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    Hi guys, I was just wondering if you'd be able to clarify which molecule shapes mean that the molecule will not have a dipole even though it's bonds are polar...tetrahedral and????

    Thank you!
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    (Original post by nicolaa)
    Hi guys, I was just wondering if you'd be able to clarify which molecule shapes mean that the molecule will not have a dipole even though it's bonds are polar...tetrahedral and????

    Thank you!
    It's not necessarily the shape of the molecule, even if you have a tetrahedral molecule, you still can have a polar molecule. For example, trichlorofluromethane is polar:
    Name:  Trichlorofluoromethane.jpg
Views: 700
Size:  7.6 KB

    Because fluorine is more electronegative than chlorine, the charges will be unbalanced across the molecule and will not cancel out. Thus, it's important you know how electronegativity values can effect the overall charge of a shape.

    If there's a symmetrical molecule, like tetrachloromethane, then that is non-polar. This is because all the negative charge can be cancelled down, as there is an equal electronegative difference between carbon and the chlorine atoms around the molecule.

    In this exam, they'll mainly focus on more simple molecule shapes, such as tetrahedral, bent, triangle planers so it'll be obvious if the charges can get canceled down or not. Try some past paper questions, it has been asked quite a few times. Hope that helped!
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    I'm confused on Q11 part C in Chemical Ideas problems for 1.5
    I know the answers my question is WHY find out step c? What is its purpose? why is it necessary?

    "The concentration of an acid solution can be found by titrating the acid solution with a solution of an alkali of a known concentration. In such a titration it was found that 19.00cm3 of 0.100mol dm-3 sodium hydroxide were necessary to react with 25.00cm3 of a hydrochloric acid solution
    NaOH(aq) +HCl(aq) --> NaCl(aq) + H2O(l)

    a) How many moles of NaOH are there in 19.00cm3 of 0.100mol dm-3 solution ? Answer = 0.0019
    b) How many moles of HCl are , therefore , in 25.00cm3 of the acid solution? Answer = 0.0019
    c) How many moles of HCl are, therefore, in 1000cm3 of the acid solution? Answer = 0.0076
    d) What is the concentration of the HCl solution? Answer = 0.0076"
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    (Original post by blondie24)
    x
    Sometimes there'll be a dilution in the process. For example Jan 2012, Q1e,v.

    So you'll have to work out the dilution factor, which in the chemical ideas question will be 1000/25. Which is 40, so 40 times the initial moles you worked out of 0.0019, equals: 0.076. Check your answer again, I've checked the official answers, and it's 0.076 mol.

    This is an important concept that is especially developed as you go into F334, where the calculations are far trickier and horrible!

    Hope that helped
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    hi guys, do we have to know how to do the practical? in my revision guide it has things like 'preperation of halogenoalkanes'???
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    (Original post by abzy1234)
    Sometimes there'll be a dilution in the process. For example Jan 2012, Q1e,v.

    So you'll have to work out the dilution factor, which in the chemical ideas question will be 1000/25. Which is 40, so 40 times the initial moles you worked out of 0.0019, equals: 0.076. Check your answer again, I've checked the official answers, and it's 0.076 mol.

    This is an important concept that is especially developed as you go into F334, where the calculations are far trickier and horrible!

    Hope that helped
    so you find out the moles in 1000cm3 because its the dilution factor
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    (Original post by blondie24)
    so you find out the moles in 1000cm3 because its the dilution factor
    Yes. Have you done some titration practicals?

    Think about it. If I have an original solution of 1000cm3, it won't be wise for me to use all of it for my titrations. So I'll have to scale the volume down, and we do this by diluting it to a lower value of around 20 to 30 cm3 in most cases. This will obviously change the number of moles, so after finishing this modified practical, I'll have to scale it back up to reflect my original volume.

    Practice the calculations, it'll help a lot
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    (Original post by ashxx)
    hi guys, do we have to know how to do the practical? in my revision guide it has things like 'preperation of halogenoalkanes'???
    Yeah it's on the specification:

    (v) describe and explain the principal stages in the
    purification of an organic liquid product:
    (i) shaking with sodium hydrogencarbonate solution to
    remove acidic impurities,
    (ii) separating from other immiscible liquids using a
    separating funnel,
    (iii) drying with anhydrous sodium sulfate,
    (iv) simple distillation to allow collection of the pure
    product
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    (Original post by abzy1234)
    Yes. Have you done some titration practicals?

    Think about it. If I have an original solution of 1000cm3, it won't be wise for me to use all of it for my titrations. So I'll have to scale the volume down, and we do this by diluting it to a lower value of around 20 to 30 cm3 in most cases. This will obviously change the number of moles, so after finishing this modified practical, I'll have to scale it back up to my original volume.

