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    (Original post by CERC)
    It really depends if the substance you are adding is in excess or not.

    If only a few drops of OH ions are added then some of the H20 ligands will be replaced, if however, the OH ions are in excess, then all the ligands will be replaced.

    Eg. In [Fe(H20)6] the charge is +2. Therefore you add 2OH, and you end up with [Fe(H20)4(OH)2] which is a precipitate and 2H20 is also produced. (You add the same amount of OH ions as the charge of the ion and you make this number of water molecules also)

    As the Iron precipitate does not dissolve in excess OH ions, then this precipitate will remain even in excess.

    However, the likes of Cobalt produces a blue precipitate in OH ions and dissolves in excess giving a yellow solution. The equations look like this:

    [Co(H20)6] +2 + 2OH ------------- [Co(h20)4(oh)2] (solid) + 2H20

    in excess the equation is:

    [Co(H20)4(oh)2] + 4OH -------the-----[Co(OH)6] -4 charge + 4H20

    The overall charge is -4 as Cobalt has a charge of +2 and we added 6OH- ions and so +2-6 = -4!

    Hope i helped!
    Thanks a lot that was very helpful. So do I just need to learn the individual complex ions that will cause the ppt to dissolve with excess OH-? If the overall charge or the complex ion is neutral is that when I know theres a ppt? Do I need to know about the NH3 ligand and when it causes things to dissolve?
    thanks
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    Hey guys i was wondering if someone could help me on this quick question:
    Q11
    https://110d0b1a7b584219c71ba4cfe896...0Chemistry.pdf

    And this one (they're pretty much similar i just don't get how to to do it - i know there must be a simple way of doing it)
    Q11
    https://110d0b1a7b584219c71ba4cfe896...0Chemistry.pdf
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    https://110d0b1a7b584219c71ba4cfe896...0Chemistry.pdf

    can someone help with enthalpy change 21ii and percentage uncertainty 21 c
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    Attachment 537535I am currently stuck on question 5 of the photo. I was hoping someone could help. The answers C but i cant work out why. Thanks in advance.
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    (Original post by ayvaak)
    Attachment 537535I am currently stuck on question 5 of the photo. I was hoping someone could help. The answers C but i cant work out why. Thanks in advance.

    Look at this data and look for the big jumps. C is fairly 'flat.'

    https://en.wikipedia.org/wiki/Molar_...f_the_elements

    Elements are Na Ca Mn Ne

    Tricky question though.
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    (Original post by youreanutter)
    https://110d0b1a7b584219c71ba4cfe896...0Chemistry.pdf

    can someone help with enthalpy change 21ii and percentage uncertainty 21 c
    for 21ii) you get the value you worked out above which should be 1149.5 now remember this is the heat of the SURROUNDINGS so if energy has been released in the atmosphere so its + then from the SOLLUTION energy has been lost so its negative

    now remember standard enthalpy states that the enthaply change is the saem whatever route is taken so the for the solution the value will be -1149.5/0.010 so its -114950J but it needs to be in Kj so dived by 1000 and you get -114.95Kj and so 3 significant will be -115kJmol-1
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    Attachment 537535I am currently stuck on question 5 of the photo. I was hoping someone could help. The answers C but i cant work out why. Thanks in advance.
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    (Original post by n2697)
    Hey guys i was wondering if someone could help me on this quick question:
    Q11
    https://110d0b1a7b584219c71ba4cfe896...0Chemistry.pdf

    And this one (they're pretty much similar i just don't get how to to do it - i know there must be a simple way of doing it)
    Q11
    https://110d0b1a7b584219c71ba4cfe896...0Chemistry.pdf
    10% reacted, so 30cm^3 had been converted in to ozone, leaving 270cm^3 of oxygen. Since there is a 3:2 molar ratio, the volume of ozone produced from 30cm^3 oxygen = 30 x (2/3) = 20
    Therefore, the total volume of gas left in the mixture is now 270 + 20 = 290cm^3.

    With the second question, 100cm^3 decomposed, so 50cm^3 of nitrogen and 150cm^3 must have been formed, respectively. (Due to the molar ratios). This leaves you with 400 + 150 + 50 =600cm^3.
    Hope this made sense, and just remember, the ratio of the volumes in which gases react is the same as how they react in terms of moles!!!


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    (Original post by Marco1000)
    Thanks a lot that was very helpful. So do I just need to learn the individual complex ions that will cause the ppt to dissolve with excess OH-? If the overall charge or the complex ion is neutral is that when I know theres a ppt? Do I need to know about the NH3 ligand and when it causes things to dissolve?
    thanks
    Yes, well depending on your exam board.

    I do CCEA, so you need to know what elements dissolve in excess and what don't.
    You also need to know ligands with ammonia as well and colours of precipitates and their formula of the ion.

    And yes, if the overall charge is neutral then it is likely it will be a precipitate!

