Paper 5 Tips !! :)

[pearl angel]

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

thanxx alot

 

[Starry_night]

please could somebody help me with Q. 1 d) i) on how to find the concentration of H2O2 when volumes of H2O2 and H2O are chosen.
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_53.pdf
Thank youu!

 

[Ashique]

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

For the diagram:

Replace the Copper with Lead though!

This is how you will heat the Lead oxides, with the excess hydrogen burning at the end of the tube. And instead of pipe to the gas tap, you should attach a syringe. And in another diagram on the same page, you should draw a a test tube, with a rubber bung, attached to the same syringe. In the contents of the tube, mention a gorup II metal (For eg: Mg) + a suitable acid, like HCl. This will liberate H2 gas into the syringe, which you will then attach to the apparatus shown above.

 

[Msbh22]

please could somebody help me with Q. 1 d) i) on how to find the concentration of H2O2 when volumes of H2O2 and H2O are chosen.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_53.pdf
Thank youu!

let the total volume.. 50cm3
for 2.00mol/dm3.. you need 50 cm3 of aq hydrogen per oxide...
for 1.8mol/dm3 .. you need x cm3 of aq hydrogen per oxide..

2.00 ------> 50
1.8 -------> X

X x 2 = 50 x 1.8
find the value of X>. you will get the volume of H2O2 for 1.8mol/dm3 concentration
subract 50-X to get the volume of water

 

[pearl angel]

could anyone upload a picture or help me understand what type of diagram we have to draw in Q1(c)
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s10_qp_52.pdf

 

[Starry_night]

let the total volume.. 50cm3
for 2.00mol/dm3.. you need 50 cm3 of aq hydrogen per oxide...
for 1.8mol/dm3 .. you need x cm3 of aq hydrogen per oxide..

2.00 ------> 50
1.8 -------> X

X x 2 = 50 x 1.8
find the value of X>. you will get the volume of H2O2 for 1.8mol/dm3 concentration
subract 50-X to get the volume of water

Thanks a lot!

 

[iKhaled]

For the diagram:

Replace the Copper with Lead though!

This is how you will heat the Lead oxides, with the excess hydrogen burning at the end of the tube. And instead of pipe to the gas tap, you should attach a syringe. And in another diagram on the same page, you should draw a a test tube, with a rubber bung, attached to the same syringe. In the contents of the tube, mention a gorup II metal (For eg: Mg) + a suitable acid, like HCl. This will liberate H2 gas into the syringe, which you will then attach to the apparatus shown above.

thankssss!

 

[lavanyamane]

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.

Thank you so much! I was looking for the little connecting bits and you explained it quite nicely

 

[railey]

Omg.. HELP HELP HELP! Please october november 2011 1 d) omg!

 

[Ashique]

Omg.. HELP HELP HELP! Please october november 2011 1 d) omg!

Apparatus is hot, so handle with heat proof gloves.

 

[railey]

Apparatus is hot, so handle with heat proof gloves.

Sorry bro i meant c not d! please if u can help

 

[Ashique]

Sorry bro i meant c not d! please if u can help

It's not that hard. You simply have to follow the given guidelines. Anyway, I will list down the major points.

In this experiment, we are to show how the solubility of K(NO3)2 varies with temperature. From the previous part of the question you must already know that the temperature is the independent variable (since it's in our control), and the solubility is the dependent variable.

• The first bullet point asks you that you should "ensure a wide range of results suitable for analysis by graph". They're basically asking what range are you going to provide for you independent variable. Basically when we conduct these experiments, we usually start from 0-100 degrees. The ms wants the candidates to give a range of at least 40 degrees.
• The second bullet point wants to know what mass of K(NO3)2 and volume of what you should use. Remember the independent variable in this experiment is the temperature. So temperature is only variable you're allowed to change. ALL the other variables are to be kept in control (needs to be constant). So you need a constant mass of K(NO3)2 (say 2 g) and a constant volume of water. (around 100 cm^3)
• The third one wants you to describe how you will "measure the amounts of the two reagents". This is an easy one. Measure the K(NO3)2 using a balance measuring up to 0.02g, and volume of water by a measruing cylinder measruing up to 1 cm^3
• The fourth point: "heat the apparatus". You could describe how you could use a thermostatically controlled water bath. You also have to mention that you have to stir the solution (they already gave you a wire stirrer- a BIG clue)
• In the fifth point, you need to describe what you would do. So say you heated the solution to 10 degrees- and after about 2-5 mins you took your boiling tube out, and filtered the solution and dried it (water and propanone), then you reweigh the remaining salt and subtract it from the initial mass. And you'll have your solubility of the solution for that particular temp. Do this for the rest of the values in your temp range.

Hope this helped!

 

[nidanas]

anybdy?!
cud sme1 plz tell me d apparatus whch z to be drawn for ques 1d in oct-nov year 12 variant 2 ?!
here z d link ----> http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_52.pdf

 

[railey]

Thank you very much. It helped a lot but could you please explain in more detail the last bullet point on how to decide when the point of the temperature of the solution is to be taken please

It's not that hard. You simply have to follow the given guidelines. Anyway, I will list down the major points.

In this experiment, we are to show how the solubility of K(NO3)2 varies with temperature. From the previous part of the question you must already know that the temperature is the independent variable (since it's in our control), and the solubility is the dependent variable.

