeeeeee (sed in dr nick voice)
Originally posted by 04er
Well, the next question does tell you that its an exothermic reaction, so I guess 1(b) is correct. I just don't understand how this can be concluded from the graph :confused:
thats wut i was thinking..the graph isnt exactly a numerounoinformationextruding piece of work is it???
 

xiao1985

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Originally posted by 04er
Well, the next question does tell you that its an exothermic reaction, so I guess 1(b) is correct. I just don't understand how this can be concluded from the graph :confused:
at higher temperature, more heat is around... the equilibrium can be written as:

CO2 + H2O <=> H2CO3 + heat (exorthemic)

if heat is applied, there is more heat in the equilibrium... hence the equilibruim will shift to minimise the stress on the system...

don't use this reasoning in ur proper answer though...say something like since its exorthermic, the increase in temperature will cause the equilibrium to the left to absorb more heat... and the graph shows this relationship, hence (etc etc etc) ...
 
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