The Student Room Group

Advanced Higher Chemistry 2012-2013

Scroll to see replies

Reply 180
Original post by I am Ace
Glasgow is pretty small to be fair haha, as is Scotland itself.
Polomint city buddy, full of roundabouts :wink:


Oh that it is!

Tell me about it >.> every time I turn a corner hahaha! The whirlies is a horrible roundabout aswell:frown:!


This was posted from The Student Room's iPhone/iPad App
Reply 181
Original post by Reece:)
Oh that it is!

Tell me about it >.> every time I turn a corner hahaha! The whirlies is a horrible roundabout aswell:frown:!


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


lmao I can't believe you know the Whirlies!
I live literally a 4 minute drive from there
Reply 182
Original post by I am Ace
lmao I can't believe you know the Whirlies!
I live literally a 4 minute drive from there


Hahaha, really? I'm in East Kilbride alot :tongue: whirlies is horrible:frown: hate driving round it hahaha!


This was posted from The Student Room's iPhone/iPad App
Reply 183
pm me your facebook. I can't send you any pm's as you've exceeded a storage limit

Posted from TSR Mobile
Reply 184
Original post by I am Ace
pm me your facebook. I can't send you any pm's as you've exceeded a storage limit

Posted from TSR Mobile


Yeah I just sorted that:wink: silly limit!


This was posted from The Student Room's iPhone/iPad App
Original post by TheFOMaster
Doing Wine Analysis for my investigation, not what I wanted but that hardly matters, it seems quite easy :smile: but dull :frown:


Hey, I am also doing a wine\spirits analysis. What are your methods? I am doing a simple gravimetric weighing and my second is a back titration with potassium dichromate. :smile:


Posted from TSR Mobile
Original post by GremlinFace
Hey, I am also doing a wine\spirits analysis. What are your methods? I am doing a simple gravimetric weighing and my second is a back titration with potassium dichromate. :smile:


Posted from TSR Mobile


Um I can't really check just now, but I decolourise the red wine (stir with activated charcoal), titrate the wines (with phenylalanine as the indicator) with Sodium Hydroxide, Do a thin layer chromatography, then another titration with the wine,Sodium Hydroxide, starch, and sulphuric acid against iodine.
Original post by TheFOMaster
Um I can't really check just now, but I decolourise the red wine (stir with activated charcoal), titrate the wines (with phenylalanine as the indicator) with Sodium Hydroxide, Do a thin layer chromatography, then another titration with the wine,Sodium Hydroxide, starch, and sulphuric acid against iodine.


Sounds thrilling... There is a girl in my class who keeps drinking the whisky I use. Might add some extra pure ethanol to spice things up... :colonhash:
Original post by GremlinFace
Sounds thrilling... There is a girl in my class who keeps drinking the whisky I use. Might add some extra pure ethanol to spice things up... :colonhash:


It's pretty easy though! I don't have time for lots of practical :tongue: (also a note, don't leave you decolourised wine lying in a cupboard for a month. It goes mouldy :tongue:)

She drinks your whisky o.O you should say to her that you've added 85% Phosphoric acid to it :tongue: she'll stay away then!
Reply 189
Hey guys I'm back with TSR :biggrin: I was doing some past papers on chemistry and I've got some questions that I'm not sure about. Can anyone help me with these questions? Answers are the bold ones :rolleyes:

2011 Multiple Choice Questions:

5. The ratio of the ionic radii in sodium chloride is approximately 1:2, whereas in caesium chloride it is approximately 1:1. A compound XY contains X+ ions with a radius of 133pm and Y– ions with a radius of 220 pm. In a crystal of XY, how many Y– ions surround each X+ ion as its nearest neighbour?

A 1
B 2
C 6
D 8


20. In the presence of bright light, hydrogen and chlorine react explosively. One step in the reaction is shown below.

H2(g) + Cl(g) HCl(g) + H(g)
The enthalpy change for this step can be
represented as the bond enthalpy of

A (H—H) + (Cl—Cl)
B (H—H) (Cl—Cl)
C (H—H) + (H—Cl)
D (H—H) (H—Cl).



23. Which of the following reactions would show the greatest decrease in entropy?

A H2(g) + F2(g) 2HF(g)
B KNO3(s) KNO2(s) + ½O2(g)
C CO3 2–(aq) + 2H+(aq) H2O(l) + CO2(g)
D CO3^ 2– (aq) + CO2(g) + H2O(l) 2HCO3^– (aq)


Thanks!:smile:
Original post by SQA
Hey guys I'm back with TSR :biggrin: I was doing some past papers on chemistry and I've got some questions that I'm not sure about. Can anyone help me with these questions? Answers are the bold ones :rolleyes:

2011 Multiple Choice Questions:

5. The ratio of the ionic radii in sodium chloride is approximately 1:2, whereas in caesium chloride it is approximately 1:1. A compound XY contains X+ ions with a radius of 133pm and Y– ions with a radius of 220 pm. In a crystal of XY, how many Y– ions surround each X+ ion as its nearest neighbour?

A 1
B 2
C 6
D 8

Thanks!:smile:


Welcome back!

