GCSE OCR 21st Century B7 C7 P7 - W/C - Monday 20th June 2016 Watch

Protostar
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#181
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#181
(Original post by realsmilee)
Just think that you will never have to do chemistry again after tomorrow!!
Very true - thank you
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realsmilee
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#182
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#182
(Original post by pinksplodge)
idk I prefer chemistry in general to physics and bio so hoping it goes well...still not too sure about some parts though how did you find the c4-6 paper?
I actually found it okay surprisingly, but i messed up on the lead ion question and probably multiple others
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realsmilee
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#183
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#183
(Original post by NiamhM1801)
Haber process:
Ammonia (2NH₃) is made by reacting nitrogen (N₂) and hydrogen (3H₂) together at a pressure of 200 atmospheres, a temperature of 450°C and using an iron catalyst. The process by which this is done industrially is called the Haber process. The reaction is reversible, meaning that not all of the nitrogen and hydrogen will be converted into ammonia as some ammonia will also break down. However none of the reactants are wasted, as those that are not used are recycled and passed through again.

Le Chatelier's principle:
Le Chatelier's principle states that in dynamic equilibrium, the position of equilibrium shifts to try to cancel out any changes you make. For example: A + B ⇌ C + D. Increasing the concentration of A means more C and D are produced to counteract the change. If the reaction is exothermic in the forward direction and endothermic in the backwards direction, heating the mixture means the equilibrium moves to the left to counteract the change. If A + B ⇌ C, compressing (increasing the pressure of) the mixture means the equilibrium moves to the right to counteract the change. If there are more molecules in one side of the equilibrium than the other, the equilibrium will shift to the side with fewer molecules to counteract the imbalance.
You make everything sound so simple
You will be fine tomorrow!
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Protostar
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#184
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#184
(Original post by realsmilee)
You make everything sound so simple
You will be fine tomorrow!
You're welcome - and thank you, I sure hope so!
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rebbarn29
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#185
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#185
Can someone please explain gas chromatography to me, I don't get it and have a funny feeling it is going to come as a 6 marker
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sharmacka
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#186
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#186
(Original post by rebbarn29)
Can someone please explain gas chromatography to me, I don't get it and have a funny feeling it is going to come as a 6 marker
I doubt it coming up. Chromatography came up as 14marks altogether last year I believe
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MR.ANONYMOUS 786
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#187
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#187
(Original post by realsmilee)
You make everything sound so simple
You will be fine tomorrow!
Tomorrow ? My P7 OCR 21st Century exam is on Friday ?


⚔🛡⚔~Nothing happens to anyone that he is not fitted by nature to bear - Maximus Decimus Meridius~⚔🛡⚔
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Ananya._c
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#188
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#188
If someone could explain how to do this standard solution question it would be really helpful
The answer is 25 cm cubed of stock solution and make up to 20 cm cubed of water.
Thanks
edit: nevermind, understood it
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Protostar
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#189
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#189
(Original post by MR.ANONYMOUS 786)
Tomorrow ? My P7 OCR 21st Century exam is on Friday ?


