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IGCSE CHEMISTRY DOUBTS

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J.Darren:
4 a ii) Lead chloride and sodium nitrate
     iii) Silver nitrate and potassium chloride
6 b i) Add sodium hydroxide, ammonia gas given off on warming, turns damp red litmus paper blue.
7 b) Mg + 2CH3COOH ? (CH3COO)2Mg + H2

Mg - 3/24 = 1/8

CH3COOH - 12/60 = 1/5

1 : 2 mole ratio

Mg is in excess as 1/8 * 2 = 1/4, which is greater than 1/5.

Compare the mole ratio of CH3COOH to hydrogen, 2:1.

1/10 * 24 = 2.4

Ivo:

--- Quote from: gaurav95 on June 04, 2010, 12:21:12 pm ---some more doubts coming up...............

q4aii),iii)thensame q bii

q6bi its saying ammoniuum ion and ms is giving test abt nitrate ion............

q7b whole

thx in advance........

--- End quote ---

4) a) ii) Pb(NO3)2 + 2NaCl -> PbCl2 + 2NaNO3
  
        iii) Ag+ + Cl- -> AgCl

    b) ii) Please click here.

6) b) i) Test: Add aqueous sodium hydroxide.  Result: turns damp red litmus paper blue.

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?

8T:
May june 04 q 4 b 2 why until 8 mm it .. isnt there a range ?
Plus b 3 I have no idea what is going on i need help .. could any1 help?

Ivo:

--- Quote from: 8T on June 04, 2010, 05:38:40 pm ---May june 04 q 4 b 2 why until 8 mm it .. isnt there a range ?
Plus b 3 I have no idea what is going on i need help .. could any1 help?

--- End quote ---


--- Quote from: Ivo on June 03, 2010, 02:40:53 pm ---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!  ;D

--- End quote ---

jellybeans:
i've got a huge doubttt again :( :( :(
please help ! :P tttthankyou.

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