Chemistry P3 7/6/2010

[NidZ- Hero]

I shall be responsible for the remainder of Organic Chemistry, Air and Water, Carbonates and Sulphates.

hey tell me de topic dat i can do

[J.Darren]

hey tell me de topic dat i can do

The remainder of the topics ...

[gaurav95]

Download this file............

then pls anss my q

can somebody list the imp.reactions that would probably come on 7th plssssssssssssss

or give the link where it is stated......

[JD46]

HEy guys ,i'm really in trouble, the problem is that my chemistry guide(bob berry) is missing an important page(Pg 33 to 34) ,about excess reagents and  coincidentally ,i always mess up that question and lose marks,so could someone  scan the page off the guide and post it,or even a picture  would do, i'm sure you''ll wil understand ,thanks!

[destructor]

Download this file............

then pls anss my q

can somebody list the imp.reactions that would probably come on 7th plssssssssssssss

or give the link where it is stated......

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

[gaurav95]

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

Thx but will still be looking for some input...........

[J.Darren]

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

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]

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!

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):

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 = ------
               M_r

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.

s08, Q7) b)

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

w08, Q7) a)

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 = ------
               M_r

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.

s04, Q7) b):

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?

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?

[elemis]

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?

If its an alkene it will have the general formula   C_n H_2n   where n is the number of carbon atoms

[$H00t!N& $t@r]

If its an alkene it will have the general formula   C_n H_2n   where n is the number of carbon atoms

um can you please check the question... do i count all the carbon atoms or just the ones on the first line?

[elemis]

um can you please check the question... do i count all the carbon atoms or just the ones on the first line?

Hang on. 

[J.Darren]

um can you please check the question... do i count all the carbon atoms or just the ones on the first line?

C17H33 has two fewer hydrogen atoms compared to C17H35, we can deduce that C17H33 is an alkene straightaway ...

[elemis]

C17H33 has two fewer hydrogen atoms compared to C17H35, we can deduce that C17H33 is an alkene straightaway ...

How does that prove anything ?