Hey there! Sign in to join this conversationNew here? Join for free

Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread Watch

    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by Tomasio)
    koh+c2h5cooh>>>>> c2h5cook + h2o
    and how would the carboxylic salt interact with any h+ ions
    Offline

    2
    ReputationRep:
    Has anyone finished unit 5 revision?
    Offline

    2
    ReputationRep:
    (Original post by samb1234)
    and how would the carboxylic salt interact with any h+ ions
    reforms the oh and propanoic acid, so the [OH-]> [H+]
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by Tomasio)
    reforms the oh and propanoic acid, so the [OH-]> [H+]
    So if it is reacting with h+ ions it is therefore reducing tje conc of h+ ajd hence the ph is greater than 7
    Offline

    6
    ReputationRep:
    Hi guys - was just doing a paper and I have a question about a multiple choice question that I can't quite get my head around.

    I tried to work out number of volume of carbon by 12/44 x 150 and then subtract this from 50cm3 to work out the volume of hydrogen?? Not really sure how to work out the question to be honest. Would really appreciate a clear solution! Thanks very much

    Offline

    3
    ReputationRep:
    (Original post by Funky_Giraffe)
    Hi guys - was just doing a paper and I have a question about a multiple choice question that I can't quite get my head around.

    I tried to work out number of volume of carbon by 12/44 x 150 and then subtract this from 50cm3 to work out the volume of hydrogen?? Not really sure how to work out the question to be honest. Would really appreciate a clear solution! Thanks very much

    Is the answer B?
    The ratio of the hydrocarbon to co2 is 1:3

    Posted from TSR Mobile
    Offline

    13
    ReputationRep:
    (Original post by Neurology)
    Is the answer B?
    The ratio of the hydrocarbon to co2 is 1:3

    Posted from TSR Mobile
    Yeah I got B as well for the same reason, as well as the fact that there are 4 lots of H2O left over.
    Offline

    3
    ReputationRep:
    Any idea on how to do this?

    Posted from TSR Mobile
    Attached Images
     
    Offline

    14
    ReputationRep:
    (Original post by Neurology)
    Any idea on how to do this?

    Posted from TSR Mobile
    is the answer D ?
    Offline

    6
    ReputationRep:
    (Original post by Neurology)
    Is the answer B?
    The ratio of the hydrocarbon to co2 is 1:3

    Posted from TSR Mobile
    Thank you, yes the answer is B. Not sure how you get it from knowing the ratio of hydrocarbon to CO2 produced though. Could you explain?
    Offline

    12
    ReputationRep:
    Hi Everyone! How are you all doing?
    How's prep coming along?
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by Funky_Giraffe)
    Thank you, yes the answer is B. Not sure how you get it from knowing the ratio of hydrocarbon to CO2 produced though. Could you explain?
    Well let's look at the equation. You should hopefully know that a hydrocarbon burns fully in oxygen to make co2 and water. So the equation would be:

    AX + BO2 ----->CH2O +DCO2, where X is the unknown hydrocarbon and A, B, C and D are the numbers needed to balance the equation. We know that the volume of the gas is proportional to the number of moles of that gas, so if we let the coefficient of X be 1 then we know there are 5o2 and 3CO2 (as their volumes are 5 and 3 times that of the hydrocarbon respectively). So:

    X + 5O2 ----> CH2O +3CO2.

    We can now balance the oxygens, there are 10 oxygens on the LHS and 6 on the RHS, so there must be 4 waters:

    X+5O2 ----> 4H2O +3CO2.

    We can then just balance the equation to work out X, there are 3 Carbons on the RHS and 8 hydrogens and therefore x must be c3h8
    Offline

    6
    ReputationRep:
    (Original post by samb1234)
    Well let's look at the equation. You should hopefully know that a hydrocarbon burns fully in oxygen to make co2 and water. So the equation would be:

    AX + BO2 ----->CH2O +DCO2, where X is the unknown hydrocarbon and A, B, C and D are the numbers needed to balance the equation. We know that the volume of the gas is proportional to the number of moles of that gas, so if we let the coefficient of X be 1 then we know there are 5o2 and 3CO2 (as their volumes are 5 and 3 times that of the hydrocarbon respectively). So:

    X + 5O2 ----> CH2O +3CO2.

    We can now balance the oxygens, there are 10 oxygens on the LHS and 6 on the RHS, so there must be 4 waters:

    X+5O2 ----> 4H2O +3CO2.

    We can then just balance the equation to work out X, there are 3 Carbons on the RHS and 8 hydrogens and therefore x must be c3h8
    Got it. Many thanks!
    Offline

    3
    ReputationRep:
    Have people started doing papers in exam conditions? How's everyone finding them? What kind of scores is everyone getting?
    Offline

    3
    ReputationRep:
    (Original post by PlayerBB)
    is the answer D ?
    The Answer is A

    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by Funky_Giraffe)
    Thank you, yes the answer is B. Not sure how you get it from knowing the ratio of hydrocarbon to CO2 produced though. Could you explain?
    Sorry I couldn't reply earlier. Was busy the entire day. Anyway I hope you understood from the reply provided by samb

    Posted from TSR Mobile
    Offline

    14
    ReputationRep:
    (Original post by Neurology)
    The Answer is A

    Posted from TSR Mobile
    That's so confusing!
    Offline

    3
    ReputationRep:
    (Original post by PlayerBB)
    That's so confusing!
    Try imagining/re-drawing the functional groups in the same rotation as the one given. They have been rotated to make it confusing.
    Offline

    14
    ReputationRep:
    (Original post by dpoojaraa)
    Try imagining/re-drawing the functional groups in the same rotation as the one given. They have been rotated to make it confusing.
    Thank you, it definitely made the question easier

    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by PlayerBB)
    Thank you, it definitely made the question easier

    Posted from TSR Mobile
    No problem! Have you started doing full papers in timed conditions? I seem so intimidated whenever I try.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Has a teacher ever helped you cheat?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.