Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    14
    ReputationRep:
    (Original post by BrainJuice)
    I was thinking to use this type of method but the questions in my book weren't that complex and I didn't know how. Thanks!
    No problem. I think I did it slightly more complicated than needed though.

    This was another method used:
    "I can understand how you worked it out. I used another method by working out the number of moles using moles=mass/mr of baso4 then i wrote the equation bacl2+h2so4=baso4+2hcl and everything is a 1:1 ratio so the same moles of bacl2 were used. I worked out the mass of bacl2 from this using the mass=mr x moles. Then i took this away from the mass of both bacl2 and mgcl2 to work out how much mgcl2 there was. My answer was also 35.5%"
    Offline

    0
    ReputationRep:
    Our teacher wrote the exam for AQA (no he didn't give anything away) but he did say that if the grade boundaries were going to go anywhere, they would most likely be lower than 80% for an A
    Offline

    14
    ReputationRep:
    (Original post by hann990)
    Our teacher wrote the exam for AQA (no he didn't give anything away) but he did say that if the grade boundaries were going to go anywhere, they would most likely be lower than 80% for an A
    Sweet - I would also concur that from taking the exam.
    Offline

    15
    ReputationRep:
    (Original post by britishtf2)
    No problem. I think I did it slightly more complicated than needed though.

    This was another method used:
    "I can understand how you worked it out. I used another method by working out the number of moles using moles=mass/mr....
    See that was actually what I was trying to do at first but I didn't know the equation. Sadly I thought I only needed to know the observations.
    Offline

    6
    ReputationRep:
    If you use 1.434g as the weight then how would you get 149 as Mr? Also what is the actual answer?
    Offline

    0
    ReputationRep:
    (Original post by britishtf2)
    Sweet - I would also concur that from taking the exam.
    Technically he also wrote the next one and he loves works of hatred so I wouldn't hope for anything easier...!
    Offline

    14
    ReputationRep:
    (Original post by SM-)
    If you use 1.434g as the weight then how would you get 149 as Mr? Also what is the actual answer?
    Er, cos you get like 8.9*10^-4 (not actual, but ballpark) for mol HCl in 25cm^3.
    Then 8.9*10^-3 mol of MHCO3 in 250cm^3.
    Divide mass/mol and you get 149. With 8.9*10^-3 you get 161 but you get my drift.
    Offline

    14
    ReputationRep:
    (Original post by hann990)
    Technically he also wrote the next one and he loves works of hatred so I wouldn't hope for anything easier...!
    Good! Harder papers are better - usually lower boundaries and I like a challenge
    Offline

    9
    ReputationRep:
    (Original post by DaVinciGirl)
    If you know anymore questions or answers please reply and the number of marks.
    What is the full electronic configuration of Al? 1s2,2s2,2p6,3s2,3p1

    What is the full electronic configuration for Cr+3? 1s2,2s2,2p6,3s2,3p6,3d3

    What is the formula of the ion with same electronic configuration of Krypton?
    Sr2+ (strontium)

    What compound has a +2 and -3 ions and has the same number electrons as Argon?
    Ca3P2

    Which ion has a smaller atomic radius Mg2+ or Na+?
    Mg 2+
    because has the same number of electrons, same shielding but more protons. So greater nuclear charge and attraction for outer electrons.

    What is the first ionisation energy of Na?
    Na (g) --> Na+ (g) + e-
    Diagram-Mark where the Mg or was it Si?
    Was and then mark the point where Sulphur was on the First ionisation graph.
    Mg/Si should have been above so the line was increasing.
    Sulphur should have been below Phosphorus but above Silicon.

