Paper 5 Tips !! :)

[lavanyamane]

Hello there! Can anyone please help me with Q.1 (i)? May seem silly, but I absolutely cannot manage to understand.
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s10_qp_51.pdf

 

[Wanderer]

use a divided flask. this type of flask has two sections. in one section we have catalyst and in another h202 so set the whole apparatus and for reaction to occur shake the flask so that the reagents come into contact!

Thanks a lot

 

[knowitall10]

Hello there! Can anyone please help me with Q.1 (i)? May seem silly, but I absolutely cannot manage to understand.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_51.pdf

Which question, what part?

 

[lavanyamane]

Which question, what part?

Q.1 (i) , already mentioned that, thanks a tonne!

 

[Ashique]

it is oct/nov 2009 variant 52.......nd plz tell me now what u think

It is because we're making a solution of 250 cm^3. The solid itself has volume. If you add a solid AND 250 cm^3 of water the total volume ends up being more than 250 cm^3.
In the s12 qp 51, we were dealing with concentrations- not volume! Concentration is a certain moles (or mass) of solid in a certain volume of water.

Hope you understood the difference!

 

[Ashique]

Q.1 (i) , already mentioned that, thanks a tonne!

Which part as in which part of the question b (i), c(i) or f(i)??

 

[lavanyamane]

Which part as in which part of the question b (i), c(i) or f(i)??

It's 1 (i) as in the alphabet "I", comes after (h), not the roman numeral for "1"

 

[Msbh22]

It's 1 (i) as in the alphabet "I", comes after (h), not the roman numeral for "1"

write a sensible value of the volume.. (within the range) example 50cm3, that could be easily measured and titrated in a lab.

 

[stella jazz]

ummmm well im havin ALOT of probs in dilutions, i just dont get it?!!!!

can u repost any that u found........there r wayy 2 many pages 2 go through...thanks

 

[pearl angel]

hello could anyone plz help me with Q1 (h),(i) and Q3 (c)
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s10_qp_51.pdf

 

[railey]

S/a, can someone please show me also the 2011 may/june paper 51 q1 c) diagram please.
Shukran.

 

[stella jazz]

@kazi umayer alim

In this experiment we get a precipitate. The prediction that we're putting into test is that the moles of of the precipitated copper hydroxide increases with the increasing conc. of CuSO4.

• So keeping our hypothesis in our mind, we should know that the conc. of CuSO4 is the independent variable, which means it is in our control. Hence, you need to prepare a range of concentrations of CuSO4. You're also aware of the fact that the solutions saturates at 1.39 mol/dm3. So the range of concentrations that you prepare must not exceed 1.39 mol/dm^3. So your range is going to be 0-1.39 mol/dm^3. You need to prepare at least five diluted solutions.
• Say you're going to prepare 100 cm^3 of each of the following conc. range: 0.01, 0.05, 1.00 1.10, 1.20 and 1.39. You need to tell the examiners exactly how you're going to prepare them. Since you're given SOLID hydrated CuSO4, you're going to have to dissolve a certain mass in water. So say you want to make the 0.01 mol/dm^3 solution- you need to know how many moles are are to dissolved in a 100 cm^3 solution. So Moles= CV= 0.01*100*10^-3= 0.01 mol. So you need to dissolve 0.01 mol of hydrated CuSo4 in 100 cm^3 of water. To find out the mass= Moles*Mr= 0.01*249.6= 2.50 g. So you dissolve 2.50 g of the salt into 100 cm^3 of water to make a 0.01 mol/dm^3 solution. The rest of the concentrations are to be prepared in the same way- but you don't need to show the working. One calculation for conc. should be enough.
• After you've added the NaOH into the solution a precipitation of Cu(OH)2 will occur. You need to filter this (Or centrifuge it,and then decant).You could leave it out to dry, or add water and propanone.

Thats about it! Hope it helped!

omg this helped a lot...finally i got it ...thankyouu

 

[Msbh22]

hello could anyone plz help me with Q1 (h),(i) and Q3 (c)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_51.pdf

question1 (H) Q=mc x change in temp.. where m is the total volume of acid and base ( as we assume 1cm3 as 1g)

this Q is the enthalpy change for the no. of moles of NaOH you used in the experiment. so to find enthalpy change neutralization, i,e part (i) find the enthalpy change for 1 mole using the above expression ( of part h)

 

[ultraviolet]

For collecting a volume of gas, can we simply use a syringe or do we have to collect it in a tube over water?
And can someone please tell me if these values are right:
beaker 250cm3
syringe 250 cm3
conical flask 200cm3
thermometer -10 - 110
testue 16 cm3

and can someone tell me if there are any other apparatus that i should know of. Thankyou

 

[Msbh22]

Can anybody tell me the capacities of different apparatus like burette pipette measuring cylinder, volumetric flask, simple beaker, conical flask???

 

[Ashique]

It's 1 (i) as in the alphabet "I", comes after (h), not the roman numeral for "1"

In part i, you're asked to derive an expression for enthalpy change of neutralization. This energy is going to be given off by the acid and the base reaction. You already know that the energy given by the reacition is (vol/mass NaOH + vol/mass H2SO4) × 4.3 × ∆T.
Look at the reaction. the ratio between NaOH and water is 1:1 (REMEMBER: The enthalpy change of neutralization is when 1 mole of water forms). So for the energy released by 1 mole of NaOH, 1 mole of H2O is produced. So if 'n' moles of NaOH were present, 'n' moles of water would form. Now, since enthalpy change of neutralization is the ENERGY when ONE MOLE of water is formed, we can derive our expression now. Energy was (vol/mass NaOH + vol/mass H2SO4) × 4.3 × ∆T per 'n' number of moles.
So the expression will be [(vol/mass NaOH + vol/mass H2SO4) × 4.3 × ∆T]/n
And this this is an EXOTHERMIC reaction, it will be -ve.

 

[iKhaled]

pls can someone show or explain how to do part 1. (e) and (d) of this question ?

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_52.pdf

 

[pearl angel]

question1 (H) Q=mc x change in temp.. where m is the total volume of acid and base ( as we assume 1cm3 as 1g)

this Q is the enthalpy change for the no. of moles of NaOH you used in the experiment. so to find enthalpy change neutralization, i,e part (i) find the enthalpy change for 1 mole using the above expression ( of part h)

thanx could u also help me with this same paper's Q3 (c)

 

[Ashique]

hello could anyone plz help me with Q1 (h),(i) and Q3 (c)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_51.pdf

For Q1 i, see my post above.
For 3 (c)
You have take any coordinate from the graph (except for the anomalous point). Say you took the coordinate (1.50, 2.45)
You have to calculate the no. of moles of both the Mg and MgO
So for Mg= 1.50/24.3= 0.06 mol
For MgO= 2.45/40.3= 0.06 mol

ratio 1:1

If we see the equation for the reaction of Mg with O2:
Mg + 1/2 O2 ----> MgO

You can see that the Mg and the MgO are in a 1:1 ratio. And we calculated the moles of a point, which was also 1:1, hence it corresponds to the formula of MgO

 

[Msbh22]

pls can someone show or explain how to do part 1. (e) and (d) of this question ?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_52.pdf

xMg+ y/2O2 = MgxOy

mass of O2 ... (2.45-1.5(from the graph)) = 0.95g
no. of moles of O2 0.03 moles
mass of Mg(from the graph) 1.5g
no. of moles of Mg 0.06moles

Mole Ratio ( O2;Mg) - 1;2.... so x=2 and y/2=1 so y=2

Mg2O2... simlplified form MgO