Cambridge Chemistry Challenge

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Please; Any help with Gold? It´s very difficult...

With Gold, I have the mass of graphene per m^2 in kg and the surface area of the Earth, also in m^2. I'm multiplying them together, which isn't returning the correct result. What am I doing wrong?

Reply 602

loopy786

Original post by Lutski

With Gold, I have the mass of graphene per m^2 in kg and the surface area of the Earth, also in m^2. I'm multiplying them together, which isn't returning the correct result. What am I doing wrong?

Hey, I'm doing this too. :frown:

GRR Silly question... My eyes hurt. I give in, time to sleep and revise in the morning. :cry2:

Reply 603

uniflegram

Original post by chemistinthemaking

I'm using my molecular modelling set, It's getting fiddly! PM me if you have the figures :biggrin:

thanks, I'll love you forever :tongue:

Yeah i'm getting bored of this endless isomerism puzzle. someone put me out of my misery again.

Reply 604

Bello08

Original post by Lutski

With Gold, I have the mass of graphene per m^2 in kg and the surface area of the Earth, also in m^2. I'm multiplying them together, which isn't returning the correct result. What am I doing wrong?

What is the value of mass per meter squared you have?

Reply 605

chemistinthemaking

someone help with Rg pleaseeeeeee pm me

Reply 606

david2457

Original post by chemistinthemaking

someone help with Rg pleaseeeeeee pm me

Same, on this one I really have not got a clue

This was posted from The Student Room's iPhone/iPad App

Reply 607

Maar

Help in Gold; I know that SA of Earth is 510064471.9 km^2 and of graphene: 2.675*10^-3 km^2/kg. But I can´t get the correct answer :s-smilie:

Reply 608

Conan

For those of you stuck on Rg. A clue is that there can only be one element in the centre and the shape of the molecule is drawn in the back ground. For 2nd one, it might help to find the different combinations of the two "poles" of the molecule as they have different bond angle. Then assess the different combinations for the 3 elements in the middle.

Reply 609

Lutski

Can someone give me the answer to Rg? I'm not sure I actually have the knowledge to complete this.

Reply 610

chemistinthemaking

Original post by Lutski

Can someone give me the answer to Rg? I'm not sure I actually have the knowledge to complete this.

same

Reply 611

distroyer

can somebody private message me with some help with the Rg! seriously struggling with it! :frown:

Reply 612

david2457

Original post by distroyer

can somebody private message me with some help with the Rg! seriously struggling with it! :frown:

Same, I have holes in my knowledge to complete this

This was posted from The Student Room's iPhone/iPad App

Reply 613

chemistinthemaking

someone tell me Rg and I will be soooo grateful. pm me pleaseeeee I need to study for my exam as well

Reply 614

Conan

For the first compound, only Si can be in the middle, for the second one, only P in the middle and for the last one, only S in the middle. From there, it's just the number of different permutations you can have which would result in optical isomers.
This is a massive clue for the last molecule:
http://en.wikipedia.org/wiki/Octahedral_molecular_geometry

Reply 615

uniflegram

so it's 1 + something^2 plus 30^3?

Reply 616

Lutski

This makes me sad, I feel as if I can't do anything at all.

Reply 617

warpsinski

I still dont get it with Rg... :/ I thought abt 30^3 + 1 + something^2 too.....

Reply 618

chemicangel

There's more than one isomer for the first one - it's chiral...

As for the second one, use the fact that you know the 3rd is 30. Compare the number of different ligands to the number if isomers.

Reply 619

warpsinski

Original post by chemicangel

There's more than one isomer for the first one - it's chiral...

As for the second one, use the fact that you know the 3rd is 30. Compare the number of different ligands to the number if isomers.