A-level Chemistry Revision Squad!
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Original post by jyyl
For pentanoic acid, write like this will do 
CH3CH2CH2CH2COOH
CH3CH2CH2CH2COOHRight and if you oxidise then immediately distil butan-1-ol would the product be butanal or butan-1-al?
Original post by kiiten
Right and if you oxidise then immediately distil butan-1-ol would the product be butanal or butan-1-al?
Butanal and butan-1-al are the same thing but when naming aldehydes you don't need write the number to show which carbon the functional group is on because the aldehyde group is always on carbon one, so you can just write Butanal
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Original post by LThomas694
Butanal and butan-1-al are the same thing but when naming aldehydes you don't need write the number to show which carbon the functional group is on because the aldehyde group is always on carbon one, so you can just write Butanal
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Oh yeah, of course its alcohols that can be on different carbons not aldehydes. Thanks
Original post by FemaleBo55
Lol I really like your mind maps, what other ones have you made for chemistry?
All that I have! (so far anyway)
There are some more things I need to add one though, like in kinetics when a question says 'equimolar' it's a hint to say pH=pKa .
And obviously the mechanisms one isnt finished becuase AS mechanisms can also come up.
so yh, quite a bit left to do!
do you have anything that i could use??
Original post by P____P
All that I have! (so far anyway)
There are some more things I need to add one though, like in kinetics when a question says 'equimolar' it's a hint to say pH=pKa .
And obviously the mechanisms one isnt finished becuase AS mechanisms can also come up.
so yh, quite a bit left to do!
do you have anything that i could use??
There are some more things I need to add one though, like in kinetics when a question says 'equimolar' it's a hint to say pH=pKa .
And obviously the mechanisms one isnt finished becuase AS mechanisms can also come up.
so yh, quite a bit left to do!
do you have anything that i could use??
They look a bit like AS notes? Btw TSR resources are free. Have you checked them out?
Please could someone explain why the answer is 1.5 and not 3?
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Posted from TSR Mobile
Original post by kiiten
In the question it says 50%
Hi could someone please help me with this question...
A 1.575g sample of ethanedioic acid crystals, H2C2O4.nH2O, was dissolved in water and made up to 250cm3. One mole of the acid reacts with two moles of NaOH. In a titration, 25.0cm3 of this solution of acid reacted with exactly 15.6cm3 of 0.160 mol dm-3 NaOH. Calculate the value of n.
Any help would be appreciated
A 1.575g sample of ethanedioic acid crystals, H2C2O4.nH2O, was dissolved in water and made up to 250cm3. One mole of the acid reacts with two moles of NaOH. In a titration, 25.0cm3 of this solution of acid reacted with exactly 15.6cm3 of 0.160 mol dm-3 NaOH. Calculate the value of n.
Any help would be appreciated
Also could someone please help with this question...
3.88g of a monoprotic acid was dissolved in water and the solution was made up to 250cm3. 25.0cm3 of this solution was titrated with 0.095 mol dm-3 NaOH solution, requiring 46.5 cm3. Calculate the relative molecular mass of the acid.
The answer is 87.8 but I got 878...?
3.88g of a monoprotic acid was dissolved in water and the solution was made up to 250cm3. 25.0cm3 of this solution was titrated with 0.095 mol dm-3 NaOH solution, requiring 46.5 cm3. Calculate the relative molecular mass of the acid.
The answer is 87.8 but I got 878...?
Original post by P____P
All that I have! (so far anyway)
There are some more things I need to add one though, like in kinetics when a question says 'equimolar' it's a hint to say pH=pKa .
And obviously the mechanisms one isnt finished becuase AS mechanisms can also come up.
so yh, quite a bit left to do!
do you have anything that i could use??
There are some more things I need to add one though, like in kinetics when a question says 'equimolar' it's a hint to say pH=pKa .
And obviously the mechanisms one isnt finished becuase AS mechanisms can also come up.
so yh, quite a bit left to do!
do you have anything that i could use??
do you have any AS revision notes, similar to the mind maps??
If so , could I possibly use them?, they seem amazing sources of revision
Original post by Samistrawberry
They look a bit like AS notes? Btw TSR resources are free. Have you checked them out?
