Qualification > Sciences
Need help on CIE Chemistry Paper 5!
Deadly_king:
--- Quote from: Greed444 on October 03, 2010, 12:16:19 pm ---i understand now why we have to use <0.06 moles.
does the [acid] depends mainly only on 1 mole of H+ but not 2 moles of H+?
is that why we have to divide by 2?
how bout for triprotic acid like phosphoric acid?
H3PO4 --> 3H+ + PO4^3- (am i wrong? correct me if im wrong plz ^_^'')
so does that mean we must use less than 0.06/3? since 1mole of the acid provides 3 moles of H+?
--- End quote ---
Yupz........it usually depends on that :)
Basically this is the principle!
I can see that I have been a good tutor :P
Deadly_king:
--- Quote from: Greed444 on October 03, 2010, 12:30:06 pm ---Thanks Bro!!
I finally get it!
so it doesn't matter if i used diffrent volumes and conc and get half of 63g or twice the 63g?
when i use 20cm^3 and 1.0 mol/dm^3, n=0.02 which is <0.06/2,
so to produce 1,0mol/dm3 in 250cm3, 0.25moles needed
and i get mass=31.5g
--- End quote ---
You're welcome pal :)
You are a very good student.......... :D
Yeah it's alright! But it would be best to calculate a solution using the minimum mass possible ;)
Greed444:
need helps on May/June 2010 paper52 Q1(c) and (d)
Q1(c)
what do we have to do draw here? i simply drew a solution-filled beaker on top of a stand with a bunsen burner underneath it for boiling. and also a thermometer inserted into the beaker. Is this a good approach or the other way around? and are there better ways to answer this?
Q1(d)
can someone explain why the limited supply of 100g of d.water?
do we have to divide them into beakers
(e.g 10g in one beaker, 20g in the 2nd beaker while 30g and 40g in the 3rd and 4th one)?
what do we have to write to gain a mark of three? :-\
can someone guide me to answer this type of question? i can't really answer this very well.
Thanks :)
Deadly_king:
--- Quote from: Greed444 on October 04, 2010, 10:11:48 am ---need helps on May/June 2010 paper52 Q1(c) and (d)
Q1(c)
what do we have to do draw here? i simply drew a solution-filled beaker on top of a stand with a bunsen burner underneath it for boiling. and also a thermometer inserted into the beaker. Is this a good approach or the other way around? and are there better ways to answer this?
Q1(d)
can someone explain why the limited supply of 100g of d.water?
do we have to divide them into beakers
(e.g 10g in one beaker, 20g in the 2nd beaker while 30g and 40g in the 3rd and 4th one)?
what do we have to write to gain a mark of three? :-\
can someone guide me to answer this type of question? i can't really answer this very well.
Thanks :)
--- End quote ---
(c) I guess your diagram should be ok. But you need to add some details.
The question ask to label all the apparatus used and its capacity.
Hence you should state the capacity of the thermometer ( Ex : 100oC) and also that of the beaker used (Ex : 500cm3).
For the measurement of temperature, I would suggest you to make use of two thermometers.
One which is just immersed in the solution, that is its bulb is just in the water. Another one which is just on the surface of the liquid without touching it though. Make use of retort stands and clamps to hold your thermometers in position.
NOTE : Both thermometers should have same range and same sensitivity. This also should be stated in your diagram.
If you add all these, all the three marks should be yours ;)
(d) Yeah, you should divide the water accurately so that you will not need more than 100g. You should be preparing solutions of different concentrations each time using some water so as to keep total mass used constant.
But you should not in any way exceed the use of 100g of deionised water. This is a limitation which you must precisely make sure that you cope with.
Then you need to devise a way such that you obtain at least 5 solutions of different concentration in an appropriate table.
The last mark is for correct calculation of the molality of the different solutions that you need to calculate.
moon:
For diprotic acids number of moles should still not exceed 0.06/2 since 1 mole of acod already provides 2 moles of H+.
[/quote]
Why do we have to divide by 2 and not multiplying??????????? 1mole NaOH reacts with 1mole of sulphuric acid and hence reacts with 2H+
so 0.06moles NaOH reacts with 0.06*2 of H+ Isn't it???????????????? I still didn't understand this point and how can we keep the total final volumes the same in each experiment?????????????????? I really need help urgently cuz the markschemes and the examiner's reports are not very useful............PLZ HELP!!!! thanx in advance :)
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