    Why does Sulphur have a different first ionisation energy compared to Phosphorus?
    -Both have same shielding and in the same orbital the 3p orbital
    -Sulphur is paired with an electron in the 3p orbital whilst Phosphorus isn’t
    -There’s repulsion between the negatively charged electrons
    -So electrons are easier to remove

    What is the method of making a standardised solution of 250cm^3? (6 marks)
    -Weigh out known mass on weighing boat using a mass scale
    -Remember to deduct the mass of the weighing boat to find the exact mass of the solid. (or zero the scale mass)
    EDIT: You may not need this part as you've been told the known mass
    -Then transfer the contents into a beaker. Use distilled water to make sure all solids are in the beaker.
    -Then add bit more distilled water
    -Stir with a glass rod to make sure it's all mixed
    -Then pour solution into a Volumetric flask using a funnel
    -Clean the funnel with distilled water
    -Then fill the volumetric flask to the 250cm^3 mark. At near the mark use a pipette to add water dropwise and until meniscus is on the mark line.
    Then place stopper on it. Twist the volumetric flask a few times to ensure it's all mixed.

    Find mean titre value.
    Ignore the first number as it was the rough one when calculating mean.
    EDIT: You were only supposed to use 2 values that are 0.1 part when calculating mean.

    What was the concentration of the solution.

    Student titrated, what should she to improve her methods so there's a smaller percentage uncertainty?
    -Titrate a greater volume of solution
    -Why? So there's a greater value as percentage uncertainty= uncertainty/value *100. Bigger volume means there's a greater titre value.

    Why does Bromine have lower boiling point than Magnesium?
    -Bromine has a simple molecular structure.
    -There’s weak Van der Waals forces between molecules.
    -So smaller amount of energy needed to break these forces.
    -Magnesium has a metallic lattice.
    -It has many strong metallic bonds.
    -Bonds formed from the strong attraction between positiveions and delocalised electrons.
    -The metallic bonds are stronger than the Van der Waals forces

    How do the ions get accelerated, detected by ToF and how abundanceis measured?
    - Accelerated by negative electric field
    -The ions reach the detector and generate a small current, which is fed to a computer for analysis. The current is produced by electrons transferring from the detector to the positive ions. The size of the current is proportional to the abundance of the species.

    Calculation to find mass of one ion.
    1/Avogadro constant * mass. Then divide by 1000 to put it in Kg.
    Using kinetic energy, find velocity.
    Starts with 1........ Quite a large number.

    Draw the spectrum of Br2 with two isotopes of equal abundance.
    Attachment 539511

    Suggest why Magnesium has a longer melting point than Bromine. (Something like that?)
    -It has much more bonds than Bromine. So more bonds are needed to be broken when melting.
    a) Calculate the mass of Ammonia in Q?
    Basically use the ideal gas equation and usethe temp given, multiply the pressure by 1000 to convert it to SI Units, divide the volume by a million to convert to m^3 . Then use everything given to find the number of moles in Ammonia. Then find the mass of Ammonia using the mr.
    b) Calculate the volume in P given that the temperature decreased by 5 degrees and the total pressure is 75kPa?
    -Find volume of the whole system with the new values and use moles from previous question
    -Then subtract from the volume of Q given.

    Draw NH3 and AlCl3 and write the bonding angle.
    NH3- 1 Lone pair and 3 bonding pair. Should have drawn a trigonal pyrimidal.
    Bond angle- 107

    AlCl3- 3 bonding pair so Trigonal Planar
    Bond angle- 120

    What type of bond between NH3 and AlCl3 to make 3HNAlCl3?
    Dative covalent
    Electron pair donated from NH2 to AlCl3.

    What is the half equation of Chlorate(I) into Chlorine?
    2ClO- + 4H+ +2e- ---> Cl2 + 2H20

    What is the redox equation of Sodium Chlorate(I) and HCl?
    I have no idea.

    What's the observation when colourless Potassium iodide and Sodium Chloride are mixed and what reaction occurred for this to be seen?
    There was a brown solution seen, as Iodine ion was reduced to Iodine.

