Join TSR now and get all your revision questions answeredSign up now
    Offline

    0
    ReputationRep:
    Guys I think I'm having a stupid moment but doing this empa past paper from june 2010 and can't seem to get the answer right for this question! The MS says the answer's 17.6, could anyone explain to me how they got that?

    "A saturated solution of ammonia contains 300 g of ammonia in 1.00dm3 of solution.

    Calculate the concentration, in mol dm–3, of ammonia in this solution."

    Offline

    14
    ReputationRep:
    Posted from TSR Mobile

    (Original post by sophiegashman)
    Guys I think I'm having a stupid moment but doing this empa past paper from june 2010 and can't seem to get the answer right for this question! The MS says the answer's 17.6, could anyone explain to me how they got that?

    "A saturated solution of ammonia contains 300 g of ammonia in 1.00dm3 of solution.

    Calculate the concentration, in mol dm–3, of ammonia in this solution."

    Pretty straightforward.

    First find the moles of NH3 using N = Mass/Mr ---------- 300g/17 gmol-1 = 17.647 mol

    Then use the equation C = N/V so 17.647 mol/1 dm3 volume (no convertion of units required = 17.6 mol dm-3 to 3 sig figs.

    Make sure you understand that I got the Mr from adding the Ar (or mass number) of Nitrogen which is 14 on the OCR table. With 3 x 1.0 (1.0 is the mass number of hydrogen, since it's Ammonia NH3. We multiply by 3). 14 + 3 = 17 g mol-1

    Remember the equations N = Mass/Mr for moles. N = C x V for solutions and N = V/Vm for gas volumes.

    Practice loads and you'll get better
    Offline

    5
    ReputationRep:
    Welp incoming mod to delete these posts, you guys know talking about the EMPA is against the rules so stop plserino thankerino.
    Offline

    14
    ReputationRep:
    (Original post by ozzie2)
    Welp incoming mod to delete these posts, you guys know talking about the EMPA is against the rules so stop plserino thankerino.
    Actually, it is only against the rules if you talk about EMPA/ISA specifics. That is, the questions that are on the paper. Talking about TOPICS is not against the rules unless explicitly stated that they were on the paper AND that significant help was administered to warrant a correct answer on that question.
    Offline

    3
    ReputationRep:
    This is a question from 2014 paper unit 1 I'm confused.

    The equation is 4Fe(NO3)3 ---> 2FeO3 + 12NO2 + 3O2

    4Fe(NO3)3 mass is 2.16g and Mr is 241.8

    it it said to work out moles of iron nitrate I got it as 0.00893n which was right

    then it said work out the moles of oxygen gas produced?
    I got the mole ratio 4:3 but how do you work out the moles in 3O2 so confused
    • Peer Support Volunteers
    • Clearing and Applications Advisor
    • Study Helper
    Offline

    18
    (Original post by samwillettsxxx)
    This is a question from 2014 paper unit 1 I'm confused.

    The equation is 4Fe(NO3)3 ---> 2FeO3 + 12NO2 + 3O2

    4Fe(NO3)3 mass is 2.16g and Mr is 241.8

    it it said to work out moles of iron nitrate I got it as 0.00893n which was right

    then it said work out the moles of oxygen gas produced?
    I got the mole ratio 4:3 but how do you work out the moles in 3O2 so confused
    Looking at the mole ratio - for every 4 moles of iron(III) nitrate, there are 3 moles of oxygen (or if it's easier, you will have 1 mole of iron(III) nitrate for 3/4 mole of oxygen). That means you need to multiply the number of moles of iron(III) nitrate by 3/4 to get the number of moles of oxygen.
    Offline

    3
    ReputationRep:
    (Original post by usycool1)
    Looking at the mole ratio - for every 4 moles of iron(III) nitrate, there are 3 moles of oxygen (or if it's easier, you will have 1 mole of iron(III) nitrate for 3/4 mole of oxygen). That means you need to multiply the number of moles of iron(III) nitrate by 3/4 to get the number of moles of oxygen.
    Oh I see thank you that makes sense now! I was dividing 4/3 instead of 3/4 lmao
    • Peer Support Volunteers
    • Clearing and Applications Advisor
    • Study Helper
    Offline

    18
    (Original post by samwillettsxxx)
    Oh I see thank you that makes sense now! I was dividing 4/3 instead of 3/4 lmao
    No worries.
    Offline

    3
    ReputationRep:
    (Original post by usycool1)
    No worries.
    So for the nitrogen dioxide 4/12 = 3
    3x 0.00893 is moles in nitrogen dioxide
    • Peer Support Volunteers
    • Clearing and Applications Advisor
    • Study Helper
    Offline

    18
    (Original post by samwillettsxxx)
    So for the nitrogen dioxide 4/12 = 3
    3x 0.00893 is moles in nitrogen dioxide
    Yup :yy:
    Offline

    3
    ReputationRep:
    (Original post by usycool1)
    Yup :yy:
    Thank you so much, very helpful!
    Offline

    2
    ReputationRep:
    (Original post by C0balt)
    The carbon which has hydroxyl group attached to it has three carbon atoms so there is no H atom on its own attached to the said carbon. Draw a tertiary alcohol and you will see. But I don't think the explanation is on any syllabus.

