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    (Original post by shhh123)
    Isnt it just that there are equal amount of moles on each side so there will be no visible change ?
    No. It is nothing to do with equilibrium as far as I'm aware

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    (Original post by samb1234)
    No. It is nothing to do with equilibrium as far as I'm aware

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    Im sure it does cus of you first have look at the number of moles on each side then see if eqilibria will shift to side with a side of fewer moles in this case there is two moles on the LHS and RHS so there will be no overall effect
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    (Original post by shhh123)
    Isnt it just that there are equal amount of moles on each side so there will be no visible change ?
    Yes exactly there are equal amount of moles on each side so there will be no effect on equilibrium.
    Therefore when compressed the colour will be darker as the coloured particles are closer together.


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    (Original post by shhh123)
    Im sure it does cus of you first have look at the number of moles on each side then see if eqilibria will shift to side with a side of fewer moles in this case there is two moles on the LHS and RHS so there will be no overall effect
    Correct equilibrium will not move. However reducing the volume has the effect of 'concentrating' the colour as you still have the same number of coloured I2 molecules but in a smaller space, therefore they are closer and the colour appears stronger

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    Anyone have any notes for either F331 or F332 that follows all the points on the spec?
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    Also how are you guys revising?
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    Has anyone done the evaluative practical task for OCR A about ketones? if so can you please give an idea of the things I need to know/revise as ive got mine in 2 days. thank you
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    Can somebody check my answer to a titration question please?

    20.45cm3 of H2SO4, 0.05moldm-3 was required to neutralise 25cm3 of NH3. What is the concentration of NH3?

    Equation is H2SO4 + 2NH3 --> (NH4)2S04
    Moles of H2S04 = 0.05 x 20.45 / 1000 = 1.0225x10-3 mol
    1:2 ratio so moles of NH3 = 2.045x10-3 mol
    Concentration = moles / volume
    2.045x10-3 / 0.025 = 0.0818 moldm-3
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    (Original post by prophetkid)
    Can somebody check my answer to a titration question please?

    20.45cm3 of H2SO4, 0.05moldm-3 was required to neutralise 25cm3 of NH3. What is the concentration of NH3?

    Equation is H2SO4 + 2NH3 --> (NH4)2S04
    Moles of H2S04 = 0.05 x 20.45 / 1000 = 1.0225x10-3 mol
    1:2 ratio so moles of NH3 = 2.045x10-3 mol
    Concentration = moles / volume
    2.045x10-3 / 0.025 = 0.0818 moldm-3
    Looks right

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    Right can someone help me out it is to do with analytical techniques.

    1). What is meant by the molecular ion

    The molecular ion is a molecule that has lost an electron in the mass spectrometer forming a positive ion with a radical.

    2). What is the M peak?

    The M peak is the peak that represents the molecular ion with the biggest M/Z value

    3). How do fragments get formed?
    Not entirely sure, possibly something to do with the bombarding of the electrons.

    4). Which parts of a molecule absorb infra-red?
    The covalent bonds

    5). Why do most infra- red spectra of organic molecules have a strong, sharp peak at around 3000cm^-1?
    Most organic molecules have a C-H bond.

    6). On an infra-red spectrum, what is meant by the 'fingerprint region'
    It is the region which is specific to the compound. Can someone explain this bit to me a bit more. Do we ever use it and if we do how so?

    Thanks
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    (Original post by Super199)
    Right can someone help me out it is to do with analytical techniques.

    1). What is meant by the molecular ion

    The molecular ion is a molecule that has lost an electron in the mass spectrometer forming a positive ion with a radical.

    2). What is the M peak?

    The M peak is the peak that represents the molecular ion with the biggest M/Z value

    3). How do fragments get formed?
    Not entirely sure, possibly something to do with the bombarding of the electrons.

    4). Which parts of a molecule absorb infra-red?
    The covalent bonds

    5). Why do most infra- red spectra of organic molecules have a strong, sharp peak at around 3000cm^-1?
    Most organic molecules have a C-H bond.

