Qualification > Sciences
Chemistry P3 7/6/2010
gaurav95:
--- Quote from: destructor on June 04, 2010, 06:40:27 am ---Gaurav..
so u really cant ask for an easy way out...because there is a possibility of anything arising
but I feel u should study important processes such as extraction of metals especially iron,aluminium,zinc and copper
then also the haber's process and contact process.
Study chromatography thoroughly as it could come in ppr 3...
Oh yeah also study the production of nitric acid..hasnt really come before so it could now
also yeah study the basic reactions (acid+base,acid+metal etc)
Electrolysis too
good luck
--- End quote ---
Thx but will still be looking for some input...........
J.Darren:
--- Quote from: destructor on June 04, 2010, 06:40:27 am ---Gaurav..
so u really cant ask for an easy way out...because there is a possibility of anything arising
but I feel u should study important processes such as extraction of metals especially iron,aluminium,zinc and copper
then also the haber's process and contact process.
Study chromatography thoroughly as it could come in ppr 3...
Oh yeah also study the production of nitric acid..hasnt really come before so it could now
also yeah study the basic reactions (acid+base,acid+metal etc)
Electrolysis too
good luck
--- End quote ---
Extraction of copper is omitted.
Jea:
Can anyone help me with moles? i never got their head or toe........tym is almost up n i kno as much moles as i did three years ago when i did them them fr the first tym! HELP! any tricks....tips......anYthing! :'( :'( :'(
Ivo:
--- Quote from: Jea on June 04, 2010, 12:37:23 pm ---Can anyone help me with moles? i never got their head or toe........tym is almost up n i kno as much moles as i did three years ago when i did them them fr the first tym! HELP! any tricks....tips......anYthing! :'( :'( :'(
--- End quote ---
It's hard to 'teach' you moles. My advice is go through past paper questions, and then you'll find it is well easy!
I've written very thorough explanations for some questions and doubts people have asked. Here they are:
s02, Q5) c) and d):
--- Quote from: Ivo on May 30, 2010, 01:00:49 pm ---The equation you need to use is:
Volume of gas (in dm3)
Moles of gas = --------------------------
24
Note to convert cm3 to dm3, divide quantity by 1000 (e.g. 20cm3/1000=0.02dm3)
This is the proper theory and equation behind my calculations below.
Here are the answers for 5) c) as requested:
i) Moles of C4H6: 0.02/24=0.000833
From equation, we know 2 moles of C4H6 reacts with 11 moles of O2
Therefore, moles of O2: 0.000833*5.5=0.00458
Therefore, volume of O2: 4.58*24*1000=110cm3
ii) From equation, we know 2 moles of C4H6 produces 8 moles of CO2
Therefore, moles of CO2: 0.000833*4=0.00333
Therefore, volume of CO2: 0.00333*24*1000=80cm3
iii) From equation, we know 2 moles of C4H6 produces 6 moles of H2O
Therefore, moles of H2O: 0.000833*3=0.0025
Therefore, volume of H2O: 0.0025*24*1000=60
Therefore, total volume of gases is simply volume of CO2 + volume of H2O:
80+60=140cm3
Alternatively, 8+6=14, so 2 moles of C4H6 produces 14 moles of products
Therefore, moles of products: 0.000833*7=0.00583
Therefore, volume of products: 0.00583*24*1000=140cm3
For part d), you use this formula:
Mass
Moles = ------
Mr
d) From the above equation, 1 mole of butyne forms 3 moles of water
Number of moles of butyne reacted: 9/54=0.167
Number of moles of water formed: 0.167*3=0.5
Mass of water formed: 0.5*(2+16)=9g
I hope this has helped. :D
--- End quote ---
s08, Q7) b)
--- Quote from: Ivo on May 31, 2010, 11:38:26 am ---OK, hopefully this clears your doubt.
This time, you'll need to also apply this formula:
Concentration (mol/dm3) = Moles / Volume (dm3)
So:
i) Number of moles of NaOH used = 2.24*0.025 = 0.056 mols
ii) Maximum number of moles of Na2SO4.10H2O that could be formed = From the equation given, you can see 2 moles of NaOH gives 1 mole of Na2SO4.10H2O. So it is simply: 0.056/2 = 0.028 mols
iii) Mass of one mole of Na2SO4.10H2O = 322g
iv) Maximum yield of sodium sulphate-10-water = 0.028*322 = 9.02g
v) Percentage yield = (3.86/9.02)*100 = 42.8%
If you don't understand any of this, I'll be happy to explain ;)
--- End quote ---
w08, Q7) a)
--- Quote from: Ivo on May 30, 2010, 01:00:49 pm ---The equation you need to use is:
Volume of gas (in dm3)
Moles of gas = --------------------------
24
Note to convert cm3 to dm3, divide quantity by 1000 (e.g. 20cm3/1000=0.02dm3)
This is the proper theory and equation behind my calculations below.
Here are the answers for 5) c) as requested:
i) Moles of C4H6: 0.02/24=0.000833
From equation, we know 2 moles of C4H6 reacts with 11 moles of O2
Therefore, moles of O2: 0.000833*5.5=0.00458
Therefore, volume of O2: 4.58*24*1000=110cm3
ii) From equation, we know 2 moles of C4H6 produces 8 moles of CO2
Therefore, moles of CO2: 0.000833*4=0.00333
Therefore, volume of CO2: 0.00333*24*1000=80cm3
iii) From equation, we know 2 moles of C4H6 produces 6 moles of H2O
Therefore, moles of H2O: 0.000833*3=0.0025
Therefore, volume of H2O: 0.0025*24*1000=60
Therefore, total volume of gases is simply volume of CO2 + volume of H2O:
80+60=140cm3
Alternatively, 8+6=14, so 2 moles of C4H6 produces 14 moles of products
Therefore, moles of products: 0.000833*7=0.00583
Therefore, volume of products: 0.00583*24*1000=140cm3
For part d), you use this formula:
Mass
Moles = ------
Mr
d) From the above equation, 1 mole of butyne forms 3 moles of water
Number of moles of butyne reacted: 9/54=0.167
Number of moles of water formed: 0.167*3=0.5
Mass of water formed: 0.5*(2+16)=9g
I hope this has helped. :D
--- End quote ---
s04, Q7) b):
--- Quote from: Ivo on June 04, 2010, 12:40:43 pm ---7) b) i) Moles of Mg = 3/24 = 0.125
Moles of CH3COOH = 12/60 = 0.200
Using ratios given in the equation, to compare, divide 0.2/2 = 0.100 moles of CH3COOH.
Therefore, magnesium is in excess.
ii) The one not in excess: ie. 0.1
iii) 0.1*24 =2.4dm3
Got it?
--- End quote ---
I hope that's enough questions for you to follow! ;P
$H00t!N& $t@r:
i have a question.... w06 q8 c(iii)
how do i deduce the formula? i need to know that to ans the question
ms says the test is bromine water and the result for the first one is brown to colourless this means that it is an alkene right? bt hw do i knw that the first one is alkene from the fromula given?
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