AS Chemistry- helping each other out!
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Original post by Dinaa
Hah they died.
Wut you don't remember the famous dina photoshop of you?
I COULD ALWAYS POST AND SHOW YOU?
Wut you don't remember the famous dina photoshop of you?
I COULD ALWAYS POST AND SHOW YOU?
Just like me in 3 weeks, and oh I might remember lol but that don't even look like me now,
Pls don't post it I will cri and be sad
Original post by Gladiatorsword
Ok. Say you're given a reaction with an negative enthalpy change - its exothermic. But how do you know if the forward reaction is exothermic or the backwards? ??
Posted from TSR Mobile
Posted from TSR Mobile
It will always be the forwards reaction when given in an exam question, or otherwise it will specify. Don't forget that the ΔH in endothermic reactions is POSITIVE and the ΔH in exothermic reactions is NEGATIVE.
Original post by ozzie2
Just like me in 3 weeks, and oh I might remember lol but that don't even look like me now,
Pls don't post it I will cri and be sad
Pls don't post it I will cri and be sad
What does ozzie look like now then?
Original post by ozzie2
Incoming 300 private messages from people asking for answers. But I'm sure you did fine and if you didn't do as well as you hoped, it is only worth 20% as far as I know and if you do well in the exams you should still be able to get a good grade
lol yep i got loads. yeah i heard somewhere you can get a very low mark and still get an A
Chemistry aqa empa help/ tips anyone?
Original post by Dinaa
What does ozzie look like now then?
I am beautiful swan now
Original post by thatgr
lol yep i got loads. yeah i heard somewhere you can get a very low mark and still get an A
Yeah it is pretty possible, the big one is unit 2 paper which is worth almost 50% so make sure you do well in that and you should be ok
Original post by ozzie2
I am beautiful swan now
Yeah it is pretty possible, the big one is unit 2 paper which is worth almost 50% so make sure you do well in that and you should be ok
Yeah it is pretty possible, the big one is unit 2 paper which is worth almost 50% so make sure you do well in that and you should be ok
i lold
pics or gtfo
Original post by frozo123
Is it awkward having a surname as Boner?
No but it is annoying when I encounter immature people like you
Original post by Dinaa
i lold
pics or gtfo
pics or gtfo
I still gat my insecurities kid soz
Original post by frozo123
go sort your boner out
The banter is unreal, I am so jealous
Original post by ozzie2
I still gat my insecurities kid soz
The banter is unreal, I am so jealous
The banter is unreal, I am so jealous
I gave you some rep so you don't feel too insecure.
I've had a fair number of reports complaining about people trying to cheat on their ISA/EMPA exams on here. Trying to cheat on your exams is strictly against the rules on TSR and attempting to do so can lead to a ban from the site. More importantly, AQA do seem to be monitoring the site and have said that they are well within their rights to take action against any student who is involved in malpractice like this. I would therefore urge all of you to be careful in what you post. Disclosing information using Private Messages (PMs) is not allowed either.
If you do see anyone breaking the rules (either on here or if they send you a PM), you can report them to the moderation team using the report function and one of us will try to answer it ASAP.
And while I'm here - many people also seem to be giving away their emails to request help with their ISA/EMPA tasks. Please be very careful with giving away your email address to strangers on here unless you're sure that they are who they say they are. We have had some instances of TSR users getting abusive or threatening emails from people because of revealing their email addresses (either publicly or by PM) so again, please be careful.
Thanks
If you do see anyone breaking the rules (either on here or if they send you a PM), you can report them to the moderation team using the report function and one of us will try to answer it ASAP.
And while I'm here - many people also seem to be giving away their emails to request help with their ISA/EMPA tasks. Please be very careful with giving away your email address to strangers on here unless you're sure that they are who they say they are. We have had some instances of TSR users getting abusive or threatening emails from people because of revealing their email addresses (either publicly or by PM) so again, please be careful.
