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Author Topic: IGCSE CHEMISTRY DOUBTS  (Read 28349 times)

Offline gaurav95

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Re: IGCSE CHEMISTRY DOUBTS
« Reply #120 on: June 05, 2010, 04:53:33 pm »
I mean 6bii

and u stii havent given ans to 4biii
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Offline gaurav95

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Re: IGCSE CHEMISTRY DOUBTS
« Reply #121 on: June 05, 2010, 04:55:56 pm »
Some more doubts
4bi reason

7b again i want a reason.like what was the reaction and all the stuff thx
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Offline Saladin

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Re: IGCSE CHEMISTRY DOUBTS
« Reply #122 on: June 05, 2010, 04:58:35 pm »
Quote from: gaurav95 on June 05, 2010, 04:55:56 pm
Some more doubts
4bi reason

7b again i want a reason.like what was the reaction and all the stuff thx

4B(i) A metal like Sodium or Potassium. Because these metals are from group 1 and 2.

7(b) It would release Nitrogen Dioxide, so it would be a brown gas....
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Offline Ivo

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Re: IGCSE CHEMISTRY DOUBTS
« Reply #123 on: June 05, 2010, 04:58:59 pm »
Quote from: gaurav95 on June 05, 2010, 04:53:33 pm
I mean 6bii

and u stii havent given ans to 4biii

s04

4) b) ii):

Quote from: Ivo on June 04, 2010, 07:08:28 pm
Well, as I explained before.  The sodium hydroxide is in excess, so the maximum needed to fully react with the iron(III) chloride is 12cm3.  So, simplify the ratio down to 12:4 = 3:1/  So 3 moles of sodium hydroxide reacted with 1 mole of iron(III) chloride.  So the balanced equation is:

3NaOH + FeCl3 -> 3NaCl + Fe(OH)3

OK, now if iron (II) chloride:

2NaOH + FeCl2 -> 2NaCl + Fe(OH)2

So 2 moles instead of 3 was required.  So if 3 moles was 12cm3, then 1 mole is 4cm3, then 2 moles must be 8cm3.  Done!  :)

4) b) iii):

Quote from: Ivo on June 04, 2010, 06:39:30 pm
What specifically did you not get, I can try and elaborate.

And for iii) The graph would rise to maximum height, then decrease to zero again.  Why?  Well because aluminium hydroxide (the product formed) dissolves in excess.  Think of the test for aluminium cation:  add aqueous sodium hydroxide, soluble in excess giving a colourless solution.  Same applies here, when a small volume was added, there was precipitate, as you add more and more, it goes colourless, therefore no precipitate.  The common mistake is to think of reactivity.

I hope that clears the issue somewhat.  :D

What don't you get for Q7) b).  I thought i made it quite clear.  What don't you get?

Any more?
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Offline Ivo

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Re: IGCSE CHEMISTRY DOUBTS
« Reply #124 on: June 05, 2010, 05:02:34 pm »
Quote from: gaurav95 on June 05, 2010, 04:55:56 pm
Some more doubts
4bi reason

7b again i want a reason.like what was the reaction and all the stuff thx

w04?

4) b) i)  It would increase because silver is less reactive than tin, so now the difference in reactivity between zinc and silver is greater, therefore the voltage would increase.

Next time, please state which session and year, because I think you're asking me doubts from both s04 and w04, I'm getting confused!
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Offline J.Darren

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Re: IGCSE CHEMISTRY DOUBTS
« Reply #125 on: June 05, 2010, 06:23:51 pm »
By the way the more reactive metal is a better reudcing agent becuase it has a greater tendency to form positive ions, whereas the relatively weaker metal is a better oxidising agent as it is the best at accepting electrons.
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Offline Allyza1n

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Re: IGCSE CHEMISTRY DOUBTS
« Reply #126 on: June 05, 2010, 09:19:20 pm »
hey...
 Question 5 b from chemistry may june 2009 variant 1
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Offline Vin

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Re: IGCSE CHEMISTRY DOUBTS
« Reply #127 on: June 05, 2010, 10:43:20 pm »
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



Hey Ivo, can you tell me in detail how did you get the part in bold^^. I'm totally confused with this chapter. Also do you have any rules/notes based on this chapter. Your help is much appreciated! :)
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Offline Ivo

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Re: IGCSE CHEMISTRY DOUBTS
« Reply #128 on: June 05, 2010, 10:48:50 pm »
Quote from: ~VIN1094~ on June 05, 2010, 10:43:20 pm
Hey Ivo, can you tell me in detail how did you get the part in bold^^. I'm totally confused with this chapter. Also do you have any rules/notes based on this chapter. Your help is much appreciated! :)

OK, we know from the equation given, the mole ratio: 1 mole of magnesium reacted with 2 moles of ethanoic acid.

