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Size:  142.5 KBGuys i couldn't understand how to answer part (ii)
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    (Original post by Snowie9)
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Size:  142.5 KBGuys i couldn't understand how to answer part (ii)
    How many moles is 1 gram of isomer?

    therefore how many moles of Cl- precipitate out of each sample?

    therefore how many Cl- are not bonded to the Chromium for each sample?

    therefore how many Cl- are bonded to the Chromium for each sample?
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    (Original post by MexicanKeith)
    How many moles is 1 gram of isomer?

    therefore how many moles of Cl- precipitate out of each sample?

    therefore how many Cl- are not bonded to the Chromium for each sample?

    therefore how many Cl- are bonded to the Chromium for each sample?
    For example, isomer A, 1 mol of chromium produces 1 mol of AgCl. So i guess i need one Cl?
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    (Original post by Snowie9)
    For example, isomer A, 1 mol of chromium produces 1 mol of AgCl. So i guess i need one Cl?
    So that means 1 of the chloride ions wasnt bound to the Chromium, so n=2 for that compound you've got it!
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    (Original post by MexicanKeith)
    How many moles is 1 gram of isomer?

    therefore how many moles of Cl- precipitate out of each sample?

    therefore how many Cl- are not bonded to the Chromium for each sample?

    therefore how many Cl- are bonded to the Chromium for each sample?
    (Original post by Snowie9)
    Name:  IMG_8771.PNG
Views: 18
Size:  142.5 KBGuys i couldn't understand how to answer part (ii)

    A quicker way to do this - if you grasp from the question what the three possible isomers are. From the information given you can put the isomers in order of most to least AgCl they can form in this reaction.
    No mole calculations necessary using this logic-based approach.
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    (Original post by MexicanKeith)
    So that means 1 of the chloride ions wasnt bound to the Chromium, so n=2 for that compound you've got it!
    Thank you (:
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    (Original post by TutorsChemistry)
    A quicker way to do this - if you grasp from the question what the three possible isomers are. From the information given you can put the isomers in order of most to least AgCl they can form in this reaction.
    No mole calculations necessary using this logic-based approach.
    I see it now. Thanks for the help (:
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    (Original post by Snowie9)
    I see it now. Thanks for the help (:
    glad you've worked it out


    TutorsChemistry is right, an easier method would be to notice that there are only four possible isomers (only three of which would precipitate any AgCl) so it must be those three, then go from there.
 
 
 
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