    Question on Anhydrous Magnesium Chloride and Anhydrous Barium Chloride mixture. Water was added and Sulphuric acid too. White precipitate was formed. What was this an find percentage of Magnesium in mixture?
    Barium Sulphate.
    Calculation:

    Multiple Choices:
    -Brown Colour is from the Reactant and forward is exothermic.
    -Empirical of hydrocarbon with 9/11th Carbon- C3H8
    -Doubts about biggest dipole. I put down BF3 and checked in my CGP book and states that bonds with electronegativity differences would make a dipole. Others argue that its Carbon one because BF3 is not polar.
    -Group 2's ionisation energy decreases downwards
    -There's 50cm^3 of gas in the end.
    -Gold is 19.3
    -SiO2 was Macromolecular
    -Balance was 4:7
    -The reducing agent was C204
    -How much sulphuric acid needed? D
    -Heavier isotope took longer in ToF
    -H2 Molecules was 1.81* 10^24
    good mark scheme and i think its bf3 as its not because the dipoles all balance out but it still has dipoles its just that those dipoles are on each side
    Offline

    14
    ReputationRep:
    (Original post by GabbytheGreek_48)
    good mark scheme and i think its bf3 as its not because the dipoles all balance out but it still has dipoles its just that those dipoles are on each side
    I also put BF3. CF4 will have balanced dipoles and the others I cannot remember, but I think had similar e.negativities.
    Offline

    9
    ReputationRep:
    (Original post by Cinna21)
    Are you sure its Mg2+ I'm sure It's Na+ which has the smaller ion
    same number of electrons so same amount of energy shells but magnesium has more protons causeing electron shells to be drawn in more
    Offline

    6
    ReputationRep:
    I just used a calculator and keep getting Mr as 145 or 146 when using three or the two titres and using the mass of 1.434g
    Offline

    9
    ReputationRep:
    (Original post by britishtf2)
    I also put BF3. CF4 will have balanced dipoles and the others I cannot remember, but I think had similar e.negativities.
    plus even if permamnent dipoles dont form temporary dipoles always do and in bf3 the temporary dipoles would be strongest
    Offline

    14
    ReputationRep:
    (Original post by SM-)
    I just used a calculator and keep getting Mr as 145 or 146 when using three or the two titres and using the mass of 1.434g
    I used 3 of the titres. I think you are supposed to use 2 (to get 9.70cm^3 avg). I don't know what you get if you use 2 of them. I suspect you are right - it's in the ballpark area so you'll get method marks if not all the marks.
    Offline

    9
    ReputationRep:
    (Original post by liziepie)
    I did the PQ question differently too, not sure if it was right. Use the moles from the previous part to work out the new number of moles in Q given the new temp and pressure. Then, take the new moles away from the old moles value to give you the moles in P, then work out V=nrt/p with the new value. Got something like 377cm3 which was less than 1000cm3 in Q so it sounded right
    probably right because my classmates were like its in a vacccummn so that means the same moles
    Offline

    9
    ReputationRep:
    (Original post by Science_help)
    for the potassium iodide question, isn't a black solid produced - iodine
    either theyll accept iodine solid is black but then u get some scientists saying brownish so theyll probably accept both
    Offline

    4
    ReputationRep:
    Didn't the kinetic energy question come up in the set 2 specimen? Isn't it the exact same method?
    Attached Images
     
    Offline

    9
    ReputationRep:
    ok i think i lost about 15 marks but giving myself some extra just incase i misss calculated so about 20 but 80 ums might not be 80% so hopefullyyyyy its either 15 marks lost or less than 80% for 80 ums
    Offline

    14
    ReputationRep:
    (Original post by GabbytheGreek_48)
    ok i think i lost about 15 marks but giving myself some extra just incase i misss calculated so about 20 but 80 ums might not be 80% so hopefullyyyyy its either 15 marks lost or less than 80% for 80 ums
    It will be no more than 80% for an A, for sure, and I wouldn't be surprised if 70-75% for an A.
    Offline

    9
    ReputationRep:
    (Original post by britishtf2)
    It will be no more than 80% for an A, for sure, and I wouldn't be surprised if 70-75% for an A.
    i hope but even so ill just expect the worst so i get my stuff together and smash paper 2
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?

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

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