The amounts of substance actually used was: moles of magnesium = 0.125, moles of ethanoic acid = 0.2.  But from above, you see the ratio 1:2.  So here, we either divide 0.2/2=0.1 or multiply 0.125 by 2 = 0.25.

Either way, you'll find that the magnesium is in excess and the ethanoic acid the limiting reagent.  Has this helped?

For the second part, we are also using the mole ratio from the equation where 2 moles of ethanoic acid gives 1 mole of hydrogen.  We can't use the calculations for magnesium as this was in excess, so we use the limiting reagent (ie. ethanoic acid), so 0.2/2=0.1.

And no sorry, don't have any notes.  I guess, as long as you can handle this question, I think you'll be fine.  I don't think it gets any harder than this.
« Last Edit: June 05, 2010, 10:52:05 pm by Ivo »
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Offline Ivo

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Re: IGCSE CHEMISTRY DOUBTS
« Reply #129 on: June 05, 2010, 10:53:13 pm »
Also, can someone please explain the colour changes and what happens for the potassium manganate and potassium iodide, I have no idea what they mean.  Thanks in advance.
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Offline Vin

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Re: IGCSE CHEMISTRY DOUBTS
« Reply #130 on: June 05, 2010, 11:05:51 pm »
okay!!! so, we just divide 0.2/2 to get into 1:1 ratio, got it. If yes, then cool +rep. If you say no and start another explanation then imma -rep you >:( .. CHILL just joking ;) Its just that we were never taught such stuff(even if I were I don't remember). THANK YOU SO MUCH IVO, YOU'VE BEEN GREAT HELP! :D

 
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Offline Ivo

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Re: IGCSE CHEMISTRY DOUBTS
« Reply #131 on: June 05, 2010, 11:14:41 pm »
Quote from: ~VIN1094~ on June 05, 2010, 11:05:51 pm
okay!!! so, we just divide 0.2/2 to get into 1:1 ratio, got it. If yes, then cool +rep. If you say no and start another explanation then imma -rep you >:( .. CHILL just joking ;) Its just that we were never taught such stuff(even if I were I don't remember). THANK YOU SO MUCH IVO, YOU'VE BEEN GREAT HELP! :D

Yep, I guess the top tip for working out limiting reagents and which is in excess is first to convert the substances to moles.  Then for comparison, change to 1:1 ratio. 

You got it!
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Offline Vin

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Re: IGCSE CHEMISTRY DOUBTS
« Reply #132 on: June 05, 2010, 11:17:36 pm »
Quote from: Ivo on June 05, 2010, 10:53:13 pm
Also, can someone please explain the colour changes and what happens for the potassium manganate and potassium iodide, I have no idea what they mean.  Thanks in advance.

Potassium manganate (VII) is an oxidising agent, purple compound. The oxidation state of manganese is +VII. BUt it is much more stable in oxi. state +II. So it is strongly driven to gain electrons and reduce its oxi. state to +II.
So it takes electrons from other substances, in the presence of a little acid. It itself is reduced causing a colour change..

                     (reduction)
MnO4-         ---------->           Mn2+
manganate(VII)ion                  manganese ion(II)
purple                                    colourless

This means KMnO4 can be used to test for a reducing agent.
« Last Edit: June 05, 2010, 11:19:14 pm by ~VIN1094~ »
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Offline Vin

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Re: IGCSE CHEMISTRY DOUBTS
« Reply #133 on: June 05, 2010, 11:25:50 pm »
Quote from: Ivo on June 05, 2010, 11:14:41 pm
Yep, I guess the top tip for working out limiting reagents and which is in excess is first to convert the substances to moles.  Then for comparison, change to 1:1 ratio.  

You got it!

And buddy you got some rep ;) Thanks again.. :)


About Potassium Iodide, a reducing agent.

H2O2  +  2KI   +  H2SO4  ----->  I2  K2SO4  +  2H2O

H2O2 here is loses O2, therefore reduced. KI helps this to happen also it itself is oxidised to iodine which brings a colour change::

       (oxidation)
2I-1  ------------ > I2
colourless             brown

 
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Offline anonymous7

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Re: IGCSE CHEMISTRY DOUBTS
« Reply #134 on: June 06, 2010, 05:56:48 am »
Question:

Products at the electrodes. What's the difference for a dilute aqueous electrolyte and concentrated aqueous electrolyte? Take sodium chloride for example.
« Last Edit: June 06, 2010, 06:16:36 am by anonymous7 »
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