A-level Chemistry Revision Squad!
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Original post by alow
Sack up.
What does that mean
Original post by NimbleNeil
What does that mean
http://www.urbandictionary.com/define.php?term=sack%20up
Why is the electrode postive at the site of reduction in an electrochemical cell?
I'm not sure id be able to answer that well if they asked that in an exam :/
I'm not sure id be able to answer that well if they asked that in an exam :/
Original post by wwwwwwwwwwwf
Why is the electrode postive at the site of reduction in an electrochemical cell?
I'm not sure id be able to answer that well if they asked that in an exam :/
I'm not sure id be able to answer that well if they asked that in an exam :/
Reduction is gain... so I'm guessing all the negative species are attracted by the positive electrode to become oxidised?!
Original post by K3001N
Reduction is gain... so I'm guessing all the negative species are attracted by the positive electrode to become oxidised?!
This might make it a bit more clear, reduction is happening at the Cu electrode (Cu2+ > Cu)
Cu electrode is postive, so the electrons are flowing from the Fe to the Cu electrode, but I just don't really understand why the Cu electrode is the postive electrode
Original post by wwwwwwwwwwwf
This might make it a bit more clear, reduction is happening at the Cu electrode (Cu2+ > Cu)
Cu electrode is postive, so the electrons are flowing from the Fe to the Cu electrode, but I just don't really understand why the Cu electrode is the postive electrode
This might make it a bit more clear, reduction is happening at the Cu electrode (Cu2+ > Cu)
Cu electrode is postive, so the electrons are flowing from the Fe to the Cu electrode, but I just don't really understand why the Cu electrode is the postive electrode
Opposites attract.
Original post by wwwwwwwwwwwf
This might make it a bit more clear, reduction is happening at the Cu electrode (Cu2+ > Cu)
Cu electrode is positive, so the electrons are flowing from the Fe to the Cu electrode, but I just don't really understand why the Cu electrode is the positive electrode
This might make it a bit more clear, reduction is happening at the Cu electrode (Cu2+ > Cu)
Cu electrode is positive, so the electrons are flowing from the Fe to the Cu electrode, but I just don't really understand why the Cu electrode is the positive electrode
Ahhhh, it's to do with EMF! Cu has a more positive EMF, this means it gains a positive electrode potential and its reduced. Where as Fe has a negative EMF, this means that it gains a negative electrode potential. So it donates electrons via the external circuit to the Cu.
basically the Fe reduces the Cu and the Fe gets oxidised.
More negative the emf the more it donates electrons (becomes oxidised)
the more positive the emf the more it accepts electrons (becomes reduced)
I think this should help!
This is so nice!
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Hey guys could you help me interpret proton NMR spectra?
Hey guys could you help me interpret proton NMR spectra?
Hey can anyone explain 23c to me?
Its from the june 2014 ial Chem 5 paper
Thanks x
Original post by the-anonymous-me
Hey can anyone explain 23c to me?
Its from the june 2014 ial Chem 5 paper
Thanks x
Hey can anyone explain 23c to me?
Its from the june 2014 ial Chem 5 paper
Thanks x
CxHy (10) + (x + y/4)O2 (v) -> (y/2)H20 (0) + xC02 (40)
The bold are the volumes.
Use ratios:
C02:CxHy = x = 40/10 = 4
To find the volume of Oxygen used, use this:
Change in volume = volume of gas produced - volume of gas used
Then again use ratios to find the value of (x + y/4) and then solve for y.
Original post by demotivated
CxHy (10) + (x + y/4)O2 (v) -> (y/2)H20 (0) + xC02 (40)
The bold are the volumes.
Use ratios:
C02:CxHy = x = 40/10 = 4
To find the volume of Oxygen used, use this:
Change in volume = volume of gas produced - volume of gas used
Then again use ratios to find the value of (x + y/4) and then solve for y.
The bold are the volumes.
Use ratios:
C02:CxHy = x = 40/10 = 4
To find the volume of Oxygen used, use this:
Change in volume = volume of gas produced - volume of gas used
Then again use ratios to find the value of (x + y/4) and then solve for y.
Thanks xx - I think I get it now
Original post by the-anonymous-me
Thanks xx - I think I get it now
No problem, though i'm having trouble with a similar question 18 on June 2014 R
"15 cm^3 of a gaseous hydrocarbon requires 90cm^3 of oxygen for complete combustion, both volumes being measured at 15C and 1 atm. The formula of the hydrocarbon is:"
a) C4H6
b) C4H8 (this is the answer)
c) C4H10
d) impossible to determine
anyone have any idea?
