Advanced Higher Chemistry 2012-2013

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Reply 784

Anybody got this one right? struggling to see how the answer is A
Q14 2009.png

Reply 785

Original post by Thuglife

Anybody got this one right? struggling to see how the answer is A
Q14 2009.png

what is the equation for the partition coefficent? find that out and see if it helps you.

Reply 786

Original post by RKurd

Anyone help me with these two questions from the 2009 .mc. numbers 34 and 37. I thought that for an amine to dissolve in water you would pick the one with the lowest rrelative formula mass, being option A but the answer is D.

I have no idea on how to approach number 37.

Thanks in advance

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34: I think its because it has to react with sodium as well- a carboxylic acid reacts with sodium? :hmmmm2:

and i didn't get 37 either when i did it.

Reply 787

Original post by deedee123

what is the equation for the partition coefficent? find that out and see if it helps you.

Thats the problem. I know in general for equilibrium constants, its meant to be K=[products]/[reactants] but for partition coefficients I was taught it was K=[upper]/[lower]?

Reply 788

Original post by Thuglife

Thats the problem. I know in general for equilibrium constants, its meant to be K=[products]/[reactants] but for partition coefficients I was taught it was K=[upper]/[lower]?

It's K=products/reactants for partition coefficent as well i'm sure. It gives you the correct answer when you do that anyway. I think its normally done so that the product is the upper layer and the reactant is the lower layer, it's maybe to trick you.

(edited 10 years ago)

Reply 789

Original post by deedee123

It's K=products/reactants for partition coefficent as well i'm sure. It gives you the correct answer when you do that anyway. I think its normally done so that the product is the upper layer and the reactant is the lower layer, it's maybe to trick you.

Ah sneaky! thanks, I dont suppose you could help me with this one too? haha

Q27 2009.png

the answer is D but I have no idea how Im meant to know that combination is formed?

Reply 790

Original post by Thuglife

Ah sneaky! thanks, I dont suppose you could help me with this one too? haha

Q27 2009.png

the answer is D but I have no idea how Im meant to know that combination is formed?

I didn't get that either, i got it wrong :/

Reply 791

DragonWalnut157

Original post by Thuglife

Ah sneaky! thanks, I dont suppose you could help me with this one too? haha

Q27 2009.png

the answer is D but I have no idea how Im meant to know that combination is formed?

I think it goes something like this, but I'm not certain:
Chlorine displaces the Iodine to produce Potassium Chloride and Iodine in solution. Whilst that's happening some of the chlorine begins to react with the but-2-ene, producing 2,3-dichlorobutane. Then a few moments later Iodine will also be available to react, forming 2-iodo-3-chlorobutane because a chlorine will have already have been able to attach itself to the carbon.

Reply 792

Does anyone know how to work out the enthalpy if solution, Im stuck on 2011 question 22 (multiple choice).
I keep getting option B, but its A, can anyone help?

Thanks in advance

Reply 793

Bonzo10

Original post by SanahAhmed

Does anyone know how to work out the enthalpy if solution, Im stuck on 2011 question 22 (multiple choice).
I keep getting option B, but its A, can anyone help?

Thanks in advance

Enthalpy of solution is the energy required to dissolve one mole of ionic crystal in water.

First step is breaking the lattice, which is just the opposite sign of the lattice enthalpy for CaCl₂.

So 2223kJ required to turn one mole of ionic lattice into its constituent ions in gaseous form. The next step is to hydrate the ions - essentially the energy released when one mole of gaseous ions is hydrated. Look up your values: Ca = -1650kJ and Cl = -364kJ

What you've probably done is forgotten to multiply the value for the hydration of chlorine by 2.

2223 +(-1650) +2(-364) = .155kJ per mol

Reply 794

Original post by Bonzo10

Enthalpy of solution is the energy required to dissolve one mole of ionic crystal in water.

First step is breaking the lattice, which is just the opposite sign of the lattice enthalpy for CaCl₂.

So 2223kJ required to turn one mole of ionic lattice into its constituent ions in gaseous form. The next step is to hydrate the ions - essentially the energy released when one mole of gaseous ions is hydrated. Look up your values: Ca = -1650kJ and Cl = -364kJ

What you've probably done is forgotten to multiply the value for the hydration of chlorine by 2.

2223 +(-1650) +2(-364) = .155kJ per mol

Thanks a lot! Really helped !

Reply 795

Can someone help me with 2012 q14 (m.c).
My working so far is:

-396+-210+74 = -532

Thanks in advance

Screen Shot 2013-05-27 at 17.21.40.png

(edited 10 years ago)

Reply 796

I am Ace

4

Original post by SanahAhmed

Can someone help me with 2012 q14 (m.c).
My working so far is:

-396+-210+74 = -532

Thanks in advance

Posting a link increases your chances of getting an answer

Reply 797

emma_1995

Can anyone help me with this question. I don't know where to start

And for questions like this:

Do you have to know what colour corresponds to what wavelength or is there a way of working it out?

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Reply 798

Bonzo10

Original post by SanahAhmed

Can someone help me with 2012 q14 (m.c).
My working so far is:

-396+-210+74 = -532

Thanks in advance

Screen Shot 2013-05-27 at 17.21.40.png

This ones a hard one to type up. Basically, you need to look at your target equation and flip the equations to fit. If you flip eqns 2 and 3, which would work chemically, and cancel out as appropriate, you should get -396+210+74 = -112?

Reply 799