IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: Sue T on December 23, 2009, 01:18:38 pm
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november 2003 (ppr2)
in question 3(bii)they ask us 2 calculate th heat released
n they gave th temp change as 12.2°C
now i used e=mc?t
n ?t should be in kelvin, so ?t=(12.2 + 273) K
but in the ms, they jus kept it as 12.2 !
why? >:(
& (d) also - i don get it
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3b(ii)
In case of water
The specific heat capacity of water is:
o cwater = 1 cal/(g°C) = 1 cal/(g?K)
o cwater = 4.184 J/(g°C) = 4.184 J/(g?K)
o cwater = 4184 J/(kg°C) = 4184 J/(kg?K)
So u shouldn't change the rise in temp to Kelvin...A change in temperature of 12.2
oC is of the same magnitude as a change in temperature of 12.2 K.
For the (d) bit of it
The reaction is Ca+ 2H2O---> Ca(OH)2+ H2
1 mole=40g of Ca give 24dm3 of H2(1 mole of H2...1 mole of any gas at rtp is 24dm3)
therefore 1g of Ca will give 24/40dm3 of hydrogen 0.6dm3 or 600cm3
Hope u've got it
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got it! thank you so much! ;D
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got it! thank you so much! ;D
ur welcome :)
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can any1 help me in solving que.6 paper1 ( AS chem.)????
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can any1 help me in solving que.6 paper1 ( AS chem.)????
could u specify which year plz?
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here to help too
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oh, am sooo sorry
its M/J 2003
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PV=nRT is the ideal gas equation
Ideal gases r thosr which have a negligible size and no intermolecular forces
If you need a close-to accurate value of Mr of a real gas....it can tend to ideal when pressure is low..so molecules are further apart and occupy negligible space and when temp is high..so again the molecules are far apart with hardly any interactions(thus satisfying the conditions of an ideal gas and hence the equation)..
So the ans is C
Hope you understood
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thanx alot :-*
cud u also help me in Q3 nov/04 p1???????????
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CH3SH + 3O2 ? CO2 + SO2 + 2H2O
A sample of 10 cm3 of methanethiol was exploded with 60 cm3 of oxygen.
The first task is to find the limiting reagent...
Calculate the number of moles of CH3SH
n = 10/240000 = 0.00041666666667
Calculate number of moles of O2
n = 60/ 24000 = 0.0025
divide by three to get in correct ratio:
0.0008333333
And from this you can see that the oxygen gas was added in excess and so methanethiol was the limiting reagent.
From the equation we know that three moles of oxygen react with one mole of methanethiol...
If we multiply the number of moles of methanethiol by 3, we get 0.00125 moles... this is the number of moles of oxygen that actually reacted...
therefore,
0.00125 moles = 30 cm³
only 30 cm³ of the 60cm³ reacted....
means that 30 was left over...
TOTAL VOLUME OF REMAINING GASES:
10cm³ of carbon dioxide was produced
10cm³ of sulphur dioxide was produced
30cm³ of oxygen was left unreacted...
Hence the answer is
50cm³
C
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thank u sooo much :-*
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hey ppl,
in da may06 paper1 Q36
why is ammonia not a reducing agent??????
and also Q32 of da same year is a bit difficult
thanx in advance 4 any help...
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Question 36
The answer is B. That is both statements 1 and 2 are correct.
1 is true because Cl2 which has an oxidation state of 0, becomes Cl-. Thus as reduction is gain of electrons, chlorine is reduced by Ammonia.
2 is true because NH3 becomes NH4+. It gains a hydrogen ion i.e. a proton. As we know bases are proton acceptors.
Question 32
The answer is B. That is both statements 1 and 2 are correct.
1 is true because you need to isolate an atom of sodium, before you can ionize it, i.e. by removing an electron.
2 is true because, in order for an atom of sodium to gain a charge you need to ionize it. As it has only a charge of 1+, you only take into account the first ionization energy.
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Thank You sooooooo much
;)
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No problem, it was my pleasure and gain.
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hey,,,
i really need ur help guys in que.2.c.i of Nov02 paper2
Thank You in advance :-*
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Will look now.
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Working on it.
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okay......
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Here u go
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Please tell me if u fully understand.
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ya i got it... :)
Thank You ;)
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wait am stuck....
where did u get da 0.06 moles from????????????
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it is explained in the ethanol table!
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ok.... i guess i got it
thanx 4 da help :)
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Just doing my job.
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hey dude,
could u help me with this?
may07 ,Q9,16,18 -p1
nov02, Q1 - p1
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November 2002,
Q1 P1
1 A mixture of 10 cm3 of methane and 10 cm3 of ethane was sparked with an excess of oxygen.
After cooling to room temperature, the residual gas was passed through aqueous potassium
hydroxide.
What volume of gas was absorbed by the alkali?
A 15 cm3
B 20 cm3
C 30 cm3
D 40 cm3
The equation for the reaction is:
CH4(g) + C2H6(g) 5.5 O2(g) ? 3CO2(g) + 5H2O (l)
Avogadro's law states that Equal volumes of ideal or perfect gases, at the same temperature and pressure, contain the same number of particles, or molecules
Therefore,
10cm³ of Methane
10cm³ of Ethane
and 30cm³ of Carbon Dioxide
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thanx :D