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AS redox reaction MCQ

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    Guys, how do we solve the MCQ (attached) using oxdiation numbers?

    I have already found out the oxidation numbers of iodine, it's +1 in HIO, 0 in I2 and +5 in HIO3. But I'm stuck after that. Any help?
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    The redox needs to balance for each element i.e. the total oxidation change on the left = total oxidation change on the right

    m x 1 = (n x 0) + (p x 5) [the 1, 0 and 5 relate to the oxidation states of I on the left and right]

    m = 5p

    so p = 1 and m = 5

    then use these values to get n
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    (Original post by EierVonSatan)
    The redox needs to balance for each element i.e. the total oxidation change on the left = total oxidation change on the right

    m x 1 = (n x 0) + (p x 5) [the 1, 0 and 5 relate to the oxidation states of I on the left and right]

    m = 5p

    so p = 1 and m = 5

    then use these values to get n
    You made it sound so easy, thank you so much!

    I didn't know the mole value affects the value of the oxidation numbers.

    BTW - Would it be okay if I posted a few more MCQs here if I encounter any other problems? I have my Chemistry MCQ paper tomorrow and while my preparation is good, I might have problems with other MCQs.
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    (Original post by leosco1995)
    You made it sound so easy, thank you so much!

    I didn't know the mole value affects the value of the oxidation numbers.
    It doesn't, it affects the balancing Welcome of course

    BTW - Would it be okay if I posted a few more MCQs here if I encounter any other problems? I have my Chemistry MCQ paper tomorrow and while my preparation is good, I might have problems with other MCQs.
    Fell free to post as many questions as you like, someone (if not me) will get around to answering them :yes:
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    OK, here's another one:

    How many different substitution products are possible, in principle, when a mixture of bromine and ethane is allowed to react?

    A) 3
    B) 5
    C) 7
    D) 9

    I know this involves free radical substitution, but I'm not very good at it, I thought a list of products would be this:

    CH3CH2Br
    CH3CHBr2
    CH3CBr3
    CH2BrCBr3
    CHBr2CBr3
    CBr3CBr3

    This is only 6.. but the answer is actually 9, what am I missing?
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    (Original post by leosco1995)
    OK, here's another one:

    How many different substitution products are possible, in principle, when a mixture of bromine and ethane is allowed to react?

    A) 3
    B) 5
    C) 7
    D) 9

    I know this involves free radical substitution, but I'm not very good at it, I thought a list of products would be this:

    CH3CH2Br
    CH3CHBr2
    CH3CBr3
    CH2BrCBr3
    CHBr2CBr3
    CBr3CBr3

    This is only 6.. but the answer is actually 9, what am I missing?
    Theres only one way to get one Br on:

    CH3CH2Br

    There's two ways to get 2 Br's on:

    BrCH2CH2Br and CH3CHBr2

    keep going, see if you can find more :yes:
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    (Original post by leosco1995)
    OK, here's another one:

    How many different substitution products are possible, in principle, when a mixture of bromine and ethane is allowed to react?

    A) 3
    B) 5
    C) 7
    D) 9

    I know this involves free radical substitution, but I'm not very good at it, I thought a list of products would be this:

    CH3CH2Br
    CH3CHBr2
    CH3CBr3
    CH2BrCBr3
    CHBr2CBr3
    CBr3CBr3

    This is only 6.. but the answer is actually 9, what am I missing?
    you have missed a few isomers
    BrCH2CH2Br
    BrCH2CHBr2
    so on.......
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    Oh, right. Of course..

    Another one:

    In a car engine, non-metallic element X forms a pollutant oxide Y.

    Further oxidiation of Y to Z occurs in the atmosphere. In this further oxidation, 1 mol of Y reacts with 0.5 mol of gaseous oxygen.

    What can X be?

    1) Carbon
    2) Nitrogen
    3) Sulfur

    2 and 3 are both right, but why can't 1 also be right? Aren't these equations both possible?

    C + 1/2 O2 -> CO (in a car engine)
    CO + 1/2 O2 -> CO2 (further oxidation)

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