    Hope i helped and if you have any more questions, im happy to help, if i can!
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    (Original post by PlayerBB)
    Hey, can you please explain a little further on the M+1peak, like for Cl2, there is Cl(35) and Cl(37) so which one would be for the M+1 peak ?

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    Ok, so i'm just going to type out what my revision guide says about this:


    1) The M+1 peak is a small peak one m/e value above the M+ peak. It is caused by the presence of ONE Carbon-13 atom in the molecular ion.

    2) A MAJOR peak at M+2 is caused by the presence of either Cl or Br atom in the molecule. Both Chlorine and Bromine have isotopes that differ by two mass units. Chlorine has 35 and 37 whilst Bromine has 79 and 81.

    IMPORTANT: If the ration of the M+ peak to the M+2 peak is 3:1 i.e. the M+ peak is three times as abundant as the M+2 peak, then a chlorine atom is present.

    However, the Bromine 79 and Bromine 81 have the same abundance, in a ration of 1:1. So, if the M+ peak has the same abundance as the M+2 peak then a Bromine atom is present in the molecule!


    Really hope this helps you!
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    (Original post by itsastudentyo)
    i was just wondering if we have to learn organic chemistry for paper 1 AS level which is on the 27th may?
    Look up your exam's syllabus and they will break down the topics into Unit e.g Unit 1 and Unit 2. and Unit 3 which is the practical paper.

    If you mean the practical is on the 27th May, then yes, you need to learn all this year's organic preparations like making an ester for example.


    Learn unit 1 for your first exam and unit 2 for your second.

    hope this helps!
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    (Original post by Junaidc122)
    10% reacted, so 30cm^3 had been converted in to ozone, leaving 270cm^3 of oxygen. Since there is a 3:2 molar ratio, the volume of ozone produced from 30cm^3 oxygen = 30 x (2/3) = 20
    Therefore, the total volume of gas left in the mixture is now 270 + 20 = 290cm^3.

    With the second question, 100cm^3 decomposed, so 50cm^3 of nitrogen and 150cm^3 must have been formed, respectively. (Due to the molar ratios). This leaves you with 400 + 150 + 50 =600cm^3.
    Hope this made sense, and just remember, the ratio of the volumes in which gases react is the same as how they react in terms of moles!!!


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    Thank you so much this really helped !!
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    Hey can someone help me with this please?
    14.3g of hydrated sodium carbonate, Na2CO3.10H2O is dissolved in water and made up to 250cm3 of solution.
    What is the concentration of sodium ions in the solution, in Moldm-3?
    The answers 0.400 but I'm not sure why! Thanks 😊
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    Can someone help me out with this question? I think the answer is B. What do you think? :
    1. A 27.0 g sample of an unknown hydrocarbon, CxHy, was burned completely in excessoxygen to form 88.0 g of carbon dioxide and 27.0 g of water. [Molar masses / g mol–1: CO2 = 44; H2O = 18]
      Which of the following is a possible formula of the unknown hydrocarbon?
    A CH4
    B C2H6
    C C4H6
    D C6H6
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    (Original post by Farmerjj)
    Hey can someone help me with this please?
    14.3g of hydrated sodium carbonate, Na2CO3.10H2O is dissolved in water and made up to 250cm3 of solution.
    What is the concentration of sodium ions in the solution, in Moldm-3?
    The answers 0.400 but I'm not sure why! Thanks 😊
    Firstly work out moles of hydrated sodium carbonate which is Mass/Mr = 14.3/286 = 0.05 moles of Na2CO3.10H2O.
    When the hydrated sodium carbonate dissolves, the ions dissociate from the compound so 2Na+ ions are in solution.
    So the molar ratio of Na2CO3.10H2O to the Na+ ions is 1:2
    Thus you multiply your moles by 2:
    0.05 x 2 = 0.1 moles of Na+ ions

    Then using the formula (C x V)/1000 = moles, you can work out the concentration which is 0.4 mol dm-3
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    If 1.45% of 1 dm3 of air is CO2, how many parts per million is CO2?
    I'm so stuck on this question so if you know how you do this please could you explain it. Thank you

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    Also could someone explain how you get the answer for this question. Thanks

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    (Original post by TeaAndTextbooks)
    Also could someone explain how you get the answer for this question. Thanks

    Posted from TSR Mobile
    Mole H2O = mass / Mr = 1.8/18 =0.1
    Therefore to find the number of molecules of H2O, multiply by Avagadro's:
    0.1 x 6.0 x10^23 = 6 x 10^22
    Next times by three , because there are 3 atoms per molecule, giving 1.8 x 10^23
    Hope this helped 😇


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    What marks did you guys get on the Physical and Inorganic spec paper for AS Chemistry? Just want to get an idea as to how difficult it is in reality
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    Name:  20160525_191448.jpg
Views: 186
Size:  495.8 KBsorry I don't know why you multiply by 2?? I don't remember having to do this in other questions?
 
 
 
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