Original post by demotivated
No problem, though i'm having trouble with a similar question 18 on June 2014 R
"15 cm^3 of a gaseous hydrocarbon requires 90cm^3 of oxygen for complete combustion, both volumes being measured at 15C and 1 atm. The formula of the hydrocarbon is:"
a) C4H6
b) C4H8 (this is the answer)
c) C4H10
d) impossible to determine
anyone have any idea?
"15 cm^3 of a gaseous hydrocarbon requires 90cm^3 of oxygen for complete combustion, both volumes being measured at 15C and 1 atm. The formula of the hydrocarbon is:"
a) C4H6
b) C4H8 (this is the answer)
c) C4H10
d) impossible to determine
anyone have any idea?
There may be a quicker way but this is how I did it:
The hydrocarbon: oxygen ratio is 15:90 which can be simplified to 1:6
CxHy + 6 O2 ----> x CO2 + 0.5y H2O
x + 0.25y = 6 This is by looking at the coefficients of O2 on the right
The only thing that satisfied this from the options given was B as
4 + 0.25(8) = 4 + 2 = 6
(edited 7 years ago)
Original post by the-anonymous-me
There may be a quicker way but this is how I did it:
The hydrocarbon: oxygen ratio is 15:90 which can be simplified to 1:6
CxHy + 6 O2 ----> x CO2 + 0.5y H2O
x + 0.25y = 6 This is by looking at the coefficients of O2 on the right
The only thing that satisfied this from the options given was B as
4 + 0.25(8) = 4 + 2 = 6
The hydrocarbon: oxygen ratio is 15:90 which can be simplified to 1:6
CxHy + 6 O2 ----> x CO2 + 0.5y H2O
x + 0.25y = 6 This is by looking at the coefficients of O2 on the right
The only thing that satisfied this from the options given was B as
4 + 0.25(8) = 4 + 2 = 6
Okay, thank you!
How do you do this question?
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Original post by Glavien
Cr(OH)3 + 3OH- ---> Cr(OH)6
Cr(OH)3 + 3H+ ----> Cr(H2O)6
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I know this is AS but really stuck any help/jints?
Direct titration
A student has read on a packet of aspirin that each tablet contains 250 mg of aspirin, and has decided to find out if this is true.
The systematic name of aspirin is 2-ethanoylhydroxybenzoic acid. Aspirin is a carboxylic acid and neutralises a cold dilute solution of sodium hydroxide to form a salt. This reaction is used to form the basis of a titration.
CH3COOC6H4COOH + NaOH → CH3COOC6H4COONa + H2O
aspirin
1. Calculate the relative molecular mass of aspirin (2-ethanoylhydroxybenzoic acid). CALCUALTED 180 (1 mark)
The student decides to use this reaction in a titration to determine the mass of aspirin in each tablet.
She crushes four tablets, transfers them to a 100 cm3 volumetric flask, dissolving them in a 50:50 mixture of ethanol and distilled water (it is difficult to dissolve aspirin in pure water). Using a pipette she transfers 25 cm3 of this solution into a conical flask and titrates it against a 0.1 mol dm−3 standard solution of sodium hydroxide, using phenolphthalein as an indicator. The average titre is 14.70 cm3.
1. Calculate the mass of aspirin in each tablet. (4 marks)
Not sure where to start
Direct titration
A student has read on a packet of aspirin that each tablet contains 250 mg of aspirin, and has decided to find out if this is true.
The systematic name of aspirin is 2-ethanoylhydroxybenzoic acid. Aspirin is a carboxylic acid and neutralises a cold dilute solution of sodium hydroxide to form a salt. This reaction is used to form the basis of a titration.
CH3COOC6H4COOH + NaOH → CH3COOC6H4COONa + H2O
aspirin
1. Calculate the relative molecular mass of aspirin (2-ethanoylhydroxybenzoic acid). CALCUALTED 180 (1 mark)
The student decides to use this reaction in a titration to determine the mass of aspirin in each tablet.
She crushes four tablets, transfers them to a 100 cm3 volumetric flask, dissolving them in a 50:50 mixture of ethanol and distilled water (it is difficult to dissolve aspirin in pure water). Using a pipette she transfers 25 cm3 of this solution into a conical flask and titrates it against a 0.1 mol dm−3 standard solution of sodium hydroxide, using phenolphthalein as an indicator. The average titre is 14.70 cm3.
1. Calculate the mass of aspirin in each tablet. (4 marks)
Not sure where to start
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