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

    2
    ReputationRep:
    (Original post by tcameron)
    why just the moles of chromium metal? I found the moles of Cr2(SO4)3
    How do you do that?
    Offline

    2
    ReputationRep:
    (Original post by Kamara7)
    I literally just did this paper and finished marking it XD I also got the exact same answer as you :lol: Did you get to the 4th step in itsConnor_'s working out above? If so, then you don't need to divide by two afterwards because you're finding the % mass of chromium and not the chromium-containing compound (Fe(CrO2)2) and as you can see in the equation, the moles of Cr are the same on both sides of the equation (4 moles on each side), so you don't need to multiply/divide by anything as you already got the moles of Cr (0.034)! I don't know if you did the same mistake as me though, so it could be something else.
    Ah thankyou! Yeah I had divided by two because the ratio of fe(cro)2 to the sodium compound was 1:2, but now I can see that the moles of Cr are actually the same
    Offline

    2
    ReputationRep:
    (Original post by AqsaMx)
    Could you explain this please?
    Its fairly straightforward,

    First write out the Ka equation for the dissociation of methanoic acid

    Ka = [H+][HCOO-]/[HCOOH]

    and from the question you have been given the concentrations of the acid, the salt and the Ka value

    0.100 mol dm–3 with respect to sodium methanoate
    0.200 mol dm–3 with respect to methanoic
    The Ka of methanoic acid is 1.8 × 10–4 mol dm–3

    Rearrange the equation to get [H+] and plug in the given values, then use -log([H+])
    Offline

    3
    ReputationRep:
    (Original post by RetroSpectro)
    Its fairly straightforward,

    First write out the Ka equation for the dissociation of methanoic acid

    Ka = [H+][HCOO-]/[HCOOH]

    and from the question you have been given the concentrations of the acid, the salt and the Ka value

    0.100 mol dm–3 with respect to sodium methanoate
    0.200 mol dm–3 with respect to methanoic
    The Ka of methanoic acid is 1.8 × 10–4 mol dm–3

    Rearrange the equation to get [H+] and plug in the given values, then use -log([H+])
    I think I misred the question, I thought It said making a buffer solution so I started doing the excess-limiting thing oops, thanks for clearing that up

    How do you know when you're supposed to do the limiting calculation, will the question be phrased differently?
    Offline

    2
    ReputationRep:
    (Original post by AqsaMx)
    I think I misred the question, I thought It said making a buffer solution so I started doing the excess-limiting thing oops, thanks for clearing that up

    How do you know when you're supposed to do the limiting calculation, will the question be phrased differently?
    In those questions you are usually given the volume and concentration of both an acid and the base. This allows you to calculate the moles, to find that the acid moles will be in excess.

    In this question you had no information about the volume of the solution or any other reagent.

    I think the best way to go about it is to see if it is possible to calculate the moles of each reagent
    Offline

    3
    ReputationRep:
    Name:  image.jpeg
Views: 157
Size:  169.5 KB

    Could someone explain the actual answer of this question please?
    Offline

    3
    ReputationRep:
    (Original post by RetroSpectro)
    In those questions you are usually given the volume and concentration of both an acid and the base. This allows you to calculate the moles, to find that the acid moles will be in excess.

    In this question you had no information about the volume of the solution or any other reagent.

    I think the best way to go about it is to see if it is possible to calculate the moles of each reagent
    Oh okay thank you v much
    Offline

    2
    ReputationRep:
    (Original post by AqsaMx)
    Name:  image.jpeg
Views: 157
Size:  169.5 KB

    Could someone explain the actual answer of this question please?
    You are unable to determine the enthalpy changes that occur in the CO3 2- ion as it consists of more than one element

    I think thats one way you can interpret it but im sure theres more answers
    Offline

    3
    ReputationRep:
    With equilibrium questions for example - 2NO2 > N2O4
    If pressure increases, why does the conc of NO2 increase more and then decreases?

    Is it because initially when we increase pressure, the side with more moles with increase and those with the least moles decrease and the equilibrium shifting happens to increase pressure? Confused with this
    Offline

    3
    ReputationRep:
    (Original post by RetroSpectro)
    You are unable to determine the enthalpy changes that occur in the CO3 2- ion as it consists of more than one element

    I think thats one way you can interpret it but im sure theres more answers
    So would you need to work out the enthalpy change of formation of CO32- also?
    Offline

    3
    ReputationRep:
    (Original post by ImNervous)
    Cell potential increases*

    Posted from TSR Mobile
    why would oxidation increase if the cell is going in the reduction process?
    Offline

    3
    ReputationRep:
    (Original post by cr7alwayz)
    Can anyone show how you would work out the enthalpy change of neutralisation for a reaction between 50cm3 of 0.1M H2SO4 and 50cm3 of of 0.1M NaOH with specific heat capacity of 4.18 and temp change from 20 to 30?

