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    Mole ratio doubt -

    C : H : O
    1.3 : 3 : 1

    should I round up 1.3 to 2 or 1?
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    (Original post by ihaspotato)
    Mole ratio doubt -

    C : H : O
    1.3 : 3 : 1

    should I round up 1.3 to 2 or 1?
    Multiply everything by 3?
    To give r C4H9O3 I mean
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    (Original post by C0balt)
    Multiply everything by 3?
    To give r C4H9O3 I mean
    I took 1.3 as an example.. but there was a paper where the ratios were

    1.75 : 3 :1
    (x4)
    giving C7H12O4

    but I wanted to know if it was 1.3 - should I round it up to 1 or 2?
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    (Original post by _H_V)
    When calculating pH of buffer do you use the moles of the weak acid and its salt or do you use the concentration of each in the equation?
    I just did a couple of buffer questions and realised I was forgetting to change the moles of the acid and salt back into concentration to put into the equation but the answer was still correct.
    Anyone ?
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    (Original post by ihaspotato)
    I took 1.3 as an example.. but there was a paper where the ratios were

    1.75 : 3 :1
    (x4)
    giving C7H12O4

    but I wanted to know if it was 1.3 - should I round it up to 1 or 2?
    Do you mean round it up to 1.0 or 2.0? I don't think you should do that because 0.3 is quite significant as opposed to 0.08 or 0.91 or something!
    I think it depends how high or low the 0.3 is. I always go with gut feeling lol
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    (Original post by _H_V)
    Anyone ?
    Concentration
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    Data booklet
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Size:  9.0 KB

    CH ppm 1.8-3.0 how? what does it fall under?
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    (Original post by _H_V)
    Anyone ?
    Perhaps the answers were still correct because the volumes cancelled?
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    (Original post by ihaspotato)
    Data booklet
    Name:  Untitled.png
Views: 57
Size:  9.0 KB

    CH ppm 1.8-3.0 how? what does it fall under?
    Can you screenshot the question? We need the context of the CH group
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    (Original post by ihaspotato)
    Data booklet
    Name:  Untitled.png
Views: 57
Size:  9.0 KB

    CH ppm 1.8-3.0 how? what does it fall under?
    It's the H-c-c=o environment I believe if you look at the data booklet
    This is June 15 isn't it I did it this morning
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    (Original post by Funky_Giraffe)
    Can you screenshot the question? We need the context of the CH group
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    (Original post by ihaspotato)
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    It's the because it's the H in the H-C-C=O, which shows up at 1.8-3.0ppm
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    (Original post by Funky_Giraffe)
    It's the because it's the H in the H-C-C=O, which shows up at 1.8-3.0ppm
    Ahh makes sense. Thanks!
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    (Original post by _H_V)
    Anyone ?
    think of it this way, to work out conc for A- and HA u had to divide moles by volume; as the amount of volume in the solution is the same you divide both molar amounts by the same number(v), so its: A-/V divided by HA/V. As the v's cancel you are just left with the molar amounts. (A- divided by HA).

    You can skip converting to concentration essentially but ka definition is using concentration values not molar amounts
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    I think you tend to get method mark by calculating the concentration even if you don't really have to. That is often the case with Kp calculations etc, so i do every step just in case I mess up some place
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    Why a solution of CH3COONa is a stronger acid than NH3 (wouldn't the ammonia dissociate to give NH4+ ?
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    (Original post by PlayerBB)
    Why a solution of CH3COONa is a stronger acid than NH3 (wouldn't the ammonia dissociate to give NH4+ ?
    Yes but that also gives OH- ions in solution. NH3 is not an acid.
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    (Original post by Funky_Giraffe)
    Totally agree with this. Although being the last main-sit exam in this series, the potential is there for them to pull out all the stops and make this unit 4 a hard one...
    was searching for ways for them to 360noscope us to re-sit the exam next year and then I find this - May 2016 C1 news.
    • :nothing:
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    (Original post by PlayerBB)
    Why a solution of CH3COONa is a stronger acid than NH3 (wouldn't the ammonia dissociate to give NH4+ ?
    Which one's the answer?
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    (Original post by C0balt)
    Which one's the answer?
    B (3,2,1)
 
 
 

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