The spec. changed this year, and I found out they do some A2 kinetics stuff in AS like Kc, Kw , and some other stuff. But these mindmaps are for A2.
Original post by Science_help
do you have any AS revision notes, similar to the mind maps??
If so , could I possibly use them?, they seem amazing sources of revision
If so , could I possibly use them?, they seem amazing sources of revision
Sorry, I don't have any of my notes from AS.
I threw them all away. I didn't make any mindmaps last year.
Spoiler
Original post by kiiten
1- (black ink in the image) you start with 3.0 mol and a 1.0 mol of ethanoic acid and the alcohol.you end with 0.9 mol of water
2- (red ink in the image) you must have started with 0 mol on the right for all (because nothing has been formed)
3-(purple ink in the image) looking at the ratios of the products, you see that water and the ester are of the ratio 1:1, so if 0.9 mol of water has formed at equilibrium , then 0.9 mol of the ester must have formed too.
4- (mol of ester/mol of ethanoic acid) x 100= your answer
I find it a lot easier if you draw up a table. It's a lot clearer to see what you're dealing with.
Spoiler
(edited 8 years ago)
Original post by ilikechemistry
Hi could someone please help me with this question...
A 1.575g sample of ethanedioic acid crystals, H2C2O4.nH2O, was dissolved in water and made up to 250cm3. One mole of the acid reacts with two moles of NaOH. In a titration, 25.0cm3 of this solution of acid reacted with exactly 15.6cm3 of 0.160 mol dm-3 NaOH. Calculate the value of n.
Any help would be appreciated
A 1.575g sample of ethanedioic acid crystals, H2C2O4.nH2O, was dissolved in water and made up to 250cm3. One mole of the acid reacts with two moles of NaOH. In a titration, 25.0cm3 of this solution of acid reacted with exactly 15.6cm3 of 0.160 mol dm-3 NaOH. Calculate the value of n.
Any help would be appreciated
Hi- I've tried to give it a shot, forgive me if there are any mistakes...
Okay so first I calculated the moles of NaOH: which is conc x (vol/1000) to get me 2.496 x 10-3 moles
We know its a 1:2 ratio, for every one mole of acid, two moles of NaOH is reacted, so we divide the moles of NaOH we have calculated by 2, to get 1.248 x 10-3 moles of the acid (in 25cm3)
Now we have the moles of the acid in 25cm3 we need to calculate moles in 250cm3 so we multiply the moles by 10 to get 0.01248 moles of the acid
Now we have the moles of the acid in 250cm3. We can now divide the mass of the acid crystals by the moles to give us the Mr which is 1.575/0.01248= 126
Now we have the Mr of H2C2O4.nH2O we then first find out the Mr of the first part: H2C2O4= which is 90. 126 - 90 is 36 so H20 is 36. The Mr of H20 is 18, so we divide 36/18 to get us n which is 2.
So n is 2. I think.
Original post by ilikechemistry
Also could someone please help with this question...
3.88g of a monoprotic acid was dissolved in water and the solution was made up to 250cm3. 25.0cm3 of this solution was titrated with 0.095 mol dm-3 NaOH solution, requiring 46.5 cm3. Calculate the relative molecular mass of the acid.
The answer is 87.8 but I got 878...?
3.88g of a monoprotic acid was dissolved in water and the solution was made up to 250cm3. 25.0cm3 of this solution was titrated with 0.095 mol dm-3 NaOH solution, requiring 46.5 cm3. Calculate the relative molecular mass of the acid.
The answer is 87.8 but I got 878...?
For this question, we first find out the moles of NaOH which is: (46.5/1000) x 0.095= to give us 4.4175 x 10^-3. REMEMBER THIS IS IN 25 CM3. You HAVE to multiply the moles by 10 (which i think you forgot to do) to get 0.044175 moles of acid.
I assumed it's a 1:1 ratio so I then, divided the mass of the acid by the moles, so 3.88/0.044175= 87.8
Just remember to not forget the volumes of the solutions- we may have to convert the moles in 25cm3 to 250cm3 which is why we had to multiply by 10
So I've been stuck on this AS past paper chem question for too long and I would be very, very grateful if anyone could please help me with it, thanks. Here it is:
Anhydrous calcium chloride, CaCl2, can be used to dry some organic liquids. During this process, hydrated calcium chloride, CaCl2.2H20 is formed.