    I know the equation would be H2S04 + 2NaOH -> NA2SO4+ 2H20 and Q = -4180J
    What would you do next?
    Hey not sure if you got a reply:

    Q = 100 x 4.18 x 10

    Delta H nuet = -q/n

    So -4000/ (0.1 x0.05) equals the answer
    Offline

    2
    ReputationRep:
    (Original post by AqsaMx)
    So would you need to work out the enthalpy change of formation of CO32- also?
    Yea you would need to calculate the formation of the Carbonate anion
    Offline

    15
    ReputationRep:
    A quick query about question 4 e (iii) of the June 2015 F325 paper
    It tells you the Ka for propanoic acid is 1.35x10-5 and that both the concentration of the acid + its conjugate base are 1 mol dm-3

    The student then adds 6.075g of Mg to 1.00dm3 of this buffer and you have to calculate the new pH.
    If you write out the acid-metal reaction it is clear that the acid:Mg ratio is 1:2 so once the moles of Mg are calculated (6.075/24.3 =0.25) the actual moles of acid used are 0.25x2 so 0.5.

    Am I correct in assuming the salt of the acid that is formed could also be considered the conjugate base of the acid and therefore the salts concentration would also be used on top of the original salt concentration in a buffer calculation? And if so as the salt is (C2H5COO-)2Mg if there were 0.25 mol of that then in 1dm3 there would be 0.5mol dm-3 of C2H5COO- from this salt + the original 1 mol dm-3 C2H5COO- already in the reaction?
    Offline

    3
    ReputationRep:
    (Original post by RayMasterio)
    why would oxidation increase if the cell is going in the reduction process?
    Cu2+ + 2e- <-> Cu (s)
    That is the eq on the table.
    Adding H2O, decreases conc of Cu2+.
    As its eq, the eq will shift to left to compensate. Therefore its more:

    Cu (s) -> Cu2+ + 2e-

    Hence, more oxidiation.

    But the thing to understand I think is that, less Cu2+,so eq shifts to the left.

    This means, the EP value decreases and so overall cell potential between Ag and Cu will increase due to the larger difference in EP between the two.

    Hope that makes sense. (Im literally explaining this from wha I figured from nervous' answer, literally just understood all this from him, thought id give him a break though )
    Offline

    3
    ReputationRep:
    (Original post by VMD100)
    A quick query about question 4 e (iii) of the June 2015 F325 paper
    It tells you the Ka for propanoic acid is 1.35x10-5 and that both the concentration of the acid + its conjugate base are 1 mol dm-3

    The student then adds 6.075g of Mg to 1.00dm3 of this buffer and you have to calculate the new pH.
    If you write out the acid-metal reaction it is clear that the acid:Mg ratio is 1:2 so once the moles of Mg are calculated (6.075/24.3 =0.25) the actual moles of acid used are 0.25x2 so 0.5.

    Am I correct in assuming the salt of the acid that is formed could also be considered the conjugate base of the acid and therefore the salts concentration would also be used on top of the original salt concentration in a buffer calculation? And if so as the salt is (C2H5COO-)2Mg if there were 0.25 mol of that then in 1dm3 there would be 0.5mol dm-3 of C2H5COO- from this salt + the original 1 mol dm-3 C2H5COO- already in the reaction?

    I think you're right.

    (Original buffer solution): C2H5COOH -> C2H5COO- + H+
    Moles: 1[C2H5COOH] 1[C2H5COO- ]

    After addition of Mg:

    2C2h5COOH + Mg -> (C2H5COO-)2Mg2+ + H2
    moles reacted: 0.5[C2h5COOH] 0.25[Mg] 0.25[(C2H5COO-)2Mg2+]

    1: (C2H5COO-)2Mg2+ -> 2C2H5COO- + Mg2+
    Moles: 0.25[(C2H5COO-)2Mg2+] 0.5[2C2H5COO-] 0.25[Mg2+]

    2: (New buffer solution) : C2H5COOH -> C2H5COO- + H+
    Moles: 1-0.5[ C2H5COOH ] 1+0.5[C2H5COO-]
    =0.5[ C2H5COOH ] 1.5[C2H5COO-]


    edit:sorry about the terrible format
    Offline

    2
    ReputationRep:
    (Original post by AqsaMx)
    With equilibrium questions for example - 2NO2 > N2O4
    If pressure increases, why does the conc of NO2 increase more and then decreases?

    Is it because initially when we increase pressure, the side with more moles with increase and those with the least moles decrease and the equilibrium shifting happens to increase pressure? Confused with this
    You have to look at this in terms of the Kc expression

    Kc = [N2O4] / [NO2]^2

    Say we increase pressure by times 2. This means that

    Kc = 2 / 2^2
    = 2 / 4

    Clearly we can see that the bottom of the Kc expression has increased by more than the top. However, the value of Kc cannot change in regards to pressure, therefore the equilibrium shifts right to increase N2O4 and restore the value of Kc.

    Hope this helped
    Offline

    3
    ReputationRep:
    Hello peeps, could someone please explain how in electrochemical cells we decide which half cell is negative electrode and which is positive?

    So when drawing a diagram of two half cells, which one goes on left or right?

    The question is from Jan 12

    *not letting me upload pic*
    Offline

    2
    Do we need to understand how basic buffers work? The textbook only mentions acidic buffers
    Offline

    2
    ReputationRep:
    Chatzy.com/11324377761338

    Revision squad for F325, for quicker questions, did it for f324 worked well.
    Please join
 
 
 
Poll
Who is your favourite TV detective?

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.