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    (Original post by _NMcC_)
    Yes that looks fine. I'm getting the same answer.
    Thank you very much for your help!!!
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    (Original post by locket0511)
    Thank you very much for your help!!!
    No worries.
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    hils?
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    (Original post by Kadak)
    Yes, me !
    Absolutely pain

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    Hi

    I do not understand ionic half equations, was wondering if you guys could help me understand them.

    (i)So instance it asks to 'Write the ionic equation to show the oxidation of Calcium, Ca, to Calcium ions, Ca^2+'.

    Ca ---> Ca2+ + 2e^-

    (correct or not?)

    (ii) And 'Write the ionion half-equation to show the reduction of water to hydrogen, H2, and hydroxide ions, OH^-'.

    (could some one walk me through this, baby steps :blushing: )

    (iii) Combine the two ionic half-equations.

    (I think I know how to do this lol, I THINK)


    Also other questions:

    Complete and balance the following equations:

    (ii) Na2O + H2O --> ?

    (iii) Na2O + HCl ---? ?

    (idk because I've only learnt about group 2 and 7.. not group 1 :s )

    Thanks :ahee:
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    More questions :awesome:

    Sodium:

    1st ionisation energy: 496
    2nd ionisation energy: 4563

    Magnesium:

    1st .. : 738
    2nd.. : ?

    Predict a value for the second Ionisation energy of Magnesium. Explain your choice.

    So I was thinking:

    Sodium's 1st is lower than Magnesium due to the atomic size and nuclear attraction. (But wouldn't it's first be high because it does lose a whole electron shell?) It's 2nd is significantly higher in IE (Is it because of the loss of an electron shell?)

    Magnesium's 1st IE is greater due to it being higher in electron density (having a larger charge and smaller radius), which causes it to have a strong nuclear attraction. It's 2nd IE would result in it losing an electron shell. So wouldn't the jump/difference between 1st and 2nd be significantly great? (or is the jump at the 3rd IE?)

    I'm not sure if the 2nd IE of Magnesium is suppose to be greater or less than the 2nd IE of Sodium.

    Thanks :blushing:
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    Lololol more questions :awesome: (these are questions I missed out from numerous past paper booklets, I swear :blushing: )

    Cars are fitted with catalytic converters in order to reduce the pollution caused by the combustion of petrol. Potential pollutant gases include CO, NO and unburnt hydrocarbons. The first two compounds are removed by passing the hot gases over a platinum catalyst.

    CO(g) + NO(g) --> CO2(g) + 1/2N2(g) Delta H = -373kJmol-1

    (iii) Comment on the relative value of the Activation energy of this reaction compared with the much fast reaction of NO with O2.

    Lol, after typing that out I THINK I get it? Or.. maybe not.
    Would you just say that:

    The activation energy would be lower in the reaction of NO + O2 because less energy is required for the reaction to take place?

    Thanks again
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    (Original post by Dinaa)
    More questions :awesome:

    Sodium:

    1st ionisation energy: 496
    2nd ionisation energy: 4563

    Magnesium:

    1st .. : 738
    2nd.. : ?

    Predict a value for the second Ionisation energy of Magnesium. Explain your choice.

    So I was thinking:

    Sodium's 1st is lower than Magnesium due to the atomic size and nuclear attraction. (But wouldn't it's first be high because it does lose a whole electron shell?) It's 2nd is significantly higher in IE (Is it because of the loss of an electron shell?)

    Magnesium's 1st IE is greater due to it being higher in electron density (having a larger charge and smaller radius), which causes it to have a strong nuclear attraction. It's 2nd IE would result in it losing an electron shell. So wouldn't the jump/difference between 1st and 2nd be significantly great? (or is the jump at the 3rd IE?)

    I'm not sure if the 2nd IE of Magnesium is suppose to be greater or less than the 2nd IE of Sodium.

    Thanks :blushing:
    It's probably best not to think of it in terms of electron 'shells'. When you ionise a substance, the electron removed will be the outer most electron, which, on average, will be the electron with the most energy, in the case of sodium the one in the 3s orbital. After removing that electron, you are then having to remove an electron from the 2p orbitals, which on average will be closer so greater attraction and there is also less shielding between nucleus and the electron so there is a big jump in ie. With mg there are 2 electrons in the 3s so you can remove 2 before the large jump so hence 2nd ie of mg will be lower

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    Need some help with why dilute nitric acid is used when identifying metal halides.

    Is it to remove any carbonate/hydroxide ions so they do not interfere.

    I read something on chemrevise that I don't fully understand.

    It said that : "The role of nitric acid is to react with any carbonates present to prevent formation of the precipitate Ag2CO3. This would mask the desired observations".

    Then it has an equation: 2HNO3 + Na2CO3 ------> 2NaHNO3 + H2O + CO2.