Thanks
Can someone explain the whole primary and secondary alcohols thing to me. Quite a fair bit but it is appreciated.
Is it primary to aldehyde --> Carboxylic acid
Secondary to ketone
Tertiary do not oxidise - ( Can someone explain why)?
Reagent is sulphuric acid and acidified potassium dichromate is the oxidising agent.
Need some help with the whole fehlings and tollen's shiz. Sound my homies.
Is it primary to aldehyde --> Carboxylic acid
Secondary to ketone
Tertiary do not oxidise - ( Can someone explain why)?
Reagent is sulphuric acid and acidified potassium dichromate is the oxidising agent.
Need some help with the whole fehlings and tollen's shiz. Sound my homies.
Original post by Super199
Can someone explain the whole primary and secondary alcohols thing to me. Quite a fair bit but it is appreciated.
Is it primary to aldehyde --> Carboxylic acid
Secondary to ketone
Tertiary do not oxidise - ( Can someone explain why)?
Reagent is sulphuric acid and acidified potassium dichromate is the oxidising agent.
Need some help with the whole fehlings and tollen's shiz. Sound my homies.
Is it primary to aldehyde --> Carboxylic acid
Secondary to ketone
Tertiary do not oxidise - ( Can someone explain why)?
Reagent is sulphuric acid and acidified potassium dichromate is the oxidising agent.
Need some help with the whole fehlings and tollen's shiz. Sound my homies.
For Tollen's reagent, the solution forms a silver mirror only if an aldehyde is present. If no silver mirror is formed then it is a ketone or something else. Similar with Fehling's, if an aldehyde is present it goes brick red (i think) and no reaction with a ketone
Yes primary alcohols oxidise to an aldehyde then a carboxylic acid and secondary to a ketone and tertiary do not, I don't think you need to know why tertiary don't oxidise
(edited 8 years ago)
Original post by Super199
Can someone explain the whole primary and secondary alcohols thing to me. Quite a fair bit but it is appreciated.
Is it primary to aldehyde --> Carboxylic acid
Secondary to ketone
Tertiary do not oxidise - ( Can someone explain why)?
Reagent is sulphuric acid and acidified potassium dichromate is the oxidising agent.
Need some help with the whole fehlings and tollen's shiz. Sound my homies.
Is it primary to aldehyde --> Carboxylic acid
Secondary to ketone
Tertiary do not oxidise - ( Can someone explain why)?
Reagent is sulphuric acid and acidified potassium dichromate is the oxidising agent.
Need some help with the whole fehlings and tollen's shiz. Sound my homies.
Tertiary does not oxidise because the carbon with OH does not have an H
In primary and secondary the carbon with OH have H on it which is removed to allow the formation of double bond with O.
Reagent is dilute sulphuric acid and potassium/sodium dichromate (VI). You have to put oxidation state. You don't have to say acidified because by mentioning sulfuric acid it's already acidic.
Fehling and
Tollen's detect aldehyde
From quick googling. But I haven't learnt them lol
Posted from TSR Mobile
(edited 8 years ago)
Original post by C0balt
Tertiary does not oxidise because the carbon with OH does not have an H
In primary and secondary the carbon with OH have H on it which is removed to allow the formation of double bond with O.
Reagent is dilute sulphuric acid and potassium/sodium dichromate (VI). You have to put oxidation state. You don't have to say acidified because by mentioning sulfuric acid it's already acidic.
Fehling and
Tollen's detect aldehyde
From quick googling. But I haven't learnt them lol
Posted from TSR Mobile
In primary and secondary the carbon with OH have H on it which is removed to allow the formation of double bond with O.
Reagent is dilute sulphuric acid and potassium/sodium dichromate (VI). You have to put oxidation state. You don't have to say acidified because by mentioning sulfuric acid it's already acidic.