    6). On an infra-red spectrum, what is meant by the 'fingerprint region'
    It is the region which is specific to the compound. Can someone explain this bit to me a bit more. Do we ever use it and if we do how so?

    Thanks
    Fragments are indeed formed by the bombarding of electrons which takes place during ionisation. If the electron hits the molecule with enough energy it can potentially break the bonds forming a fragment. The fingerprint region is the region of an ir spec which is due to the stretching of the various bonds. The fingerprint region is formed of a large number of small absorptions and as far as I'm aware all you need to know is that this pattern is unique

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    (Original post by samb1234)
    Fragments are indeed formed by the bombarding of electrons which takes place during ionisation. If the electron hits the molecule with enough energy it can potentially break the bonds forming a fragment. The fingerprint region is the region of an ir spec which is due to the stretching of the various bonds. The fingerprint region is formed of a large number of small absorptions and as far as I'm aware all you need to know is that this pattern is unique

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    Got it! Thanks
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    I have been given a question to complete at home in preparation of an exam and I am struggling with some parts.

    A student investigated the reaction between calcium carbonate CaCO3 and nitric acid, HNO3 using the following apparatus.(Reaction mixture in a conical flask going to a gas syringe) They used a balance to measure out 2.00g of calcium carbonate lumps which was placed in the conical flask. They also added 50.0cm^3 of 0.250 moldm^-3 nitric acid to the conical flask before replacing the bung and using the gas syringe to measure the volume of gas produced.

    1)Write an equation for the reaction between calcium carbonate and nitric acid.
    I got CaCO3 + 2HNO3 -> CO2 + H20 + Ca(CO3)2

    2) Calculate the number of moles of calcium carbonate added to the flask.
    I got 2/(40.1+12+48) = 0.02mol
    3) Calculate the number of moles of nitric acid that was added to the flask.
    I got (50*0.250)/1000 = 0.0125mol
    4) Which Reagent was in excess?
    I got CaCO3 1:2 Mol ratio
    0.02:0.0125
    0.00625:0.0125
    So CaCO3 in excess (WHY?)

    5) Calculate the excess mass of CaCO3 that was added to the conical flask?
    How do i do this?
    6) How many moles of CO2 would you expect to be produced?
    Help Please?
    7) What volume of Carbon dioxide would be produced at RTP(room temp)
    Help Please.
    8) In each of the following cases state how the rate and volume of carbon dioxide would be affected by;
    increasing the mass of Calcium Carbonate.

    Using 50Cm3 of 0.5 moldm nitric acid rather than 0.25 moldm of nitric acid.

    Using the same mass of powdered caco3 rather than lumps.

    Any help I ould be greatful for sorry for such a large question but i only had one night to prepare.
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    (Original post by EwaHum)
    I have been given a question to complete at home in preparation of an exam and I am struggling with some parts.

    A student investigated the reaction between calcium carbonate CaCO3 and nitric acid, HNO3 using the following apparatus.(Reaction mixture in a conical flask going to a gas syringe) They used a balance to measure out 2.00g of calcium carbonate lumps which was placed in the conical flask. They also added 50.0cm^3 of 0.250 moldm^-3 nitric acid to the conical flask before replacing the bung and using the gas syringe to measure the volume of gas produced.

    1)Write an equation for the reaction between calcium carbonate and nitric acid.
    I got CaCO3 + 2HNO3 -> CO2 + H20 + Ca(CO3)2

    2) Calculate the number of moles of calcium carbonate added to the flask.
    I got 2/(40.1+12+48) = 0.02mol
    3) Calculate the number of moles of nitric acid that was added to the flask.
    I got (50*0.250)/1000 = 0.0125mol
    4) Which Reagent was in excess?
    I got CaCO3 1:2 Mol ratio
    0.02:0.0125
    0.00625:0.0125
    So CaCO3 in excess (WHY?)

