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# Edexcel - Chemistry Unit 2 - 4 June 2013 watch

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1. what apparatus has 2sources of error and which don't ?
2. (Original post by Masturchess)
I think you do; (0.04 x 2)/20, then times by 100 to get it in % which is 0.4%
What are you doing? The correct procedure is to find the percentage uncertainty in each measurement and then add them up since your combining the measurements. Which is 0.04/10+0.04/10=0.8%.

Clearly you were thinking along the same lines are our examiners. I don't know if they have any clue what they're doing though; I was told I couldn't use the dot multiplier for multiplication and had to use a cross just because they "might not know enough maths", which I think is frankly ridiculous.
3. (Original post by Khurc003)
Anyone know how to calculate this?? The ans is D but not sure how...?

The enthalpy change of neutralization of an acid by an alkali is measured by adding
10.0 cm3 of hydrochloric acid to 10.0 cm3 of sodium hydroxide. 10.0 cm3 pipettes with
an accuracy of ±0.04 cm3 are used to measure out both solutions.
The overall percentage error in measuring the total volume of the reaction mixture is

A ±0.04%
B ±0.08%
C ±0.4%
D ±4.0%
Uncertainty is error/total reading *100 for a % and so (0.04/20)*100*2(*2 as you use the pipette twice)=0.4%
In general it's not so easy to tell whether an electrophile was involved first or a nucleophile, you have to be able to see the mechanism in your head when you look at the reactants/products or reactants/catalysts. But for AS level I think the only time it is electrophilic is when it is addition onto a C=C double bond. And if it is addition then it will be electrophilic, if it is substitution of any atom other than H it will be nucleophilic (common nucleophiles are -OH, X-, -C triple bond N, -NH2, etc.), if it is substitution of H it will be free radical substitution.
Thank you so much , but what about if it's an elemenation reaction is it nucleophilic or electrophilic.
What are you doing? The correct procedure is to find the percentage uncertainty in each measurement and then add them up since your combining the measurements. Which is 0.04/10+0.04/10=0.8%.

Clearly you were thinking along the same lines are our examiners. I don't know if they have any clue what they're doing though; I was told I couldn't use the dot multiplier for multiplication and had to use a cross just because they "might not know enough maths", which I think is frankly ridiculous.

That's what I did, I think. I doubled the % error since you had to use the equipment twice; the 20 I got because you had to add 10cm to 10cm in a mixture, which would mean the total volume of the whole mixture was 20cm^3.
6. how do i approach this question
20 cm3 of sulfuric acid, concentration 0.25 mol dm–3, was neutralized in a titration with barium hydroxide, concentration 0.50 mol dm–3. The equation for the reaction is
Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2H2O(l)
(a) The volume of barium hydroxide required was
I seem to remember from doing this past paper that the answer is actually C? Though I can't explain why. I got 0.8% and then decided that 0.4% is closest so wrote that one down. (0.04/10.0)*2=0.8%.
Lool yes you're absolutly right.. ans is C.. I misread it... I guess this is what revision is doing to me...

D you know why its C though coz i got B by: 0.04/10 then x2 (which is the same as yours)
8. (Original post by davcha)
Uncertainty is error/total reading *100 for a % and so (0.04/20)*100*2(*2 as you use the pipette twice)=0.4%
Yes, but your still taking two separate measurements !!!!!!! You have to work out the uncertainty of each measurement

thanks
10. (Original post by bhowland1994)
Yes, but your still taking two separate measurements !!!!!!! You have to work out the uncertainty of each measurement
I am that's why i multiply by 2, it's the same as just adding it twice
11. (Original post by newyork newyork)
how do i approach this question
20 cm3 of sulfuric acid, concentration 0.25 mol dm–3, was neutralized in a titration with barium hydroxide, concentration 0.50 mol dm–3. The equation for the reaction is
Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2H2O(l)
(a) The volume of barium hydroxide required was
Work out the number of moles of H2SO4, then use that to work out how many moles of barium hydroxide reacted. Then use moles = conc/volume to get the volume of barium hydroxide.
12. (Original post by Tuya)
It's a tad late to learn about those now, coz half the paper might be based on those. They are also called London forces.
The theory is that in a molecule, electrons are likely to be on one side of the molecule. So that side is slightly negatively charged and the other end is slightly positively charged. This is an instantaneous dipole. Because of these, the electrons in other molecules of the substance are repelled or attracted ie an induced dipole. Overall, these are attractive forces which is why non polar molecules don't just scatter away.

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okay I get that so between polar molecules theres permanent dipole forcesand between non-polar molecules theres instantaneous-induced dipole forces right? and the permanent and instantaneous-induced forces are all VDWS correct?
13. (Original post by Priya08)
okay I get that so between polar molecules theres permanent dipole forcesand between non-polar molecules theres instantaneous-induced dipole forces right? and the permanent and instantaneous-induced forces are all VDWS correct?
Van der vaals or london forces are not permanent but temporary dipoles that form between electrons clouds.
14. who thinks green chemistry is gonna come up for section c???
15. I'm gonna start the jan 13 paper now, hopefully I can ace it.
16. (Original post by charlieejobson)
who thinks green chemistry is gonna come up for section c???
Dosen't it always come up ?
17. (Original post by davcha)
I am that's why i multiply by 2, it's the same as just adding it twice
Ok this question has me completely confused, first we are using a pipette not a burette so at what point are we taking two readings ..... second you cant work out the uncertainty for 20cm^3 as at no point do you measure a liquid of that volume, your measuring two 10cm^3 .....
18. (Original post by marseille_h)
Dosen't it always come up ?
not always but mostly...
19. (Original post by marseille_h)
Van der vaals or london forces are not permanent but temporary dipoles that form between electrons clouds.
thank you..okay I understand it now..one last question so is there such thing as permanent instantaneous dipole?
20. (Original post by bhowland1994)
Ok this question has me completely confused, first we are using a pipette not a burette so at what point are we taking two readings ..... second you cant work out the uncertainty for 20cm^3 as at no point do you measure a liquid of that volume, your measuring two 10cm^3 .....
Think about the worst that can happen (least accurate measurements) you'd use 10.04cm^3 of HCl and the same of NaOH. So you would have a total of 20.08cm^3 instead of the 2*10 that you wanted. You can work out the % difference ((20.08-20)/20)*100=0.4. It's a different way to look at the question and check your answers in an exam, hope it helps!

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