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    (Original post by mr.holmes)
    thanks
    here is an example if you can solve plz
    This is how I did it - I tried to make it clear I hope it helps.

    Let me know if you need further help

    EDIT: I realised I made a mistake - the number of electrons for the reduction reaction should be 10, not 6. But I hope you get the gist on how to solve those kind of question.
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    Posted from TSR Mobile

    U can simplify this and do all these steps in one
    Step 1- wrote the equation and balance all elements except hydrogen and oxygen
    SO2-->SO4^2- +H+
    step 2 - balance oxide with h20
    SO2 +2H2O ---> SO4^2- + H+
    Step 3- balance tge hydrogen with H+ ions
    SO2 +2H2O ----> SO4^2- + 4H+
    step 4 balance the charges
    SO2 + 2H2O ----> SO4^2- + 4H+ + 2e-
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    (Original post by mr.holmes)
    thanks so much
    for step number 1, you mean the addition of electrons for balancing the chargers and the numbers for the elements right?
    Yes
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    (Original post by Sumaiyya1999)
    Posted from TSR Mobile

    U can simplify this and do all these steps in one
    Step 1- wrote the equation and balance all elements except hydrogen and oxygen
    SO2-->SO4^2- +H+
    step 2 - balance oxide with h20
    SO2 +2H2O ---> SO4^2- + H+
    Step 3- balance tge hydrogen with H+ ions
    SO2 +2H2O ----> SO4^2- + 4H+
    step 4 balance the charges
    SO2 + 2H2O ----> SO4^2- + 4H+ + 2e-
    thanks so much it is just a matter of balancing
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    (Original post by rubyvalkyrie)
    This is how I did it - I tried to make it clear I hope it helps.

    Let me know if you need further help
    thank you it is clear and i finally understood it

    but do they ask about forming half equations including a metal in an acidic solutions such as this one
    MnO4¯(aq) + 8H+ (aq) + 5e ................ Mn2+(aq) + 4H2O(l)
    my teacher said memorize it but i am not sure.....
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    (Original post by mr.holmes)
    thank you it is clear and i finally understood it

    but do they ask about forming half equations including a metal in an acidic solutions such as this one
    MnO4¯(aq) + 8H+ (aq) + 5e ................ Mn2+(aq) + 4H2O(l)
    my teacher said memorize it but i am not sure.....
    I'm not quite sure but I guess better to learn it and be prepared rather than ignoring it and hope that it doesn't pop up in the exam.

    When the question mentions acidic conditions it normally means that there will be H+ on the reactant side of the equation.
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    (Original post by rubyvalkyrie)
    I'm not quite sure but I guess better to learn it and be prepared rather than ignoring it and hope that it doesn't pop up in the exam.

    When the question mentions acidic conditions it normally means that there will be H+ on the reactant side of the equation, some (if not all) of which are used to produce water.
    i wish it does not come in the exam

    thanks for helping
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    Came across this question in June 2014

    Give the equation for the reaction between chlorine and sodium hydroxide solution that forms sodium chlorate (V) as one o the products. State symbols are not required.

    How is the answer: 3Cl2 + 6NaOH = 5NaCl + NaClO3 + 3H2O ?
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    (Original post by Rahatara Sadique)
    Came across this question in June 2014

    Give the equation for the reaction between chlorine and sodium hydroxide solution that forms sodium chlorate (V) as one of the products. State symbols are not required.

    How is the answer: 3Cl2 + 6NaOH = 5NaCl + NaClO3 + 3H2O ?
    OMG I've just done that question!

    I think it's one of those reactions you just need to memorise - when cold dilute alkali are used to disproportionate chlorine it forms ClO- (like the equation they gave you to find out the oxidation numbers several questions before), but when hot conc. alkali are used it forms ClO3- instead of ClO-. This applies to other halogens as well.

    When using hot conc. NaOH, NaClO3 ((V) refers to the oxidation number of chlorine in ClO3-, which is +5) is formed instead of NaClO. The other products (NaCl and H2O) are the same, and it's just a matter of correctly balancing the equation.
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    can someone tell me how bromine is less volatile than chlorine?
    • Thread Starter
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    (Original post by Nethmioysters)
    can someone tell me how bromine is less volatile than chlorine?
    Bromine has more electrons than chlorine therefore it has more strong london forces than chlorine. This means Bromine would require more energy to break apart the bonds and form a gas. Therefore, Bromine is less volatile than chlorine,
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    (Original post by Sandy_Vega30)
    Bromine has more electrons than chlorine therefore it has more strong london forces than chlorine. This means Bromine would require more energy to break apart the bonds and form a gas. Therefore, Bromine is less volatile than chlorine,
    oh right! thanks!!
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    This question is from the june 2015 paper.

    Bromoethane reacts with conc. alcoholic ammonia to produce ethylamine. However, in this reaction mixture, the ethylamine formed further reacts with the bromoeathane to produce diethylamine.

    This further reaction of ethylamine can best be limited by carrying out the reaction with

    A iodethane instead of bromoethane
    B less conc. Ammonia
    C excess bromoethane
    D excess ammonia

    Why is the ans D?
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    (Original post by Nethmioysters)
    This question is from the june 2015 paper.

    Bromoethane reacts with conc. alcoholic ammonia to produce ethylamine. However, in this reaction mixture, the ethylamine formed further reacts with the bromoeathane to produce diethylamine.

    This further reaction of ethylamine can best be limited by carrying out the reaction with

    A iodethane instead of bromoethane
    B less conc. Ammonia
    C excess bromoethane
    D excess ammonia

    Why is the ans D?
    Because when u add excess ammonia it will react with all the bromoethane leaving no chance for the ethylamine to further react with it as it will already be used up by reacting with (excess) ammonia
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    (Original post by DDanwar)
    Because when u add excess ammonia it will react with all the bromoethane leaving no chance for the ethylamine to further react with it as it will already be used up by reacting with (excess) ammonia
    Ohk! Thanks alot!
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    For an equilibrium with an exothermic forward reaction why does increasing the temperature make the equilibrium shift to the reactant's side?
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    (Original post by rubyvalkyrie)
    For an equilibrium with an exothermic forward reaction why does increasing the temperature make the equilibrium shift to the reactant's side?
    If the forward reaction is exothermic, then the backward reaction is endothermic. Heat favours the endothermic reaction and the equilibrium shifts to the left to absorb the added heat.
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    are we safe if da question asks fr da structural formulae n instead we draw display formulae marks arent deducted ryt?
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    (Original post by saitama kun)
    are we safe if da question asks fr da structural formulae n instead we draw display formulae marks arent deducted ryt?
    i think its fine. they usually accept displayed formulae when asked for structural
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    can someone pleases explain to me the answer for Q24 a) ii) in the jan 2016 paper?!
 
 
 
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