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And activation energy is too high

Oh I forgot to put that bit, but its sort of assumed that the rate of reaction is too slow due to not as many molecules having successful collisions because the activation energy is too high for the molecules to exceed

Reply 3261

georgiaaaxo

pleaseeeee can someone tell me where the 40 comes from in 7b June 2010 working out the conc, I get everything else!! would be really appreciated

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Reply 3262

eggfriedrice

Original post by otrivine

a quick question

for the second half equation what did u put involving hydrogen and water and oxygen/

I didn't make a half equation?

Reply 3263

needtosucceed=)

Original post by otrivine

so is that how they got 4

yep, you have to divide by 4 to get a quarter of the rate for o2

Reply 3264

A-New-Start

Who was i talking about 3I2 and I2 crap, last night?

Reply 3265

A-New-Start

Original post by otrivine

so is that how they got 4

Ok i have a question about the enthalpy change of neutralisation in the book. Say the equation was HNO3 + KOH --> KNO3 + 2H20 ( i know that's not balanced, but still ) so then we first scale up to one mole by dividing our answer by 0.05 and then we divide by two, right? so that the the 2mol of water formed can become one.

Reply 3266

beccac94

Original post by georgiaaaxo

pleaseeeee can someone tell me where the 40 comes from in 7b June 2010 working out the conc, I get everything else!! would be really appreciated

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What 40??? On the mark scheme I have you times it by 34 which is the mr of H2O2

Reply 3267

A-New-Start

Original post by beccac94

What 40??? On the mark scheme I have you times it by 34 which is the mr of H2O2

you also have to time by 40 to go from 25cm3 to 1dm3

Reply 3268

_HabibaH_

Original post by Brad0440

You would need to put an actual salt of hexylresorcinal to get the mark, so for example, one of the OH groups could become O-Na+ (or something like that). I think that's what the question is looking for.

Okay thanks, that's fine then :smile:

Reply 3269

beccac94

Original post by A-New-Start

you also have to time by 40 to go from 25cm3 to 1dm3

I think the way the mark scheme has done it a different way :smile:

Reply 3270

eggfriedrice

Original post by A-New-Start

Who was i talking about 3I2 and I2 crap, last night?

Me xD

Reply 3271

otrivine

Original post by needtosucceed=)

yep, you have to divide by 4 to get a quarter of the rate for o2

can u show me so sorry

Reply 3272

otrivine

Original post by A-New-Start

No its the specimen paper :smile:

Reply 3273

eggfriedrice

Original post by A-New-Start

Who was i talking about 3I2 and I2 crap, last night?

Me xD

Original post by needtosucceed=)

its a 0.5: 2 ratio, so youve got the rate of hcoh which is 2 moles, so the rate of oxygen would be a quarter of it

The thing is, aren't we assuming the initial rate is for the 2 mols of HCHO?
I mean we could just as easily say the initial rate for O2 was 1x10-12 and hence the initial rate of HCHO would be four times that?

Reply 3274

needtosucceed=)

Original post by eggfriedrice

Me xD

The thing is, aren't we assuming the initial rate is for the 2 mols of HCHO?
I mean we could just as easily say the initial rate for O2 was 1x10-12 and hence the initial rate of HCHO would be four times that?

yep we are, so 2 moles of hcoh = rate of 1 x 10-12
so 1 mole of o2 would be half of that = 5 x 10^-13

so o.5 moles would be half of that again which = 2.5 x 10^-13

or you could just divide by 4 to do it all in one step