IGCSE CHEMISTRY DOUBTS

[Ivo]

Okay, I have a question.

May/June 2004, Paper 3, Question 4, (b)(ii)

Why the difference between the height of the precipitate between Iron (II) and Iron (III) salts?

OK, We know from the graph that 12cm³ of aqueous sodium hydroxide was needed to react with 4cm³ of aqueous iron(III) chloride.  Because they were all at 1.0 mol/dm³, we therefore realise that 1 mole of aqueous iron (III) chloride reacted with 3 moles of aqueous sodium hydroxide.  Therefore, the formula for the precipitate is Fe(OH)₃.

For iron (II) chloride, we know the formula of the precipitate to be Fe(OH)₂, and so for the same volume of aqueous iron (II) chloride - 4cm³, we know 8cm³ of sodium hydroxide was required.  Therefore, you would need to draw the maximum height of the precipitate when volume has reached 8cm³.

Hope this has helped! 

[Ivo]

i) how do you draw the structure of the polymer formed from but-2-ene?

Here we go!

[jellybeans]

Here we go!

THANKS

ii) How do you deduce the formula of the alkene which has a relative molecular mass of 168?
is it something to do with the empirical formula of... idk CnH2n hence 3n = 168 ... blahblah
lol help please! THANKS.

[Ivo]

ii) How do you deduce the formula of the alkene which has a relative molecular mass of 168?
is it something to do with the empirical formula of... idk CnH2n hence 3n = 168 ... blahblah
lol help please! THANKS.

We know C has A_r of 12 and H is 1.  So by using C_nH_2n:

12n + 2n = 168

So n = 12

So C₁₂H₂₄

[jellybeans]

We know C has A_r of 12 and H is 1.  So by using C_nH_2n:

12n + 2n = 168

So n = 12

So C₁₂H₂₄

THANKS

[gaurav95]

HI i have a pro

M/J 2002

q5 whole ms isnt there.....

can somebody pls tell me thx a lot...........

[gaurav95]

Frnds sorry but one more q

4 c i

can somebody explain me how did one get that ans..................

[Ivo]

HI i have a pro

M/J 2002

q5 whole ms isnt there.....

can somebody pls tell me thx a lot...........

Gosh, how many times do I have to say this.  Only joking, it's not you're fault for not knowing!  

Here are the answers, not mark scheme but they are perfectly acceptable.

the a) part?

Liquids take the shape of the container, they maintain their volume. The intermolecular forces in liquids are significant and therefore, they do not expand to occupy all the space available (as gases do). A given mass of liquid has a fixed volume. also molecules in liquid have less k.e as compared to that of gases.Particles in gases have very weak intermolecular forces and high kinetic energy, hence the particles move independent of each other and occupy the space available to them.

b)i) rate= time and volume. In other words, its the volume produced and the time taken.

ii) CO2 due to lower molecular mass. or less density

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.

[Ivo]

Frnds sorry but one more q

4 c i

can somebody explain me how did one get that ans..................

Right, the molecular formula is simply the formula showing the type and number of each element present in a molecule.  So very simple, just count the number of carbon, oxygen and atoms.  The answer is C₆H₈O₆.

[gaurav95]

Right, the molecular formula is simply the formula showing the type and number of each element present in a molecule.  So very simple, just count the number of carbon, oxygen and atoms.  The answer is C₆H₈O₆.

but there are OH groups also so i thought that it could be alcohol but its not like that right
and also same q paper q 7 biii.......................pls

[gaurav95]

One more q

do we have cyclic hydrocarbons i guess one q had come abt properties

can somebody give info abt topics related to this..........

[anonymous7]

OK, We know from the graph that 12cm³ of aqueous sodium hydroxide was needed to react with 4cm³ of aqueous iron(III) chloride.  Because they were all at 1.0 mol/dm³, we therefore realise that 1 mole of aqueous iron (III) chloride reacted with 3 moles of aqueous sodium hydroxide.  Therefore, the formula for the precipitate is Fe(OH)₃.

For iron (II) chloride, we know the formula of the precipitate to be Fe(OH)₂, and so for the same volume of aqueous iron (II) chloride - 4cm³, we know 8cm³ of sodium hydroxide was required.  Therefore, you would need to draw the maximum height of the precipitate when volume has reached 8cm³.

Hope this has helped! 

Ohmigod. I actually feel stupid now. Hahah. Thanks. That really did it. ^.^

[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!   

[Ivo]

but there are OH groups also so i thought that it could be alcohol but its not like that right
and also same q paper q 7 biii.......................pls

You would not need to incorporate and OH groups, because they only ask for molecular formula.  If they ask for the structual formula, then yes, but that would be complicated.

I'll give you an example:

Butane:  Molecular formula - C₄H₁₀; Structual formula - CH₃CH₂CH₂CH₃.

[Vin]

but there are OH groups also so i thought that it could be alcohol but its not like that right
and also same q paper q 7 biii.......................pls

Depends on the q, here its vitamin C. Usually all vitamins contain -OH groups. Only for the structural formula you need to specify -OH and not the molecular formula.


For Q.7 b)iii) There are 'n' number of types of chlorination. This depends on how much chlorine you use and which hydrocarbon you use.

When chlorination occurs it replaces one hydrogen atom from the hydrocarbon and forms HCl

E.g. Chlorination of methane::

    H                                                  H
    |                                                   |  
H- C - H   +    Cl₂ ------>  H- C - Cl  +  HCl
    |                                                   |
    H                                                  H

CH₄  + Cl₂  ----> CH₃Cl  +  HCl
Product:: Chloromethane (methyl chloride)
 

Similarly,
CH₃Cl + Cl₂  ----> CH₂Cl₂ + HCl

Product:: Dichloro methane (methylene chloride)


CH₂Cl2 + Cl₂ ----> CHCl₃  +  HCl

Product:: Trichloro methane (Chloroform)


CH₃Cl₃ +  Cl₂ ----> CCl₄  +  HCl

Product:: Tetrachloro methane (Carbon tetrachloride/Pylene)


I think you only need to know the names in bold. For your question, choose any of the following reaction BUT using butane and chlorine
Eg. C₄H₁₁Cl  + Cl₂ ----> C₄H₁₀Cl  +  HCl

OR to be safe give acc. to the ques

CH3-CH2-CH2-CH2-Cl  +  Cl2 ----> CH3-CH2-CH2-CHCL2  +  HCl
  
   H H H  H                             H  H  H Cl
   | |  |   |                             |  |  |  |
H-C-C-C-C-Cl   + Cl-Cl---->  H-C- C-C-C-Cl       +  H-Cl
   |  |  |  |                             |  |  |  |
   H  H H  H                            H  H  H H

Don't understand anything.. pls. ask.. Just under the concept of chlorination they may ask any other alkane next time
Draw the structures yourself you'll understand better