• The first bullet point asks you that you should "ensure a wide range of results suitable for analysis by graph". They're basically asking what range are you going to provide for you independent variable. Basically when we conduct these experiments, we usually start from 0-100 degrees. The ms wants the candidates to give a range of at least 40 degrees.
• The second bullet point wants to know what mass of K(NO3)2 and volume of what you should use. Remember the independent variable in this experiment is the temperature. So temperature is only variable you're allowed to change. ALL the other variables are to be kept in control (needs to be constant). So you need a constant mass of K(NO3)2 (say 2 g) and a constant volume of water. (around 100 cm^3)
• The third one wants you to describe how you will "measure the amounts of the two reagents". This is an easy one. Measure the K(NO3)2 using a balance measuring up to 0.02g, and volume of water by a measruing cylinder measruing up to 1 cm^3
• The fourth point: "heat the apparatus". You could describe how you could use a thermostatically controlled water bath. You also have to mention that you have to stir the solution (they already gave you a wire stirrer- a BIG clue)
• In the fifth point, you need to describe what you would do. So say you heated the solution to 10 degrees- and after about 2-5 mins you took your boiling tube out, and filtered the solution and dried it (water and propanone), then you reweigh the remaining salt and subtract it from the initial mass. And you'll have your solubility of the solution for that particular temp. Do this for the rest of the values in your temp range.

Hope this helped!

 

[Ashique]

Thank you very much. It helped a lot but could you please explain in more detail the last bullet point on how to decide when the point of the temperature of the solution is to be taken please

The mark scheme has four ways to explain it:

Alternate 1
(vi) Heat mixture to dissolve all the solute.
(vii) Cool and measure the temperature at which first crystals
appear.

OR Alternate 2
(vi) Heats mixture to a particular temperature.
(vii) Filters the solution (not cooled or decanted) and weighs
the residue.

OR Alternate 3
(vi) Heats mixture to a particular temperature.
(vii) filters the solution (not cooled or decanted) and evaporates
the filtrate and weighs solid.

OR Alternate 4
(vi) Heats mixture to dissolve the solute.
(vii) Records temperature at which the solute dissolves.

I used "Alternate 2" because I understand it the best. See, let's go back to what hypothesis that we're testing. We're testing the solubility of K(NO3)2 at different temperatures. So the fifth point asks at what point when the temperature be taken. Remember we used a thermostatically controlled water bath? What you could also do is you could heat using a bunsen burner to a certain temperature (say 10 degrees). Also remember that our initial mass of the salt was 2.00 g. So you heated the solution to 10 degrees (while stirring). And after that you filtered the solution, and you reweighed your residue, and say you had 1.80 g left. This means that only 0.20 g of the salt had dissolve. Then you try it again with 20 degrees, and the mass of your residue was 1.50g (0.50 g had dissolved).

 

[Ashique]

anybdy?!
cud sme1 plz tell me d apparatus whch z to be drawn for ques 1d in oct-nov year 12 variant 2 ?!
here z d link ----> http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_52.pdf

I posted it above:

For the diagram:

Replace the Copper with Lead though!

This is how you will heat the Lead oxides, with the excess hydrogen burning at the end of the tube. And instead of pipe to the gas tap, you should attach a syringe. And in another diagram on the same page, you should draw a a test tube, with a rubber bung, attached to the same syringe. In the contents of the tube, mention a gorup II metal (For eg: Mg) + a suitable acid, like HCl. This will liberate H2 gas into the syringe, which you will then attach to the apparatus shown above.

 

[nidanas]

I posted it above:

yeah I saw dat..thnx a lott for ur help!
and cud u plz tel me d ans for May june 2011 ques:2c ?!
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_52.pdf

 

[knowitall10]

For the diagram:

Replace the Copper with Lead though!

This is how you will heat the Lead oxides, with the excess hydrogen burning at the end of the tube. And instead of pipe to the gas tap, you should attach a syringe. And in another diagram on the same page, you should draw a a test tube, with a rubber bung, attached to the same syringe. In the contents of the tube, mention a gorup II metal (For eg: Mg) + a suitable acid, like HCl. This will liberate H2 gas into the syringe, which you will then attach to the apparatus shown above.

We can replace the test tube in which H2 is liberated with a conical flask right? because i realy don't like to draw apparatus in mid-air

 

[railey]

Ohh makes sense now. The bullet point in the question is kind of misleading. Anyway i get your point.
Thanks a lot..
Please pray for all of us. Insallah we will all be fine!

The mark scheme has four ways to explain it:


I used "Alternate 2" because I understand it the best. See, let's go back to what hypothesis that we're testing. We're testing the solubility of K(NO3)2 at different temperatures. So the fifth point asks at what point when the temperature be taken. Remember we used a thermostatically controlled water bath? What you could also do is you could heat using a bunsen burner to a certain temperature (say 10 degrees). Also remember that our initial mass of the salt was 2.00 g. So you heated the solution to 10 degrees (while stirring). And after that you filtered the solution, and you reweighed your residue, and say you had 1.80 g left. This means that only 0.20 g of the salt had dissolve. Then you try it again with 20 degrees, and the mass of your residue was 1.50g (0.50 g had dissolved).

 

[knowitall10]

The mark scheme has four ways to explain it:


I used "Alternate 2" because I understand it the best. See, let's go back to what hypothesis that we're testing. We're testing the solubility of K(NO3)2 at different temperatures. So the fifth point asks at what point when the temperature be taken. Remember we used a thermostatically controlled water bath? What you could also do is you could heat using a bunsen burner to a certain temperature (say 10 degrees). Also remember that our initial mass of the salt was 2.00 g. So you heated the solution to 10 degrees (while stirring). And after that you filtered the solution, and you reweighed your residue, and say you had 1.80 g left. This means that only 0.20 g of the salt had dissolve. Then you try it again with 20 degrees, and the mass of your residue was 1.50g (0.50 g had dissolved).

How do heat the solution to 10^C? can you please elaborate on that? Nice answers though