    Anyone care to explain the equation and the issue with silver carbonate, by that I mean how does that mask the observations?

    Thanks
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    (Original post by Super199)
    Need some help with why dilute nitric acid is used when identifying metal halides.

    Is it to remove any carbonate/hydroxide ions so they do not interfere.

    I read something on chemrevise that I don't fully understand.

    It said that : "The role of nitric acid is to react with any carbonates present to prevent formation of the precipitate Ag2CO3. This would mask the desired observations".

    Then it has an equation: 2HNO3 + Na2CO3 ------> 2NaHNO3 + H2O + CO2.

    Anyone care to explain the equation and the issue with silver carbonate, by that I mean how does that mask the observations?

    Thanks
    Silver carbonate would give you a yellowy - grey precipitate as it is poorly soluble in water. This is a problem because it then becomes pretty much impossible to tell if your precipitate is because of agco3 or confirms the presence of a halide ion because they would look basically identical.

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    Guys which molecules have permanent dipole-permanent dipole bonds? Is it only the polar ones?
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    (Original post by Dinaa)
    Hi

    I do not understand ionic half equations, was wondering if you guys could help me understand them.

    (i)So instance it asks to 'Write the ionic equation to show the oxidation of Calcium, Ca, to Calcium ions, Ca^2+'.

    Ca ---> Ca2+ + 2e^-

    (correct or not?)
    Yep, charges are balanced

    (ii) And 'Write the ionion half-equation to show the reduction of water to hydrogen, H2, and hydroxide ions, OH^-'.

    (could some one walk me through this, baby steps :blushing: )
    Reduction of Water so splitting water into Hydrogen and Hydroxide ions

    H2O --> H2 + OH-

    (this is the equation at first - unbalanced)

    I can't really explain it but you're trying to split water into hydrogen and hydroxide anions, so therefore you need to balance it now

    If we multiply the water moles by 2 so 2H2O instead of H2O, you realise that they are 2 oxygen atoms and 4 hydrogen atoms.
    Let's balance the remainder now. So let's leave the H2 for now but focus on the OH- though. Multiply by 2 so therefore

    2H2O --> H2 + 2OH-
    However the charges aren’t balanced L
    Add 2 electrons to the reactant to balance the charges on both sides. Therefore the equation becomes
    2H2O +e- --> H2 + 2OH-

    4 hydrogens on both sides, 2 oxygens on both sides. Charges are also balanced.


    (iii) Combine the two ionic half-equations.

    (I think I know how to do this lol, I THINK)
    2H2O + Ca +2e- = H2 + 2OH- + Ca2+ 2e-
    That’s the combined equation at first. Immediately you can cancel out the electrons so the equation becomes
    2H2O + Ca +2e- = H2 + 2OH- + Ca2+ 2e-
    2H2O + Ca = H2 + 2OH- + Ca2+
    I’m pretty sure that the hydroxide anions will attract the calcium cations so the final equation becomes
    2H2O + Ca = H2 + Ca(OH)2
    Also other questions:

    Complete and balance the following equations:

    (ii) Na2O + H2O --> ?

    Na2O + H2O à 2NaOH + ½H2
    Pretty much the same thing as previous. You should recall from Group 2 that dissolving a metal oxide in water produces a hydroxide and hydrogen. Balancing here is slightly different though.
    (iii) Na2O + HCl ---? ?

    (idk because I've only learnt about group 2 and 7.. not group 1 :s )

    Thanks :ahee:
    Na2O + 2HCl = 2NaCl + H2O
    Again, pretty much the same thing as Group 2 metals J




    (Original post by Dinaa)
    Lololol more questions :awesome: (these are questions I missed out from numerous past paper booklets, I swear :blushing: )

    Cars are fitted with catalytic converters in order to reduce the pollution caused by the combustion of petrol. Potential pollutant gases include CO, NO and unburnt hydrocarbons. The first two compounds are removed by passing the hot gases over a platinum catalyst.

    CO(g) + NO(g) --> CO2(g) + 1/2N2(g) Delta H = -373kJmol-1

    (iii) Comment on the relative value of the Activation energy of this reaction compared with the much fast reaction of NO with O2.

    Lol, after typing that out I THINK I get it? Or.. maybe not.
    Would you just say that:

    The activation energy would be lower in the reaction of NO + O2 because less energy is required for the reaction to take place?

