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

    2
    ReputationRep:
    (Original post by zahragoli97)
    1. In an experiment, 3.425 g of lead oxide was reduced to form 3.105 g of lead.The empirical formula of the lead oxide is A PbOB Pb3O2C Pb3O4D Pb4O3
    how to do this question??
    Find the mass of Oxygen = 3.425-3.105 = 0.32g

    Find the empirical formula. The ratio of Pb : O is 1:1.33. Therefore, multiply by a whole no. (3, according to the question), the ratio is now 3:4. Therefore, the formula is Pb3O4.
    Offline

    1
    ReputationRep:
    (Original post by katenell)
    thank youu but i didnot understand q10
    and i there a faster way for q 14 or is there any rule ?
    Q16 how the cycle should be drawn ?/
    you might be overthinking q10, it's just percentage yield from molar values. So if you work out the mol of the reactant and the product they are the same.
    In some cases you might get 0.5 mol forming 0.4 mol, so the percentage yield is 80%. But for q10, the reactant and product form the same mol value so 0.5/0/5 *100 = 100%

    Q14 think of the general formula for alkenes, its CnH2n. C2F4 follows the rule but the F has replaced all the H

    q16 okay forget about drawing it. Look at it this way, to form 1 mol of H2O(g) you need -242 kjmol-1 (from question, I just divided by 2) and to form H2O(l) its -286kjmol-1. So delta H = products (-286) - reactants (-242) = -286 - (-242) = -44
    Offline

    1
    ReputationRep:
    (Original post by :D :))
    you might be overthinking q10, it's just percentage yield from molar values. So if you work out the mol of the reactant and the product they are the same.
    In some cases you might get 0.5 mol forming 0.4 mol, so the percentage yield is 80%. But for q10, the reactant and product form the same mol value so 0.5/0/5 *100 = 100%

    Q14 think of the general formula for alkenes, its CnH2n. C2F4 follows the rule but the F has replaced all the H

    q16 okay forget about drawing it. Look at it this way, to form 1 mol of H2O(g) you need -242 kjmol-1 (from question, I just divided by 2) and to form H2O(l) its -286kjmol-1. So delta H = products (-286) - reactants (-242) = -286 - (-242) = -44
    The hess's cycle is bound to come up so learn how to do it. Go on chemguide, Jim Clark explains it well (in my opinion)
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by katenell)
    thank youu but i didnot understand q10
    and i there a faster way for q 14 or is there any rule ?
    Q16 how the cycle should be drawn ?/
    Q10) The percentage yield formula is: actual mass/theoretical mass x 100

    The theoretical mass in this case should be found according to ratios.
    Find the moles of NH3 : 8.5/17 = 0.5

    According to ratio of NH3 : NO which is 1:1, the moles of NO formed is also expected to be 0.5.

    Find the expected mass of NO : 0.5 x 30 = 15g.

    Now find % yield: 15/15 x 100 = 100%

    Q14) Yes there is, Memorize the general formula for Alkanes and Alkenes. Which CnH2n+2 and CnH2n respectively. So in this case, an Alkene ( Double Bond) is C2F4 whereby n is 2 and the H is replaced with F. That's all.

    Q16) there is no need to draw a cycle. Flip the 2nd equation and cancel the unwanted reactans and products that are equal on both sides:

    2H2 + O2 ---> 2H2O (l) = -572
    2H2O(g) ---> 2H2 + O2 = +484

    You can cancel 2H2 and O2 on both sides and you will remain with the overall equation: 2H2O (g) ----> 2H2O (g). Note that there are two moles in this equation and the equation in the question asks for only 1 mole of the reactants. The sign of the 2nd equation changed because of the exchanging the equation. Now just add the 2 enthalpies together and divide by two to get the enthalpy change for 1 mole. The answer is -572+484/2 = -44 kJmol^-1
    Offline

    2
    ReputationRep:
    Hello,

    Could anyone please help me out in Q17e from June 2015 GCE paper?
    Offline

    2
    ReputationRep:
    January 2011 Q16 b ii

    I would be thankful if someone would solve this question
    Offline

    3
    ReputationRep:
    (Original post by Rahatara Sadique)
    January 2011 Q16 b ii

    I would be thankful if someone would solve this question
    when i solved this paper, i wrote something like so: "add acid in small beaker. place the beaker into a larger beaker containing the coral. cover the beaker and shake gently to cause acid to spill and reaction to start.
    Offline

    2
    ReputationRep:
    1st and 2nd electron affinity. which one of them is endothermic and which one is exothermic?
    Offline

    2
    ReputationRep:
    (Original post by Nethmioysters)
    1st and 2nd electron affinity. which one of them is endothermic and which one is exothermic?
    1st is exothermic
    2nd is endothermic to overcome the repulsion between the electron and the negative charged ion
    hope this help you
    Offline

    2
    ReputationRep:
    (Original post by katenell)
    1st is exothermic
    2nd is endothermic to overcome the repulsion between the electron and the negative charged ion
    hope this help you
    yes it did! thanks alot!!!!
    Offline

    2
    ReputationRep:
    (Original post by katenell)
    1st is exothermic
    2nd is endothermic to overcome the repulsion between the electron and the negative charged ion
    hope this help you
    could you also state which of these are exothermic and which are endothermic?

    Combustion
    Formation
    Atomisation
    Lattice enthalpy
    Neutralization
    Ionization

    Thanks in Advance!!!
    Offline

    2
    ReputationRep:
    (Original post by Bliss_)
    when i solved this paper, i wrote something like so: "add acid in small beaker. place the beaker into a larger beaker containing the coral. cover the beaker and shake gently to cause acid to spill and reaction to start.
    Thankyou!
    Offline

    0
    ReputationRep:
    good luck everyone
    Offline

    0
    ReputationRep:
    Calculate the enthalpy change of the reaction, in kJ mol–1, to two significantfigures. Include a sign in your answer. 2NH4CNS(s) + Ba(OH)2(s) Ba(CNS)2(s) + 2H2O(l) + 2NH3(g) why you times the enthalpy value by 2?
    Offline

    5
    ReputationRep:
    (Original post by Nethmioysters)
    could you also state which of these are exothermic and which are endothermic?

    Combustion
    Formation
    Atomisation
    Lattice enthalpy
    Neutralization
    Ionization

    Thanks in Advance!!!
    Combustion is exothermic
    Formation is endo or exo
    Atomisation is endo
    Lattice enthalpy is exo
    Neutralistion is exo
    Ionization is endo
    Offline

    0
    ReputationRep:
    Ammonia is manufactured from hydrogen and nitrogen in the Haber process.3H2(g) + N2(g) 2NH3(g) If 60 tonnes of hydrogen produces 80 tonnes of ammonia, what is the percentageyield in the reaction? A 80 × 100%170 B 80 × 100%340 C 30 × 100%80 D 60 × 100%80 how to do this question|??
    Offline

    4
    ReputationRep:
    Paper went fine Alhamdulillah

    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    da best chem paper ever!!!!!!!
    😁😁😁
    hallelujah hallelujah
    guys we still have some hope to pass AS exam
    🎉🎉🎉🎉🎉🎉🎉🎉 🎉🎉🎉🎉
    Offline

    7
    ReputationRep:
    yeah it was easy af
    Offline

    3
    ReputationRep:
    It was good.
    What was the answer for the very last question about Shane gas?


    Posted from TSR Mobile
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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.