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SadCeliac

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one of them is diprotic so pretty sure it's C

you just have to check what has the most moles reacted total
 

rainbowspoon1

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im p sure its a, you can eliminate b bc ch2cooh is a weak acid, and then u look at the mols that react, c has 0.02 mols of water formed, d has 0.005, a has 0.025 therefore the heat of combustion is highest for a
 

user18181818

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don't u want it to have the least moles reacted? wouldn't diprotic mean it has mre moles of water
 

rainbowspoon1

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View attachment 41041
how about this im getting so confused with diprotic stuff
so u work out the [h+] for both solutions. for number 1, u hav 0.25 of H2SO4, so therefore 0.5 mols of h+ as its diprotic. Or so you think!!! H2SO4 ionizes in two stages, the first is h2s04 -> hso4- + h+, which happens 100% of the time (i.e h2so4 is a strong acid), but the ionisation of hso4 is not strong, it forms an equilibirum of hs04- <-> so42- + h+. so there will be less than 0.5 mols of h+ in solution! remeberign that ph is teh negative log of h+, we find that it has to be A, as the concentration is lower, the ph must be higher :)
 

carrotsss

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If you dilute the solution, concentrations are reduced so the system will aim to counteract the change and increase the number of moles by shifting the LHS
 

rainbowspoon1

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just c! addition of h20 favours the reverse reaction and therefore makes it more pinl
 

user18181818

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diluting means the conc of all species are decreased and the system shifts to the side with more moles which is rhs??
 

Unovan

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diluting means the conc of all species are decreased and the system shifts to the side with more moles which is rhs??
If you think about it although the overall conc decreases for both sides, the impact to concentration will be more significant to the side with lower moles, so it will shift to that direction to overcome it, hence diluting shifts equilibrium to side with the lower stoichiometric ratio

(I believe)
 

Unovan

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Actually looking it up online it agrees with u that it will shift to the side with more molecules, idk maybe its a flawed question
 

carrotsss

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If you think about it although the overall conc decreases for both sides, the impact to concentration will be more significant to the side with lower moles, so it will shift to that direction to overcome it, hence diluting shifts equilibrium to side with the lower stoichiometric ratio

(I believe)
that doesn’t make sense, it’s the equivalent of increasing volume and when you increase volume you shift to the side with more gas moles, same deal here
 

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