Equilibria, Energetics and Elements (F325) - June 2011 Exam.

Reply 1160

ruby321

11

Original post by Guinea Pig Lover

Can someone please tell me an easy way to convert gdm-3 into moldm-3? And is it the same for liquids and gases?

I think you divide by the RMM :smile:

Reply 1161

Eloades11

17

Original post by Killmepls

Can someone explain enthalpy change of solution to me? I don't understand why they put enthalpy change of solution with enthalpy change of hydration on the same side in that cycle in the book, page 175.

Enthalpy change of solution is the enthalpy change when 1 mole of a solute has fully dissolved until there is no further change in enthalpy(I think). What do you mean by the book put it on the same side as delta H(hyd)? The sides of good and evil? :biggrin:

Original post by Guinea Pig Lover

Can someone please tell me an easy way to convert gdm-3 into moldm-3? And is it the same for liquids and gases?

To turn moldm-3 into gdm-3 you divide by the Mr, so the opposite would be x by the Mr :smile:

(edited 12 years ago)

Reply 1162

CoventryCity

Original post by Guinea Pig Lover

Can someone please tell me an easy way to convert gdm-3 into moldm-3? And is it the same for liquids and gases?

It is the same as converting from mass to moles
So mass/ Mr = Moles

Reply 1163

intellectual1

3

Original post by science rules! :)

HaHa same here, hope tomorrow is better. F215 was horrific :frown:

it will be like jan11 or even tougher look at the examiners report...

http://www.ocr.org.uk/download/rep_11/ocr_58284_rep_11_gce_jan.pdf

Reply 1164

raj16

Original post by Guinea Pig Lover

Can someone please tell me an easy way to convert gdm-3 into moldm-3? And is it the same for liquids and gases?

g/dm3=mols/dm3 x g/mol,
by puttin a 1/ you value your units flip. so if u had g/dm3 and g/mol and wanted mols/dm3. u put 1 over the value for g/mol units flip mol/g and then the g cancel

Reply 1165

CoventryCity

Original post by ruby321

hey guys, you see in the June 2010 paper? The very last question, number 7(b), I have no idea why the moles of H2O2 in 250cm^3 are multiplied by 40?

Hope someone can help me :frown:

they have divided by 25/1000 which is the same as multiplying by 40

Reply 1166

Wor Carroll

For an answer I put 2000 and the mark scheme has it as 2 times 10^3. It's the same thing so can I just right it as 2000?

Reply 1167

CoffeeStinks

11

I know people have complained about the permed hair question in Jan 2011, I really would like to know how to tackle it.

:frown:

Reply 1168

CoventryCity

Original post by Wor Carroll

For an answer I put 2000 and the mark scheme has it as 2 times 10^3. It's the same thing so can I just right it as 2000?

Yep

Reply 1169

ruby321

11

Original post by CoventryCity

they have divided by 25/1000 which is the same as multiplying by 40

Ok, so...

1.1725x10^-2 moles are in 250 cm^3

1.1725x10^-2/0.25 = 0.0469 moldm^-3 concentration of 250cm^3

This means that the conc of the 25cm^3 is 0.0469 x 10 = 0.469 moldm^-3

so 0.469 x 34(RMM) = 15.946

Did I do it right?

Reply 1170

CoffeeStinks

11

Original post by Wor Carroll

For an answer I put 2000 and the mark scheme has it as 2 times 10^3. It's the same thing so can I just right it as 2000?

Yeah it's the same.

I'm pretty sure that you'll get the mark. Sometimes they even allow incorrect rounding so yours must be all right.

Reply 1171

iSomething

Hi, im stuck on this question from the OCR practice question section about redox equations, I'm having trouble balancing it as I cant seem to arrive at the answer they have at the back of the book:

"Aqueous thiosulfate ions, S2O3^2-, are oxidised to sulfate (VI) ions, SO4^2-, by chlorine gas, which is reduced to chloride ions." Use oxidation states to construct a redox equation."

Here's my attempt but im lost as to what I should be doing to the O atoms (I could add 2H2O but as it would solve the 2O atoms and could balance with 4H+ ions on other side but the book says its 5H2O and 10H+ on other side. How is that possible unless im being really stupid.

Reply 1172

ruby321

11

For standard lattice enthalpy, it says "enthalpy change when 1 mole of a solid ionic lattice is formed from is gaseous ions under standard conditions" surely the ions cannot exist in a gaseous form when it is 25 degrees C and 100kPa?

Reply 1173

sportycricketer

Just to confirm the colour changes for the thiosulphate, copper and iodine reaction..

Is it Brown to Colourless/Straw?

Reply 1174

ChubbyRain

Original post by ruby321

For standard lattice enthalpy, it says "enthalpy change when 1 mole of a solid ionic lattice is formed from is gaseous ions under standard conditions" surely the ions cannot exist in a gaseous form when it is 25 degrees C and 100kPa?

I think its just lattice enthalpy, not standard lattice enthalpy?

Reply 1175

viksta1000

15

Original post by ruby321

For standard lattice enthalpy, it says "enthalpy change when 1 mole of a solid ionic lattice is formed from is gaseous ions under standard conditions" surely the ions cannot exist in a gaseous form when it is 25 degrees C and 100kPa?

which one of the reasons why you have to use born haber to find LE...and because its impossible to measure directly

Reply 1176

intellectual1

3

January 2011 Examiners Report highlights some points:

candidates were often vague and muddled in the language that was used and questions on basic concepts were often poorly attempted.

some candidates wrote too much in their responses to early questions; others seemed to get bogged down trying to complete what were obviously difficult parts to questions, with a detrimental effect on later questions.

all except the weakest candidates generally showed skill on the calculation questions

http://www.ocr.org.uk/download/rep_11/ocr_58284_rep_11_gce_jan.pdf

Reply 1177

ruby321

11

Original post by ChubbyRain

I think its just lattice enthalpy, not standard lattice enthalpy?