    The chromium (VI) in dichromate has orange colour and chromium (III) is green. It is reduced in oxidation of alcohol this is where the colour change comes from.

    Posted from TSR Mobile


    Posted from TSR Mobile
    Stuck on a physics question .
    Offline

    3
    ReputationRep:
    (Original post by Kadak)
    Posted from TSR Mobile
    Stuck on a physics question .
    Which?

    Posted from TSR Mobile
    Offline

    2
    ReputationRep:


    Posted from TSR Mobile

    I made a thread about it.
    Offline

    2
    ReputationRep:
    How do you know how many lone pair of electrons an element when it's connected to a compound, eg. the Oxygen in Phenol?
    without drawing a dot and cross diagram?
    Offline

    3
    ReputationRep:
    (Original post by Kadak)
    Posted from TSR Mobile

    I made a thread about it.
    I just responded :yy:

    Posted from TSR Mobile
    Offline

    1
    ReputationRep:
    (Original post by Georgiam247)
    Thanks for replying.

    No I understand but it says the method for the reaction is to bubble hydrogen halides (gases @ rt) into liquid alkANE, but if this happened no reaction would occur. The hydrogen halide gas should be bubbled into a liquid alkENE....right?
    I can see where you're coming from, but I'm presuming it means bubbled and turned into liquid alkane product?! But yes, they are reacted with alkenes and made into alkanes, so not very well written...

    Chemguide.co.uk says:
    Conditions
    The alkenes react with gaseous hydrogen halides at room temperature. If the alkene is also a gas, you can simply mix the gases. If the alkene is a liquid, you can bubble the hydrogen halide through the liquid.
    Offline

    0
    ReputationRep:
    (Original post by _NMcC_)
    Posted from TSR Mobile



    Pretty straightforward.

    First find the moles of NH3 using N = Mass/Mr ---------- 300g/17 gmol-1 = 17.647 mol

    Then use the equation C = N/V so 17.647 mol/1 dm3 volume (no convertion of units required = 17.6 mol dm-3 to 3 sig figs.

    Make sure you understand that I got the Mr from adding the Ar (or mass number) of Nitrogen which is 14 on the OCR table. With 3 x 1.0 (1.0 is the mass number of hydrogen, since it's Ammonia NH3. We multiply by 3). 14 + 3 = 17 g mol-1

    Remember the equations N = Mass/Mr for moles. N = C x V for solutions and N = V/Vm for gas volumes.

    Practice loads and you'll get better


    Sorry I was having a totally stupid moment and using the Mr of Ammonium instead of Ammonia! Thanks for the help though A-levels are most definitely frazzling my brain!
    Offline

    3
    ReputationRep:
    Can anyone help me out with percentage abundance questions? For isotopes, I am awful with maths.
    So for example 10B and 11B are 2 isotopes with an relative atomic mass of 10.8

    i I only got ratio 10:11 I don't know what to do next
    Offline

    5
    ReputationRep:
    (Original post by frozo123)
    How do you know how many lone pair of electrons an element when it's connected to a compound, eg. the Oxygen in Phenol?
    without drawing a dot and cross diagram?
    What I do is find the amount of electrons on the outer shell of the bonding atom.
    Then I work out the electrons that the bonding atom wants - eg chlorine wants 1 (as it will be sharing the same as the electron it wants in the bond if you get me)
    Then I add these both together to get the total electrons and divide by 2 as there are 2 electrons in a pair. You can work out how many lone pairs after taking away the shared electrons in the compound

    Eg NH3 : So nitrogen has 5 outer electrons. Hydrogen wants a total of 3 electrons.
    Now add these both together and there are 8 electrons so a total of 4 bonds. As there are 3 hydrogens there are 3 bonds with atoms so lose 6 electrons (2 electrons per pair) and you are left with 2 electrons, which are a lone pair as it doesn't bond with any atoms.

    Hope this helps, dunno if I even answered the question properly haha
 
 
 
Poll
If you won £30,000, which of these would you spend it on?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.