    5) Calculate the excess mass of CaCO3 that was added to the conical flask?
    How do i do this?
    6) How many moles of CO2 would you expect to be produced?
    Help Please?
    7) What volume of Carbon dioxide would be produced at RTP(room temp)
    Help Please.
    8) In each of the following cases state how the rate and volume of carbon dioxide would be affected by;
    increasing the mass of Calcium Carbonate.

    Using 50Cm3 of 0.5 moldm nitric acid rather than 0.25 moldm of nitric acid.

    Using the same mass of powdered caco3 rather than lumps.

    Any help I ould be greatful for sorry for such a large question but i only had one night to prepare.
    1. Pretty sure it's typo but it's Ca(NO3)2 made

    4. To ensure all acid has reacted

    5. Calculate the mole that has reacted from the equation (which you have done). Subtract from 0.02 mol you calculated and multiply by the Mr of CaCO3.

    6. Ratio from the equation

    7. A mole of gas in standard condition occupies 24dm3

    8a no effect as the reaction is limited by moles of the acid
    b. Higher rate as higher concentration and more gas produced as the excess CaCO3 will now then react
    C. Higher rate as more surface area etc etc...

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    (Original post by C0balt)
    1. Pretty sure it's typo but it's Ca(NO3)2 made

    4. To ensure all acid has reacted

    5. Calculate the mole that has reacted from the equation (which you have done). Subtract from 0.02 mol you calculated and multiply by the Mr of CaCO3.

    6. Ratio from the equation

    7. A mole of gas in standard condition occupies 24dm3

    8a no effect as the reaction is limited by moles of the acid
    b. Higher rate as higher concentration and more gas produced as the excess CaCO3 will now then react
    C. Higher rate as more surface area etc etc...

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    Just awnsered in another thread

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    1) It should be Ca(NO3)2
    4) It was important that the calcium carbonate was in excess so the nitric acid fully reacted.
    5) You multiply the number of moles from the nitric acid by the molar mass of the CaCO3 to work out the exact mass required, rather than adding in excess..... 0.0125 / 2 (1:2 ratio) x 100 = 0.625g As 2g was used, 2-0.625=1.375g in excess
    6)Refer back to original equation. 2HNO3:1Co2, 2:1 ratio, meaning for every 2 moles of HNO3 used only 1 mole of CO2 is produced. 0.0125 moles of HNO3 were used, therefore 0.0125/2 = 0.00625 moles of CO2 produced.
    7) Moles x 24000 = volume in cm3, therefore 0.00625 x24000 = 150cm3
    8) Doubling the volume of HNO3 means that 50*0.5 /1000 = 0.025 moles of HNO3 used and 0.025/2 = 0.0125 moles of CO2 produced. 0.0125 x 24000 = 300cm3 of CO2 produced (doubles)
    Powdered CaCO3 would have no effect on the volume of CO2 produced, the reaction would just happen faster






    Hope this helps!!!
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    (Original post by C0balt)
    1. Pretty sure it's typo but it's Ca(NO3)2 made

    4. To ensure all acid has reacted

    5. Calculate the mole that has reacted from the equation (which you have done). Subtract from 0.02 mol you calculated and multiply by the Mr of CaCO3.

    6. Ratio from the equation

    7. A mole of gas in standard condition occupies 24dm3

    8a no effect as the reaction is limited by moles of the acid
    b. Higher rate as higher concentration and more gas produced as the excess CaCO3 will now then react
    C. Higher rate as more surface area etc etc...

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    So the moles of the CO2 would be 0.00625?
    How would i use the 24dm in question 7?
    Thanks for the help with the other questions.
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    (Original post by EwaHum)
    So the moles of the CO2 would be 0.00625?
    How would i use the 24dm in question 7?
    Thanks for the help with the other questions.
    Yes.
    24dm3/mol so just multiply by the moles of CO2

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    (Original post by mathscot)
    Just awnsered in another thread

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    Well thanks for letting me know....?

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    (Original post by C0balt)
    Well thanks for letting me know....?

    Posted from TSR Mobile
    you both answered at the same time sorry for taking both of your time up but thanks a lot to both all of you.
 
 
 
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