⚔🛡⚔~Nothing happens to anyone that he is not fitted by nature to bear - Maximus Decimus Meridius~⚔🛡⚔
C7 is tomorrow
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katxxxx
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#190
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#190
Any predictions for C7 six markers?
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KomradeKorbyn
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#191
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#191
(Original post by Ananya._c)
If someone could explain how to do this standard solution question it would be really helpful
The answer is 25 cm cubed of stock solution and make up to 20 cm cubed of water.
Thanks
edit: nevermind, understood it
Posted from TSR Mobile
I'm really confused by this, would you be able to explain it and/or post the mark scheme? What year's paper is it from?
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Ananya._c
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#192
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#192
(Original post by KomradeKorbyn)
I'm really confused by this, would you be able to explain it and/or post the mark scheme? What year's paper is it from?
It's from an old specification and the mark scheme literally says the answer I mentioned above :/
I used the formula:
final concentration / initial concentration x the volume = amount to add
therefore:
6.3/63 x 250 = 25
so add 25 cm 3 of nitric acid, to make up to 250 cm3 add more (225cm3) water to get the desired volume
essentially you make the solution 10 x more dilute
That's what I understood anyway, sorry if this is really bad explanation
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Protostar
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#193
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#193
(Original post by katxxxx)
Any predictions for C7 six markers?
I'm not sure about topics, but I have a feeling they're going to be pretty mean given the kindness of C1-6 and the absolute beauty that was last year's paper
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Alexandra00
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#194
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#194
Any C7 predictions?
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KomradeKorbyn
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#195
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#195
(Original post by Ananya._c)
It's from an old specification and the mark scheme literally says the answer I mentioned above :/
I used the formula:
final concentration / initial concentration x the volume = amount to add
therefore:
6.3/63 x 250 = 25
so add 25 cm 3 of nitric acid, to make up to 250 cm3 add more (225cm3) water to get the desired volume
essentially you make the solution 10 x more dilute
That's what I understood anyway, sorry if this is really bad explanation
Thanks! I looked at your explanation and managed to figure out a way to do it that made sense to me as to why it works, here:
  • The formula for concentration is concentration = mass / volume
  • Rearranged is concentration * volume = mass
  • You want 250 cm3 volume at a concentration of 6.3 g/dm3
  • Convert 250 cm3 to dm3, 250 / 1000 = 0.25
  • So, add the numbers into the formula, 6.3 * 0.25 = 1.575
  • So you want your end solution to have 1.575g mass of nitric acid in a volume of 250 cm3, in order to give you the 6.3 g/dm3 concentration they asked for
  • Then, you find how much of the original solution you need to get 1.575g
  • mass / concentration = volume
  • 1.575 / 63 = 0.025 dm3 = 25 cm3
  • So you have 25 cm3 of your original solution, and it contains 1.575g, now you just need to dilute it until you get the correct volume and the concentration will then be correct
  • So, you add 225 cm3 of water, giving you 250 cm3 solution (the volume the question asked for) - and it contains 1.575g
  • You can then check it is at the correct concentration by doing mass / volume = concentration
  • 1.575 / 0.25 (250cm3 / 1000 to convert to dm3) = 6.3 g/dm3 concentration, which is what they asked for, so it works!
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isabelaskey
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#196
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#196
Can someone please explain how to work out a titration calculation with an example?
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Ananya._c
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#197
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#197
(Original post by KomradeKorbyn)
Thanks! I looked at your explanation and managed to figure out a way to do it that made sense to me as to why it works, here:
  • The formula for concentration is concentration = mass / volume
  • Rearranged is concentration * volume = mass
  • You want 250 cm3 volume at a concentration of 6.3 g/dm3
  • Convert 250 cm3 to dm3, 250 / 1000 = 0.25
  • So, add the numbers into the formula, 6.3 * 0.25 = 1.575
  • So you want your end solution to have 1.575g mass of nitric acid in a volume of 250 cm3, in order to give you the 6.3 g/dm3 concentration they asked for
  • Then, you find how much of the original solution you need to get 1.575g
  • mass / concentration = volume
  • 1.575 / 63 = 0.025 dm3 = 25 cm3
  • So you have 25 cm3 of your original solution, and it contains 1.575g, now you just need to dilute it until you get the correct volume and the concentration will then be correct
  • So, you add 225 cm3 of water, giving you 250 cm3 solution (the volume the question asked for) - and it contains 1.575g
  • You can then check it is at the correct concentration by doing mass / volume = concentration
  • 1.575 / 0.25 (250cm3 / 1000 to convert to dm3) = 6.3 g/dm3 concentration, which is what they asked for, so it works!
That works nicely too! Good luck for tomorrow :yy:
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KomradeKorbyn
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#198
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#198
(Original post by Ananya._c)
That works nicely too! Good luck for tomorrow :yy:
You too, hopefully this year's grade boundaries are low!
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Protostar
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#199
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#199
What a beautiful paper, I'm shocked
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MezmorisedPotato
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#200
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#200
Wonderful paper, apart from a few stupid mistakes (in a six marker but only a few marks), everything else was shockingly pleasant!
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