[Freddy-Francis]

(Original post by Tarmu)
Do you have a rough estimate of the UMS grade boundaries?

Goto the link. http://social.ocr.org.uk/files/ocr/ocr_60649_admin_unit_level_raw_m ark_grade_boundaries0611[1].pdf
Click Ctrl + F
Then Type in C7
Press next and look for the one that’s highlighted.
This is last years UMS

[the A* guy]

(Original post by Freddy-Francis)
Goto the link. http://social.ocr.org.uk/files/ocr/ocr_60649_admin_unit_level_raw_m ark_grade_boundaries0611[1].pdf
Click Ctrl + F
Then Type in C7
Press next and look for the one that’s highlighted.
This is last years UMS

I hope the grade boundaries stay this LOW or Go even LOWER!

[Tarmu]

(Original post by Freddy-Francis)
Goto the link. http://social.ocr.org.uk/files/ocr/ocr_60649_admin_unit_level_raw_m ark_grade_boundaries0611[1].pdf
Click Ctrl + F
Then Type in C7
Press next and look for the one that’s highlighted.
This is last years UMS

Wow, thanks for that. Do you know what you would need as an overall UMS mark for an A/B?

[Freddy-Francis]

(Original post by the A* guy)
I hope the grade boundaries stay this LOW or Go even LOWER!

I hope too :'(

[Freddy-Francis]

(Original post by Tarmu)
Wow, thanks for that. Do you know what you would need as an overall UMS mark for an A/B?

http://pdf.ocr.org.uk/download/kd/oc...gcse_spec.pdf?
Press Ctrl + F
Type Grading
Then click on the Grading (which will lead you to the grade boundary page)

[Jamsebob]

Really Really easy compared to previous papers.
Prepare for high grade boundaries!!!

[07oakom]

(Original post by Freddy-Francis)
I thought it was 0.11. Not 0.49...

It was 0.495g I wrote 0.5g as well just in case

[Freddy-Francis]

(Original post by Jamsebob)
Really Really easy compared to previous papers.
Prepare for high grade boundaries!!!

Man. Don’t scare me (((((((((

[JackPayne96]

no it wasn't a trick question the RFM of sodium is 23, oxygen is 16 and hydrogen is 1 so this equals 40 for NaOH

[Coke1]

June 2011 was a straightfoward paper, and the boundary was 31 for the A*.

[Tarmu]

Ram:
Na - 23
O - 16
H - 1

23+16+1= 40

That's how I did it

[Freddy-Francis]

(Original post by Belchi)
What did everyone get for the prove NaOH = 40?
I got 64, it was a trick question wasn't it

That was any trick question. You basically have to show that the RFM of NaOH Is 40. Easy marks. You messed it up. :L
So:- Na =11
O = 16
H = 1
Therefore 23 + 16 + 1 =40
Thats what I put. I am sure that its right

[Tarmu]

(Original post by Freddy-Francis)
That was any trick question. You basically have to show that the RFM of NaOH Is 40. Easy marks. You messed it up. :L
So:- Na =11
O = 16
H = 1
Therefore 11 + 16 + 1 =40
Thats what I put. I am sure that its right

11+16+1=40, ehhhh?

- Do you mean 23 for Na?

[Coke1]

(Original post by Freddy-Francis)
That was any trick question. You basically have to show that the RFM of NaOH Is 40. Easy marks. You messed it up. :L
So:- Na =11
O = 16
H = 1
Therefore 11 + 16 + 1 =40
Thats what I put. I am sure that its right

Correct. Definatley not a trick question.

[Freddy-Francis]

(Original post by Tarmu)
11+16+1=40, ehhhh?

- Do you mean 23 for Na?

yeah. I put 23 in my paper. I just looked up Na from internet and read the wrong one.

[Freddy-Francis]

What did you guys put for the last question? Explain why its made under 37.9g?

[07oakom]

(Original post by Freddy-Francis)
What did you guys put for the last question? Explain why its made under 37.9g?

I wrote some energy is lost as heat and other types of energy etc

[Freddy-Francis]

I put that the glucose gets used up and high concentration of alcohol could kill of the yeast. Will i get any marks??

[ZakariaMar]

(Original post by Freddy-Francis)
What did you guys put for the last question? Explain why its made under 37.9g?

Is this the fermentation one? If it is the reason is because you're assuming all of the glucose will be used up. Some reactants aren't used to their full potential in terms of the amount of product that can be formed.

[ZakariaMar]

Which is why the actual yield is always less than the theoretical yield.