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AS Chemistry- helping each other out!

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Hi I'm doing OCR B salters and have no idea how to learn all the information...
It seems like when I do the past papers the information I need is completely new to me. How do I tackle this, especially for the F332 paper?
Original post by A84
I have it. is it allowed to share it here? need moderators confirmation..


I'm afraid not.
hi guys. just wondering if anyone can help me on question 6 of this worksheet pls. Have no clue how to start it.
[INDENT]hi guys. just wondering if anyone can help me on question 6 of this worksheet pls. Have no clue how to start it. [/INDENT]Attached Thumbnails
Original post by jpatel1234
hi guys. just wondering if anyone can help me on question 6 of this worksheet pls. Have no clue how to start it.


Have you got a)?
For the second part they tell you the enthalpy change per mole, so if you work out the MR of butane then divide it by the enthalpy (in kJ) then you have the mass of butane required to produce 1 kJ of energy through combustion


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(edited 9 years ago)
Original post by langlitz
Have you got a)?
For the second part they tell you the enthalpy change per mole, so if you work out the MR of butane then divide it by the enthalpy (in kJ) then you have the mass of butane required to produce 1 kJ of energy through combustion


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thanks. ye i have got A. Then for the second part, that makes so much sense:smile:. Cheers a lot mate:biggrin::tongue:
Reply 446
Original post by underestimate
Hello guys, I am on OCR A, just finished unit 1 and going on to organic chemistry. :smile:


Hellooo, I'm doing OCR A too. How are you finding organics?
Hi!! Was wondering if anyone could help out with some AS chemistry? My question is, why do halogenoalkanes combust more incompletely than alkanes/alcohols etc, and why do iodoalkanes combust less easily than for example bromoalkanes? The explanation my teacher gave was that because the halogens were heavier atoms, they are a larger proportion of the molecule as a whole, which I get, but not how that relates to the combustibility... If someone could explain that'd be great!! Thanks :smile:

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Reply 448
Original post by haemo
If you have the time, yes.

When you go to AQA's website for the latest papers (2009-2013) you'll see on the left hand side that it says 'older papers'.

When you read the older papers, there are some things that aren't on the spec for Unit 1 anymore (such as reagent tests, solubility, alkali and acids) so you should ignore them - other than that, they're excellent and I've found it a great way to revise after you've written up, or are doing, your notes.

The 2014 paper is harder than most, it's the paper that's had the most maths in for a while - other than that, it's no different and in some sections it's really nice.


just want to ask if you have got the 2014 paper on you... if so can you send it to me please.... much appreciated because i really want to get an A in AS chemistry... ive done notes and now im going through all the practice questions after each section, after that im gonna start doing some past paper....do you think i should do more and is the enough for me to get an A???
thanks in advance for all your help :h:
Original post by ellsie98
Hi!! Was wondering if anyone could help out with some AS chemistry? My question is, why do halogenoalkanes combust more incompletely than alkanes/alcohols etc, and why do iodoalkanes combust less easily than for example bromoalkanes? The explanation my teacher gave was that because the halogens were heavier atoms, they are a larger proportion of the molecule as a whole, which I get, but not how that relates to the combustibility... If someone could explain that'd be great!! Thanks :smile:

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Halogens do not easily make combustion products.

Chlorine, for example, does not form a particularly stable oxide. In the process of burning halogenated hydrocarbons the halide must either become hydrogen halides (reducing the enthalpy of reaction overall) or free halogens, which also reduces the overall enthalpy change.
Original post by ellsie98
Hi!! Was wondering if anyone could help out with some AS chemistry? My question is, why do halogenoalkanes combust more incompletely than alkanes/alcohols etc, and why do iodoalkanes combust less easily than for example bromoalkanes? The explanation my teacher gave was that because the halogens were heavier atoms, they are a larger proportion of the molecule as a whole, which I get, but not how that relates to the combustibility... If someone could explain that'd be great!! Thanks :smile:

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Don't take my word for it, but I think it's because halogenoalkanes have less branching than alkanes?? I'm pretty sure branching is a factor in combustion efficiency, something to do with it having less Van Der Wall's forces, so less bonds to break and less heat required.
The idoalkanes is because iodine has more electrons than bromine so it's more likely to induce a temporary dipole (VDW's). More VDW's, more heat energy required to break the bonds.
Original post by charco
Halogens do not easily make combustion products.

Chlorine, for example, does not form a particularly stable oxide. In the process of burning halogenated hydrocarbons the halide must either become hydrogen halides (reducing the enthalpy of reaction overall) or free halogens, which also reduces the overall enthalpy change.


Okay thanks, that makes sense! But surely reducing the enthalpy change would make it easier for the reaction to take place, as it requires less energy?

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Original post by Dani California
Don't take my word for it, but I think it's because halogenoalkanes have less branching than alkanes?? I'm pretty sure branching is a factor in combustion efficiency, something to do with it having less Van Der Wall's forces, so less bonds to break and less heat required.
The idoalkanes is because iodine has more electrons than bromine so it's more likely to induce a temporary dipole (VDW's). More VDW's, more heat energy required to break the bonds.


Yeah I'm not sure about the branching, doesn't the halogen just replace one of the hydrogens? I'll have to look into the VDWs thing, thanks :smile:

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Original post by ellsie98
Okay thanks, that makes sense! But surely reducing the enthalpy change would make it easier for the reaction to take place, as it requires less energy?

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No, combustion reactions always release energy, the more the better.
Original post by charco
No, combustion reactions always release energy, the more the better.


Okay thanks for clarifying :smile:

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Anyone doing OCR B salters chemistry??? :redface:
hi please can someone help me with this question
9 1.12 g of iron reacts with oxygen to form 1.60 g of an oxide of iron.
Use relative atomic masses: Fe = 56, O = 16.
What is the formula of this oxide of iron?
A FeO5
B Fe2O10
C Fe3O2
D Fe2O3
i originally got a but it is wrong and i don't know why
thanks
Original post by haemo
C

An s orbital holds two electrons (spins: one up, one down)
A p orbital holds six electrons (spins: three up, three down)

If you don't understand what I mean by up and down spins, research 'spin diagrams'

p orbitals fill up differently, they have to have three up spins and then they begin to fill up with down spins.

A: 1s2 contains two electrons (hence the 2) and 2s2 holds two electrons (hence the 2) (two up spins, two down spins)
Unpaired electrons: 0

B: Exactly the same as above, however the p orbital has 3/6 electrons (all up spins)
Unpaired electrons: 3

C: The p orbital has 4/6 electrons (three up spins, one down spin)
Unpaired electrons: 2

D: The p orbital has 5/6 electrons (three up spins, two down spins)
Unpaired electrons: 1


thanks i get it now
Original post by Elhamm
Anyone doing OCR B salters chemistry??? :redface:


:yep:

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Original post by Arger
The first step is to work out the moles:
1.12/56=0.02 - Iron
1.60/16=0.1 - Oxygen

Then divide by the smallest amount to get a ratio:

0.02/0.02=1 - Iron
0.10/0.02=5 - Oxygen

The ratio is 1:5

Therefore we get the formula of Fe05 so A


the mark scheme says d though?

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