****Leaving Certificate Chemistry [All Levels] Before and After Discussion****

little sis...

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

MacBizzle

Are people studying ethanal and ethanoic acid just in case they come up even though they shouldn't? :P I'm half paranoid that I should now :pac:

little sis...

Maybe have a look at the 2005 and 2007 Q2's.
If they come up it will be very similar if not the same as them

CL987

Heaney is coming up in the experiments , so I hear anyway

iWin1010

CL987 wrote: »

Heaney is coming up in the experiments , so I hear anyway

I'd nearly have a better chance of passing this if he did...

AtomicKoala

CL987 wrote: »

Heaney is coming up in the experiments , so I hear anyway

Why are “the old axles and iron hoops rusting"? What is the formula for the compound produced “outside"?

MacBizzle

Excited lads?

robman60

Thanks for the tips last night peeps, hopefully the few hours I've done will pay off.

To be honest, actually doing some study has only shown me all I don't know but sure I'll be in the pub by 5:05 this evening so happy days!

MacBizzle

robman60 wrote: »

Thanks for the tips last night peeps, hopefully the few hours I've done will pay off.

To be honest, actually doing some study has only shown me all I don't know but sure I'll be in the pub by 5:05 this evening so happy days!

Lucky you... :mad:

Good luck robman, I'm in the same boat as you :pac:

Daniel2590

Well that went a hell of a lot better than I thought!

Msci

Daniel2590 wrote: »

Well that went a hell of a lot better than I thought!

What expts came up?

Daniel2590

Msci wrote: »

What expts came up?

The two I did were preparation of soap and the effect of temperature on reaction rate, can't remember the other.

TooMuchStudy

Daniel2590 wrote: »

The two I did were preparation of soap and the effect of temperature on reaction rate, can't remember the other.

Other one was water crystallization

Inspector Coptoor

That was the nicest, most student friendly HL chemistry paper in a long time.

ave123

may i just say....THAT CHEMISTRY PAPER WAS MARVELLOUS!....didnt do a tap of study and was still able to answer most of the questions...:D

little sis...

I didn't do Q5 because the last few parts didn't look too nice.
I also Couldn't do the pH calculation but Q10 and 11 were nice so they made up for it.

WoolyAbyss

Did anyone else get 2198? for the heat of combustion?

EDIT -2198

little sis...

WoolyAbyss wrote: »

Did anyone else get 2198? for the heat of combustion?

can't remember but isn't it negative?

TooMuchStudy

WoolyAbyss wrote: »

Did anyone else get 2198? for the heat of combustion?

i got -5000 and something, can't remember exactly

little sis...

WoolyAbyss wrote: »

Did anyone else get 2198? for the heat of combustion?

EDIT -2198

Yep just checked I got that

WoolyAbyss

little sis... wrote: »

can't remember but isn't it negative?

ya it was -2198 sorry just forgot it.

chatterboxxx95

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

WoolyAbyss

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

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.

Huell

Did anyone get that ph question out?

little sis...

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

AtomicKoala

WoolyAbyss wrote: »

ya it was -2198 sorry just forgot it.

I got ΔH = -2877.3 kJ mol^-1, did anyone else?

EDIT: I checked Wikipedia. I think your value is for propane:

http://en.wikipedia.org/wiki/Butane_(data_page)

http://en.wikipedia.org/wiki/Propane_(data_page)

Wikipedia says it -2877.

little sis...

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

Wanderer41

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! :P

accountname

Grand paper, did all but one question :P

AtomicKoala wrote: »

I got ΔH = -2877.3 kJ mol^-1, did anyone else?

Yep, did it out twice and got that both times!

MacBizzle

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