    Practice the calculations, it'll help a lot
    I just cant get my head around it
    So in any question referring to titrations after i find out the number of moles of the solution its asking i then find out how many moles there would be in 100cm3? Is this the same for all titration molarity Q's? Sorry to be such a pain btw i know the calculations aren't hard i'm just having a hard time understanding why
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    (Original post by blondie24)
    I just cant get my head around it
    So in any question referring to titrations after i find out the number of moles of the solution its asking i then find out how many moles there would be in 100cm3? Is this the same for all titration molarity Q's? Sorry to be such a pain btw i know the calculations aren't hard i'm just having a hard time understanding why
    Don't worry, it's a tricky concept!

    The key is you read the question carefully. In most calculations in F332, they will not ask you about scaling things back up. It's only come up once. Try doing this question, it should help you get your head around it:

    Name:  Q.jpg
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Size:  85.5 KB

    Google dilution calculations as well, plus go onto youtube; there's some really good videos that I watched about titrations calculations. Hope this helps!
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    (Original post by abzy1234)
    It's not necessarily the shape of the molecule, even if you have a tetrahedral molecule, you still can have a polar molecule. For example, trichlorofluromethane is polar:
    Name:  Trichlorofluoromethane.jpg
Views: 700
Size:  7.6 KB

    Because fluorine is more electronegative than chlorine, the charges will be unbalanced across the molecule and will not cancel out. Thus, it's important you know how electronegativity values can effect the overall charge of a shape.

    If there's a symmetrical molecule, like tetrachloromethane, then that is non-polar. This is because all the negative charge can be cancelled down, as there is an equal electronegative difference between carbon and the chlorine atoms around the molecule.

    In this exam, they'll mainly focus on more simple molecule shapes, such as tetrahedral, bent, triangle planers so it'll be obvious if the charges can get canceled down or not. Try some past paper questions, it has been asked quite a few times. Hope that helped!
    Ahhh, right! That makes sense...thanks a lot! just one more question....so for argument's sake if the molecule was symmetrical which shapes would mean that the molecule is polar....bent....? Thank you so much!
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    (Original post by abzy1234)
    Yeah it's on the specification:

    (v) describe and explain the principal stages in the
    purification of an organic liquid product:
    (i) shaking with sodium hydrogencarbonate solution to
    remove acidic impurities,
    (ii) separating from other immiscible liquids using a
    separating funnel,
    (iii) drying with anhydrous sodium sulfate,
    (iv) simple distillation to allow collection of the pure
    product
    thank you ... gotta remember all this now
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    to produce a aldehyde, do you do reflux then distil but for carboxylic acid, do you reflux, distil and reflux or just reflux? (obviously with acidified potasium dichromate)
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    (Original post by Welbeck)
    to produce a aldehyde, do you do reflux then distil but for carboxylic acid, do you reflux, distil and reflux or just reflux? (obviously with acidified potasium dichromate)
    Just keep refluxing (is that an actual verb? ) The only reason you need to immediately distill an aldehyde is that it would oxidise further to form a carboxylic acid if you didn't.
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    (Original post by nicolaa)
    Ahhh, right! That makes sense...thanks a lot! just one more question....so for argument's sake if the molecule was symmetrical which shapes would mean that the molecule is polar....bent....? Thank you so much!
    In that instance, yes, often is bent. Have a look at the hydrogen sulphide molecule (H2S). But, it doesn't really come down to the shape, but the values of electronegativity and the different levels of electronegativity in a polar bond.

    No problem, happy to help
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    okay so im stuck on a question... about atom economy.... so...
    C2H4 + 0.5O2 ---> C2H40 (epoxyethane)
    -In one reaction, 5.6kg of ethene produced 1.32kg of epoxyethane. Calculate the % yield for this reaction....help would be much appreciated I know the answer but just not the correct method to get to it
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    (Original post by ashxx)
    okay so im stuck on a question... about atom economy.... so...
    C2H4 + 0.5O2 ---> C2H40 (epoxyethane)
    -In one reaction, 5.6kg of ethene produced 1.32kg of epoxyethane. Calculate the % yield for this reaction....help would be much appreciated I know the answer but just not the correct method to get to it
    What's the answer? (Don't wanna give you the wrong solution)
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    (Original post by super121)
    What's the answer? (Don't wanna give you the wrong solution)
    15%.... i keep on getting 17%? i probably can't add knowing me *awks*
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    (Original post by super121)
    What's the answer? (Don't wanna give you the wrong solution)
    ahh nvm, i literally just worked out how to do it, lol i got the Mr of C2H4 wrong... awks but thanks for the help
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    (Original post by super121)
    The molecule has E/Z isomers because of the carbon double bond on the right. It can't be the one on the left because there isn't 2 different groups on each carbon of the carbon double bond. You can swap the methyl groups because they are bonded to different carbon double bonds.
    It may help to draw out the full structural formula to understand it better.


    Posted from TSR Mobile
    Thanks mate I think I get it now, but also does this mean stereoisomerism can be between functional groups as well as side chains?

    Posted from TSR Mobile
 
 
 
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