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****Leaving Certificate Chemistry [All Levels] Before and After Discussion****
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accountname wrote: »To find if the bonds are polar, electronegativity all the way
But the molecule as a whole can be non polar if the poles centre of gravity cancel
eg with Boron triflouride
The B - F bonds will be polar because there's a large electronegativity difference, but the BF3 molecule is non polar because the poles have a central gravity. I can't explain it well but hopefully this kinda clears it up
https://figures.boundless.com/18492/full/polarity-boron-trifluoride.png
Ohhh I think our teacher was talking about the polarity of the molecule when she mentioned the shapes.
So to check if bonds are polar = electronegativity difference
To check if molecule is polar = shape
Right it makes sense now
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WoolyAbyss wrote: »Did anyone else get 2198? for the heat of combustion?
EDIT -2198
Yup I think so :pac: can't remember and can't be arsed looking in my paper which is about 2 metres away from me :rolleyes:
Thought it was a grand paper overall, didn't like the question on pH tho
did an extra question tho so hopefully all will go well 
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Same I got about halfway through the pH question couldn't do the calculation so I just wen't onto the next question.I think I did nine excluding the pH one so I'm pretty happy.chatterboxxx95 wrote: »Yup I think so :pac: can't remember and can't be arsed looking in my paper which is about 2 metres away from me :rolleyes:
Thought it was a grand paper overall, didn't like the question on pH tho did an extra question tho so hopefully all will go well 0 -
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WoolyAbyss wrote: »Same I got about halfway through the pH question couldn't do the calculation so I just wen't onto the next question.I think I did nine excluding the pH one so I'm pretty happy.
Yeh me too haha wasted time going halfway then realised I had no clue what was going on. It was a pitty because all the other parts were nice!
Yeh I did 9 excluding that too 0 -
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AtomicKoala wrote: »I got ΔH = -2877.3 kJ mol^-1, did anyone else?
I re-did it just now and got -2198....unless I'm doing something wrong?:rolleyes:
It ends up nice so that you don't have any decimals I believe 0 -
AtomicKoala wrote: »I got ΔH = -2877.3 kJ mol^-1, did anyone else?rainbowtrout wrote: »No, you are correct
The total charge on the S2O3 2- is a -2 charge
So that's 3(-2) = -6 for the oxygen and that means +4 charge overall for the two sulphurs which means they have a +2 charge each on the LHS
On the RHS S4O6 has a 2- charge on it
So starting with oxygen again
6 oxygens each have a 2- charge = -12 in total
So as there is an overall charge of -2 that means the four sulphurs have +10 charge in total
each of the four sulphurs have a +2.5 charge
So sulphur has a change from +2 to +2.5 which is 0.5 of an election
As an increase in oxidation number is oxidation this would indicate a loss of half an electron (in theory anyway)
As you probably worked out Iodine went from a 0 charge to -1 which was reduction and a gain of 1 electron
So the sulphurs need to lose one electron to make the loss/gain of electrons balanced
To do this you need to double the number of sulphurs on the LHS of the equation
So you need to put a 2 in front of S2O3 to balance the redox part of it and then a 2 in front of I- to make the whole thing balance.
Thank you so much for explaining, sorry for not replying sooner! I ended up not answering the oxidation question but it helped me before hand so thank you
And I got -2877.33 as well, Atomic Koala.. hope we're right! :P0 -
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TooMuchStudy wrote: »i got -5000 and something, can't remember exactly
You didn't divide by 2 probably
That was so nice, think I got a B1/A2
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