Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    Hi,
    Please help me solve these past-paper questions I have come Across.
    I'll post a question and as I get a satisfied answer, I will post the next question.
    There are 7 questions CURRENTLY.
    Here is the First Question:
    Name:  1.jpg
Views: 186
Size:  76.3 KB
    Offline

    7
    ReputationRep:
    (Original post by Kk1999)
    Hi,
    Please help me solve these past-paper questions I have come Across.
    I'll post a question and as I get a satisfied answer, I will post the next question.
    There are 7 questions CURRENTLY.
    Here is the First Question:
    Name:  1.jpg
Views: 186
Size:  76.3 KB
    Both are C
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Dowel)
    Both are C
    How? Can u explain please?
    Offline

    2
    ReputationRep:
    (Original post by Kk1999)
    Hi,
    Please help me solve these past-paper questions I have come Across.
    I'll post a question and as I get a satisfied answer, I will post the next question.
    There are 7 questions CURRENTLY.
    Here is the First Question:
    Name:  1.jpg
Views: 186
Size:  76.3 KB
    This is a very common and basic chemistry question. 1st one its C because as you go down a group the ionic radius increases because the number of shells down the group increases. The 2nd is also C because the greatest jump is btw the last two values thereofer its belong the group 4
    Offline

    7
    ReputationRep:
    (Original post by Kk1999)
    How? Can u explain please?
    So the first one is about ionic radius. They all form a 2+ ion so that doesn't effect the trend in any way, as you go down the group there are more electron shells so the radius is bigger.

    Second one the big gap is between 4-5 that means 4 electrons have been removed easily. The element must be in group 4 because this group forms 4+ ions easily but cannot form 5+ ions easily
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Sunethra)
    This is a very common and basic chemistry question. 1st one its C because as you go down a group the ionic radius increases because the number of shells down the group increases. The 2nd is also C because the greatest jump is btw the last two values thereofer its belong the group 4
    So, Sunethra..u mean the answer should be found by looking for the greatest jump.
    in question 2 the greatest jump is between 4400 (4th value) to 16100 (5th value).
    the greatest jump was from 4 to 5 so the answer is Group 4.
    Right
    Offline

    7
    ReputationRep:
    (Original post by Kk1999)
    So, Sunethra..u mean the answer should be found by looking for the greatest jump.
    in question 2 the greatest jump is between 4400 (4th value) to 16100 (5th value).
    the greatest jump was from 4 to 5 so the answer is Group 4.
    Right
    That's how you do it
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Dowel)
    That's how you do it
    Thank you Dowel and Sunethra.
    My next Question is:
    Name:  2.jpg
Views: 157
Size:  10.1 KB
    Offline

    7
    ReputationRep:
    (Original post by Kk1999)
    Thank you Dowel and Sunethra.
    My next Question is:
    Name:  2.jpg
Views: 157
Size:  10.1 KB
    Answer is B

    You have C5H12 itself
    You have 2-methyl butane
    and 2,2 methyl propane
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Dowel)
    Answer is B

    You have C5H12 itself
    You have 2-methyl butane
    and 2,2 methyl propane
    C5H12 itself should be counted as an isomer?
    and how do you know how many isomers are there??
    any formula?
    AND I HAVE PHY UNIT-1 TOMMOROW SO,
    LL RETURN TO THE THREAD AFTER MY EXAM.
    Offline

    7
    ReputationRep:
    (Original post by Kk1999)
    C5H12 itself should be counted as an isomer?
    and how do you know how many isomers are there??
    any formula?
    AND I HAVE PHY UNIT-1 TOMMOROW SO,
    LL RETURN TO THE THREAD AFTER MY EXAM.
    CHH12 is counted because it is a compound that has 5 carbons and 12 hydrogen's
    you are looking for compounds that have 5 carbon and 12 hydrogen and pentane is one of them.

    To work out how many isomers you need to try all the different possibilities, in this case their must be an isomer with a max carbon chain length of 4 and one with a max carbon length chain of 3 (two and one are impossible) and you need to add the last carbon into a position where it does not create a 5 carbon chain

    So for example on the 4 carbon chain the methyl must go on the second carbon to achieve this and for the 3 carbon chain they must go on the central carbon.

    Essentially your looking for any compounds that have the same number of carbon and hydrogen (same molecular formula) but have a different structural formula
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Dowel)
    CHH12 is counted because it is a compound that has 5 carbons and 12 hydrogen's
    you are looking for compounds that have 5 carbon and 12 hydrogen and pentane is one of them.

    To work out how many isomers you need to try all the different possibilities, in this case their must be an isomer with a max carbon chain length of 4 and one with a max carbon length chain of 3 (two and one are impossible) and you need to add the last carbon into a position where it does not create a 5 carbon chain

    So for example on the 4 carbon chain the methyl must go on the second carbon to achieve this and for the 3 carbon chain they must go on the central carbon.

    Essentially your looking for any compounds that have the same number of carbon and hydrogen (same molecular formula) but have a different structural formula
    Two is not possible because...when we draw the isomers they look as if the chain is simply bent? Right?
    Offline

    7
    ReputationRep:
    (Original post by Kk1999)
    Two is not possible because...when we draw the isomers they look as if the chain is simply bent? Right?
    Two isn't possible because if you try and draw it you find that the max chain length you can actually get is three. Remember that if you have something like this:

    C-C
    C

    If you add any more carbons to that the chain lengths become > 2 and as we have five carbons we cannot not add any more carbons.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Dowel)
    Two isn't possible because if you try and draw it you find that the max chain length you can actually get is three. Remember that if you have something like this:

    C-C
    C

    If you add any more carbons to that the chain lengths become > 2 and as we have five carbons we cannot not add any more carbons.
    Thank You Dowel...
    My 3rd question is:
    Name:  3.jpg
Views: 123
Size:  84.7 KB
    Offline

    14
    ReputationRep:
    (Original post by Kk1999)
    Thank You Dowel...
    My 3rd question is:
    Name:  3.jpg
Views: 123
Size:  84.7 KB
    What exactly do you need help with in this question ?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by PlayerBB)
    What exactly do you need help with in this question ?
    Sorry Sorry that i did not mention the part.
    ITS PART B
    Offline

    14
    ReputationRep:
    (Original post by Kk1999)
    Sorry Sorry that i did not mention the part.
    ITS PART B
    You have to learn to practice to distinguish between these two reactions of alkanes, they're tricky but they'll earn you free marks

    okay alkanes have two reactions which are similar to these two
    • Cracking: the heavy alkane (has a very long chain) is broken to produce a lighter alkane (has a shorter chain ) and an alkene
    • Reforming: The alkane is reformed to a benzene or to a cyclohexane + H2
    It's supposed that is easier now, the first involves the formation of an alkene so it is Cracking and the second involves the formation of a benzene so it is reforming
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?

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.