Fehling and
Tollen's detect aldehyde
From quick googling. But I haven't learnt them lol
Posted from TSR Mobile
I don't get the tertiary alcohol bit. The carbon with OH does not have an H?
What is the stuff about the chromium stuff orange to green or the other way round can't remember? Something about reduction???
Sound
Original post by Super199
I don't get the tertiary alcohol bit. The carbon with OH does not have an H?
What is the stuff about the chromium stuff orange to green or the other way round can't remember? Something about reduction???
Sound
What is the stuff about the chromium stuff orange to green or the other way round can't remember? Something about reduction???
Sound
The carbon which has hydroxyl group attached to it has three carbon atoms so there is no H atom on its own attached to the said carbon. Draw a tertiary alcohol and you will see. But I don't think the explanation is on any syllabus.
The chromium (VI) in dichromate has orange colour and chromium (III) is green. It is reduced in oxidation of alcohol this is where the colour change comes from.
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Guys I think I'm having a stupid moment but doing this empa past paper from june 2010 and can't seem to get the answer right for this question! The MS says the answer's 17.6, could anyone explain to me how they got that?
"A saturated solution of ammonia contains 300 g of ammonia in 1.00dm3 of solution.
Calculate the concentration, in mol dm–3, of ammonia in this solution."
"A saturated solution of ammonia contains 300 g of ammonia in 1.00dm3 of solution.
Calculate the concentration, in mol dm–3, of ammonia in this solution."
Posted from TSR Mobile
Pretty straightforward.
First find the moles of NH3 using N = Mass/Mr ---------- 300g/17 gmol-1 = 17.647 mol
Then use the equation C = N/V so 17.647 mol/1 dm3 volume (no convertion of units required = 17.6 mol dm-3 to 3 sig figs.
Make sure you understand that I got the Mr from adding the Ar (or mass number) of Nitrogen which is 14 on the OCR table. With 3 x 1.0 (1.0 is the mass number of hydrogen, since it's Ammonia NH3. We multiply by 3). 14 + 3 = 17 g mol-1
Remember the equations N = Mass/Mr for moles. N = C x V for solutions and N = V/Vm for gas volumes.
Practice loads and you'll get better
Original post by sophiegashman
Guys I think I'm having a stupid moment but doing this empa past paper from june 2010 and can't seem to get the answer right for this question! The MS says the answer's 17.6, could anyone explain to me how they got that?
"A saturated solution of ammonia contains 300 g of ammonia in 1.00dm3 of solution.
Calculate the concentration, in mol dm–3, of ammonia in this solution."
"A saturated solution of ammonia contains 300 g of ammonia in 1.00dm3 of solution.
Calculate the concentration, in mol dm–3, of ammonia in this solution."
Pretty straightforward.
First find the moles of NH3 using N = Mass/Mr ---------- 300g/17 gmol-1 = 17.647 mol
Then use the equation C = N/V so 17.647 mol/1 dm3 volume (no convertion of units required = 17.6 mol dm-3 to 3 sig figs.
Make sure you understand that I got the Mr from adding the Ar (or mass number) of Nitrogen which is 14 on the OCR table. With 3 x 1.0 (1.0 is the mass number of hydrogen, since it's Ammonia NH3. We multiply by 3). 14 + 3 = 17 g mol-1
Remember the equations N = Mass/Mr for moles. N = C x V for solutions and N = V/Vm for gas volumes.
Practice loads and you'll get better
(edited 8 years ago)
Welp incoming mod to delete these posts, you guys know talking about the EMPA is against the rules so stop plserino thankerino.
(edited 8 years ago)
Original post by ozzie2
Welp incoming mod to delete these posts, you guys know talking about the EMPA is against the rules so stop plserino thankerino.
Actually, it is only against the rules if you talk about EMPA/ISA specifics. That is, the questions that are on the paper. Talking about TOPICS is not against the rules unless explicitly stated that they were on the paper AND that significant help was administered to warrant a correct answer on that question.
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