    Thanks again
    I wouldn’t really say that about this case because you’re not given the values.
    I’m not sure, but hopefully if I’m wrong, someone will correct me
    The reaction (NO + O2) has a lower activation energy as it undergoes an alternative pathway which requires less energy.
    It may be so since the CO bond is stronger than the bond in diatomic oxygen so less energy (hence a lower activation energy) is required.
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    (Original post by EMPA2015)
    I have done AS BIOLOGY I CAN HELP GIVE YOU ANSWERS< DO YOU HAVE AS CHEMISTRY ?
    This has been said billions of times across TSR and a few other people have said on this forum already, you're not allowed to provide answers for EMPA or ISAs AT ALL!
    That would be classed as cheating which wouldn't be fair

    You're not allowed to share answers at all or shared locked/unavailable papers.
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    (Original post by Feraligatr)
    This has been said billions of times across TSR and a few other people have said on this forum already, you're not allowed to provide answers for EMPA or ISAs AT ALL!
    That would be classed as cheating which wouldn't be fair

    You're not allowed to share answers at all or shared locked/unavailable papers.
    Absolute believer in this. The amount of people pm'ing me for answers and wanting to 'trade' answers for other isa's like what is so difficult doing it fairly like the majority of us have to?!:facepalm:

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    (Original post by KeilahDeere)
    Absolute believer in this. The amount of people pm'ing me for answers and wanting to 'trade' answers for other isa's like what is so difficult doing it fairly like the majority of us have to?!:facepalm:

    Posted from TSR Mobile
    I've heard that exam boards actually moderating TSR so anyone who is identified and who does post messages like such may get disqualified.
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    (Original post by KINGYusuf)
    Guys which molecules have permanent dipole-permanent dipole bonds? Is it only the polar ones?
    Yep, pretty much any polar molecule can form permanent dipole-dipole interactions. Though just to be on the safe side, I wouldn't say bonds, I'd say interactions.
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    Hey guys, is this a wise decision? At the moment because I have so much content left- I don't think it's best to do any past papers at the moment. If I am lucky, maybe I will try the 2014 paper but since I am basically trying to learn the whole of Chem and Physics- I am planning just to go over the book, the specification and it's questions without any past papers.
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    (Original post by Feraligatr)
    Yep, charges are balanced
    Sweeet :bigsmile:


    Reduction of Water so splitting water into Hydrogen and Hydroxide ions

    H2O --> H2 + OH-

    (this is the equation at first - unbalanced)

    I can't really explain it but you're trying to split water into hydrogen and hydroxide anions, so therefore you need to balance it now

    If we multiply the water moles by 2 so 2H2O instead of H2O, you realise that they are 2 oxygen atoms and 4 hydrogen atoms.
    Let's balance the remainder now. So let's leave the H2 for now but focus on the OH- though. Multiply by 2 so therefore

    2H2O --> H2 + 2OH-
    However the charges aren’t balanced L
    Add 2 electrons to the reactant to balance the charges on both sides. Therefore the equation becomes
    2H2O +e- --> H2 + 2OH-

    4 hydrogens on both sides, 2 oxygens on both sides. Charges are also balanced.
    Ow right! I got the H2O --> H2 + OH- bit, but I don't know, just thought it looked wrong haha. So just balancing then, makes sense!


    2H2O + Ca +2e- = H2 + 2OH- + Ca2+ 2e-
    That’s the combined equation at first. Immediately you can cancel out the electrons so the equation becomes
    2H2O + Ca +2e- = H2 + 2OH- + Ca2+ 2e-
    2H2O + Ca = H2 + 2OH- + Ca2+
    I’m pretty sure that the hydroxide anions will attract the calcium cations so the final equation becomes
    2H2O + Ca = H2 + Ca(OH)2
    Nice emojis and Ye that makes sense, I would have probably just left it as 2OH- and Ca2+ but of course they attract :blushing:


    Na2O + H2O à 2NaOH + ½H2
    Pretty much the same thing as previous. You should recall from Group 2 that dissolving a metal oxide in water produces a hydroxide and hydrogen. Balancing here is slightly different though.


    Na2O + 2HCl = 2NaCl + H2O
    Again, pretty much the same thing as Group 2 metals J
    Idk, I thought maybe group 1 elements react differently? I know how the group 2 elements react but thought that Group 1 elements wouldn't react similarly due to their charge of +1?



    I wouldn’t really say that about this case because you’re not given the values.
    I’m not sure, but hopefully if I’m wrong, someone will correct me
    The reaction (NO + O2) has a lower activation energy as it undergoes an alternative pathway which requires less energy.
    It may be so since the CO bond is stronger than the bond in diatomic oxygen so less energy (hence a lower activation energy) is required.

    Ugh Idk, I don't like that question haha. Had to read it like 100 times to understand it
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    (Original post by KeilahDeere)
    Absolute believer in this. The amount of people pm'ing me for answers and wanting to 'trade' answers for other isa's like what is so difficult doing it fairly like the majority of us have to?!:facepalm:

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    Hiya :hugs:
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    (Original post by Feraligatr)
    I've heard that exam boards actually moderating TSR so anyone who is identified and who does post messages like such may get disqualified.
    I hope so! Though does make me think how they could track them down via a tsr account. :dontknow:

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