I can't help with the other 2, but for that one you have to identify the type of ionic lattice(the fact that it's a lattice eliminates A & B) the ion is, which tells you how many negative ions surrounds the positive one. You divide the ppm of the -ve ion by the +ve one, if the answer is closer to 2, it is a sodium chloride type lattice (the +ve ion is small, therefore less -ve ions can surround it.) so the answer is C.
Reply 191
Original post by TheFOMaster
Welcome back!

I can't help with the other 2, but for that one you have to identify the type of ionic lattice(the fact that it's a lattice eliminates A & B) the ion is, which tells you how many negative ions surrounds the positive one. You divide the ppm of the -ve ion by the +ve one, if the answer is closer to 2, it is a sodium chloride type lattice (the +ve ion is small, therefore less -ve ions can surround it.) so the answer is C.


Oh yes got it, thanks a lot! So it takes on the sodium chloride- type structure :smile:
Original post by SQA

20. In the presence of bright light, hydrogen and chlorine react explosively. One step in the reaction is shown below.

H2(g) + Cl(g) HCl(g) + H(g)
The enthalpy change for this step can be
represented as the bond enthalpy of

A (H—H) + (Cl—Cl)
B (H—H) (Cl—Cl)
C (H—H) + (H—Cl)
D (H—H) (H—Cl).



With this one you have to look at the bonds being broken and made: as hydrogen is present as H2, there's two hydrogen atoms which are bonded to each other so that bond has to be broken. Also, HCl is formed meaning hydrogen and chlorine have to bond as well. As bond breaking is endothermic it's +ve and the bond making in HCl is exothermic which is why it's -ve :smile:

Original post by SQA

23. Which of the following reactions would show the greatest decrease in entropy?

A H2(g) + F2(g) 2HF(g)
B KNO3(s) KNO2(s) + ½O2(g)
C CO3 2–(aq) + 2H+(aq) H2O(l) + CO2(g)
D CO3^ 2– (aq) + CO2(g) + H2O(l) 2HCO3^– (aq)



Decreasing entropy means something is becoming more ordered, which could be something changing state or a decrease in the number of molecules for example. In D, there are three reactants and then only two products so it's becoming more ordered and so a decrease in entropy :smile:


I hope I made some sense; explaining isn't one of my strong points!
Reply 193
Original post by rawragee
With this one you have to look at the bonds being broken and made: as hydrogen is present as H2, there's two hydrogen atoms which are bonded to each other so that bond has to be broken. Also, HCl is formed meaning hydrogen and chlorine have to bond as well. As bond breaking is endothermic it's +ve and the bond making in HCl is exothermic which is why it's -ve :smile:



Decreasing entropy means something is becoming more ordered, which could be something changing state or a decrease in the number of molecules for example. In D, there are three reactants and then only two products so it's becoming more ordered and so a decrease in entropy :smile:


I hope I made some sense; explaining isn't one of my strong points!


Ahah okay thanks so much :biggrin:!


Posted from TSR Mobile
Original post by rawragee
Decreasing entropy means something is becoming more ordered, which could be something changing state or a decrease in the number of molecules for example. In D, there are three reactants and then only two products so it's becoming more ordered and so a decrease in entropy :smile:


I hope I made some sense; explaining isn't one of my strong points!


Thanks so much for this. Having serious trouble with this section of the course, I understand it when my teacher goes over it or when I'm reading about it but as soon as I'm faced with a question I have no idea what it is!
Just done the 2002 Past paper without realising I only had the multiple choice answers.. it's a long shot, but does anyone happen to have a full marking scheme??
Reply 196
Original post by TheFOMaster
Just done the 2002 Past paper without realising I only had the multiple choice answers.. it's a long shot, but does anyone happen to have a full marking scheme??


Did you find that paper online? Just that I can't seem to find any papers older than 2003


Posted from TSR Mobile
Original post by ilyaasr
Did you find that paper online? Just that I can't seem to find any papers older than 2003


Posted from TSR Mobile


No, I got it off my school. Want me to pm you it?
Reply 198
I was wandering, how much emphasis do your teachers put on your arrows in mechanisms being very exact? I've been finding some differences between her teaching and the bright red book. Eg, she calls Halogenation, Bromination. During the Halogenation, she failed to mention that the nucleophillic Br- ion can only attack from the side opposite the other Br atom due to its side - she did the mechanism incorrectly according the the bright red book. I am not sure who to believe?! :s-smilie:


Posted from TSR Mobile
Original post by Belaruce
I was wandering, how much emphasis do your teachers put on your arrows in mechanisms being very exact? I've been finding some differences between her teaching and the bright red book. Eg, she calls Halogenation, Bromination. During the Halogenation, she failed to mention that the nucleophillic Br- ion can only attack from the side opposite the other Br atom due to its side - she did the mechanism incorrectly according the the bright red book. I am not sure who to believe?! :s-smilie:


Posted from TSR Mobile


Bromination is Halogenation, although more specific. Your teacher was wrong, believe the book. Br+ attacks the double bond forming a cyclic ion intermediate, then the Br- attacks from the opposite side, as it's more easily accessible.

Quick Reply

Latest

Trending

Trending