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# Ocr f321 23 may 2013~official discussion thread watch

1. (Original post by Shanze S)
Yeh that makes sense , thanks
Not a problem.. Good luck tomorrow!!
2. (Original post by yodawg321)
I get 45+ in all past papers. Watch me get like 30 tomorrow lol
i know man, i just hope it's do-able. Best of luck!
3. Can someone help me on q5aiii on -

http://www.ocr.org.uk/Images/57906-q...and-groups.pdf

http://www.ocr.org.uk/Images/61527-m...me-january.pdf

It will really help me out. Thanks.
4. (Original post by SchoolKid.)
Can someone help me on q5aiii on -

http://www.ocr.org.uk/Images/57906-q...and-groups.pdf

http://www.ocr.org.uk/Images/61527-m...me-january.pdf

It will really help me out. Thanks.
i am guessing you mean biii? you have no of mol. of Ba you see the molar ratio in equation is 1:1 so no of mol of Ba(OH)2 is also 8.0*10^-4 then you write the eq. -> mol = conc. * vol(cm3) / 1000

here conc = mol/vol(cm3)/1000 = 8.0*10^-4/0.1 = 8.0*10^-3 mol dm^-3
5. (Original post by SchoolKid.)
Can someone help me on q5aiii on -

http://www.ocr.org.uk/Images/57906-q...and-groups.pdf

http://www.ocr.org.uk/Images/61527-m...me-january.pdf

It will really help me out. Thanks.
You know the mol of Ba(OH)2 as it's the same as the Barium. Also you know the volume which is 100 cm^3.
Therefore you do n/v=c which is 0.0008mol/(100/1000) as you want volume in dm^3 which =0.008moldm^-3 as a concentration.

Hope that helped
6. Just got 51/60 in a past paper. Only doing this as I'm in for it now, as I need 90-95 UMS in F325 for an A just got 51/60 on the jan 13 paper. Hoping for loads of calculation questions
7. If it's anything like January's paper I will be super super happy

I loveeeeee this unit! As long as there's no super hard calculations, I think.. hope it'll be fine (touch wood)
8. (Original post by A-New-Start)
i am guessing you mean biii? you have no of mol. of Ba you see the molar ratio in equation is 1:1 so no of mol of Ba(OH)2 is also 8.0*10^-4 then you write the eq. -> mol = conc. * vol(cm3) / 1000

here conc = mol/vol(cm3)/1000 = 8.0*10^-4/0.1 = 8.0*10^-3 mol dm^-3
Can you please explain how you got the volume to be 0.1

I use this formula for volume = Moles/24
9. (Original post by A-New-Start)
i am guessing you mean biii? you have no of mol. of Ba you see the molar ratio in equation is 1:1 so no of mol of Ba(OH)2 is also 8.0*10^-4 then you write the eq. -> mol = conc. * vol(cm3) / 1000

here conc = mol/vol(cm3)/1000 = 8.0*10^-4/0.1 = 8.0*10^-3 mol dm^-3
Sorry, I didn't see you submitting your reply until after I did mine. Nice to see we got the same answer though
10. (Original post by SchoolKid.)
Can you please explain how you got the volume to be 0.1

I use this formula for volume = Moles/24
If they ask it in cm^3 it's you times by 24,000 mate

you do moles times 24,000 to get volume in cm^3
11. (Original post by SchoolKid.)
Can you please explain how you got the volume to be 0.1

I use this formula for volume = Moles/24
That's volume for gases as opposed to solutions which is what you want here. We already know that the volume is 100 cm^3 but we want in dm^3 for the concentration so 100 cm^3 = 100/1000 dm^3 = 0.1 dm^3
12. (Original post by yodawg321)
If they ask it in cm^3 it's you times by 24,000 mate
13. (Original post by yodawg321)
Why do you guys start past papers a day before? :O
idk. Had an exam on monday so was revising. I mean "started" revising on tuesday..

My 2nd past paper : C
Now on my 3rd one

I learned "Practise makes Perfect"
14. I made flashcards, which you can do activities with, that cover pretty much everything for F321: http://quizlet.com/23659907/f321-flash-cards/

Let's hope OCR go easy on us tomorrow!
15. (Original post by SchoolKid.)
You can using 2 equations depending if its solutions or gases
16. (Original post by niniesta96)
That's volume for gases as opposed to solutions which is what you want here. We already know that the volume is 100 cm^3 but we want in dm^3 for the concentration so 100 cm^3 = 100/1000 dm^3 = 0.1 dm^3

(Original post by yodawg321)
If they ask it in cm^3 it's you times by 24,000 mate

you do moles times 24,000 to get volume in cm^3

(Original post by A-New-Start)
i am guessing you mean biii? you have no of mol. of Ba you see the molar ratio in equation is 1:1 so no of mol of Ba(OH)2 is also 8.0*10^-4 then you write the eq. -> mol = conc. * vol(cm3) / 1000

here conc = mol/vol(cm3)/1000 = 8.0*10^-4/0.1 = 8.0*10^-3 mol dm^-3
Thanks guys, I get it now. I just kept muddling up the formulas and didn't realise we had to use the 100cm^3 they gave us at the beginning of the question
17. Could someone please explain why there are 7 full orbitals in an atom of Sulfur and not 8? As there are 16 electrons and each orbital can hold up to two?

Is it because each of the 3p orbitals fill up separately with one in each?

This is a question from Jan 2012 paper.

Cheers
18. (Original post by kimsiclez)
Could someone please explain why there are 7 full orbitals in an atom of Sulfur and not 8? As there are 16 electrons and each orbital can hold up to two?

Is it because each of the 3p orbitals fill up separately with one in each?

This is a question from Jan 2012 paper.

Cheers
Sulfur has an atomic number of 16, thus meaning it has 16 electrons as electrons are equal to protons. Completing the electronic configuration would give you 1s2 - 2s2 - 2p6 - 3s2 - 3p4.

As you can see from the first four subshells are full. S holds 1 orbital which holds two electrons thus meaning 1s2, 2s2 and 3s2 have 3 full oribitals collectively. A p subsell contains 3 orbitals thus meaning 2p6 contains 3 orbitals. So far that is 6 orbitals.

Now the slightly tricky part, most people would assume 3p4 has two full orbitals, but you must remember that electrons if possible will fill out subsequent orbitals if possible, so it will look something like this:

[] [ ] [ ]. As you can see there is only one full orbital in the 3p4 subshell not two. Making a total collection of 7.
19. Ive only just found it on a 2011 paper but its not in my revision guide, do we have to know the reaction of halogens with cyclohexane? if so can anyone tell me what they are
20. (Original post by kimsiclez)
Could someone please explain why there are 7 full orbitals in an atom of Sulfur and not 8? As there are 16 electrons and each orbital can hold up to two?

Is it because each of the 3p orbitals fill up separately with one in each?

This is a question from Jan 2012 paper.

Cheers
Yeah that's pretty much it. Each electron has a "spin", up or down. In the p orbitals, they tend to fill up with one in each sub-orbital first. So P would have 6 full orbitals as there are no full orbitals in the 3p orbital, but with one more electrn in S, it will fill up the 3p x sub orbital, so you get an extra full orbital. Hope this clears it up

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