CaCl2 + 2H20 --> CaCl2.2H20
Mr 111
In a drying process, 5.55g of anhydrous calcium chloride, CaCl2, is used. Calculate the amount of water that can be removed from the organic liquid. (2 marks)
Thank you!!
Anhydrous calcium chloride, CaCl2, can be used to dry some organic liquids. During this process, hydrated calcium chloride, CaCl2.2H20 is formed.
CaCl2 + 2H20 --> CaCl2.2H20
Mr 111
In a drying process, 5.55g of anhydrous calcium chloride, CaCl2, is used. Calculate the amount of water that can be removed from the organic liquid. (2 marks)
Thank you!!
Original post by Gwenog_quidditch
So I've been stuck on this AS past paper chem question for too long and I would be very, very grateful if anyone could please help me with it, thanks. Here it is:
Anhydrous calcium chloride, CaCl2, can be used to dry some organic liquids. During this process, hydrated calcium chloride, CaCl2.2H20 is formed.
CaCl2 + 2H20 --> CaCl2.2H20
Mr 111
In a drying process, 5.55g of anhydrous calcium chloride, CaCl2, is used. Calculate the amount of water that can be removed from the organic liquid. (2 marks)
Thank you!!
Anhydrous calcium chloride, CaCl2, can be used to dry some organic liquids. During this process, hydrated calcium chloride, CaCl2.2H20 is formed.
CaCl2 + 2H20 --> CaCl2.2H20
Mr 111
In a drying process, 5.55g of anhydrous calcium chloride, CaCl2, is used. Calculate the amount of water that can be removed from the organic liquid. (2 marks)
Thank you!!
What do you mean by the amount of water like the volume or the number of moles of water that can be removed ? If the number of moles, then it is 0.1 moles, because 5.55g of anhydrous CaCl2 is used, it Mr is 111, the number of moles of CaCl2 is 0.05, the moles ratio of the CaCl2 to water is 1:2 so the number of moles of water used 0.1 which is the number of moles that will be removed from the organic liquid
Original post by haj101
Hi- I've tried to give it a shot, forgive me if there are any mistakes...
Okay so first I calculated the moles of NaOH: which is conc x (vol/1000) to get me 2.496 x 10-3 moles
We know its a 1:2 ratio, for every one mole of acid, two moles of NaOH is reacted, so we divide the moles of NaOH we have calculated by 2, to get 1.248 x 10-3 moles of the acid (in 25cm3)
Now we have the moles of the acid in 25cm3 we need to calculate moles in 250cm3 so we multiply the moles by 10 to get 0.01248 moles of the acid
Now we have the moles of the acid in 250cm3. We can now divide the mass of the acid crystals by the moles to give us the Mr which is 1.575/0.01248= 126
Now we have the Mr of H2C2O4.nH2O we then first find out the Mr of the first part: H2C2O4= which is 90. 126 - 90 is 36 so H20 is 36. The Mr of H20 is 18, so we divide 36/18 to get us n which is 2.
So n is 2. I think.
Okay so first I calculated the moles of NaOH: which is conc x (vol/1000) to get me 2.496 x 10-3 moles
We know its a 1:2 ratio, for every one mole of acid, two moles of NaOH is reacted, so we divide the moles of NaOH we have calculated by 2, to get 1.248 x 10-3 moles of the acid (in 25cm3)
Now we have the moles of the acid in 25cm3 we need to calculate moles in 250cm3 so we multiply the moles by 10 to get 0.01248 moles of the acid
Now we have the moles of the acid in 250cm3. We can now divide the mass of the acid crystals by the moles to give us the Mr which is 1.575/0.01248= 126
Now we have the Mr of H2C2O4.nH2O we then first find out the Mr of the first part: H2C2O4= which is 90. 126 - 90 is 36 so H20 is 36. The Mr of H20 is 18, so we divide 36/18 to get us n which is 2.
So n is 2. I think.
I got the same answer
Original post by PlayerBB
I got the same answer
Yay!! Thank god!
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