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# AQA - Unit 5 - Energetics, Redox and Inorganic Chemistry watch

1. Aluminium and Sodium Carbonate forms

2[Al(H2O)^3(OH)^3] + CO2

What kind of reactions with HCl? With the transition metals or with Period 3?
2. (Original post by GoodOl'CharlieB)
Hi has anyone done the last question of the CHEM5 Jan 11 paper. After you work out moles of H2O2, why do you multiply by 10 and then by 1000/5?
Because I am so confused -->
You multiply by 10 because the sample of H2O2 is 25cm3 from a solution of 250cm3.

you work out how many moles are in 25cm3, then multiply it by 10 to get the number of moles in the whole thing.

It's like working out how much fat it in a slice of cake which amounts to a sixth of the cake and multiplying it by 6 to finding out how much fat is in the whole thing.

You then have to work out the concentration of the original solution. Remember that concentration = moles / volume. Volume must be in dm3 but starts in cm3 so it becomes concentration = moles / (volume/1000).

Now for the tricky part. It doesn't ask for the concentration in 250cm3. It asks for the concentration from 5cm3. What you've got to realise is that the number of moles doesn't change when you dilute a substance.

So if you have 1.14 x 10^(-2 in 5cm3 and add 245cm3 to make 250cm3, the number of moles is still 1.14 x 10^(-2. Only the concentration is affected.

It's like having a spoonful of sugar in a glass of water compared to a spoonful of sugar in a bucket of water. The amount of sugar is the same but it is sweeter in the glass of water because it is more concentrated.

Then you just plug it into the calculation of concentration = moles / (volume / 1000)

= 1.14 x 10^(-2 / (5/1000)
= 2.28 moldm-3

Hope this helps
3. (Original post by Sparkly-Star)
(Original post by Veem)
x
Ah yeah I suck at that part of the topic >.<

I think you just have to be able to work out the emf and be able to spout out the advantages/disadvantages of using one over the others.
4. Just wondering how to work out the cell representation of the hydrogen fuel cell in the Feb 10 paper? Hardest one ive done.. and i didnt get the right anwser as well..
Many thanks and + rep for anyone who can help!
5. (Original post by jimmy303)
The first one is a metal (2) carbonate and the second is a metal (3) carbonate but the second one doesn't exist.

A carbonate ion exists as CO3(2-) so has a charge of -2. Thus logically to form neutrally charged molecules with a M2+ ion it would form MCO3 (such as CuCO3) and with a M3+ ion you would think it would be:

2M3+ + 3CO3(2-) --> M2(CO3)3

However this doesn't happen because Metal 3+ ions are more acidic than Metal 2+ ions so rather than going through a precipitation reaction with CO3(2-) (which metal 2+ ions do because they're not acidic enough) it has an acid-base reaction (just like when a carboxylic acid reacts with a carbonate).

You end up with a similar reaction to the ones where aqueous M3+ ions react with dilute NH3 or NaOH as these are acid-base reactions, where 3 hydrogen atoms are removed from the water molecules in the complex and a hydroxide solid is formed.

The 6 reactions you need to know for carbonates can be split into M2+ and M3+ reactions:

[Cu(H2O)6]2+ +CO3(2-) --> CuCO3 + 6H2O
Blue solution to blue-green precipitate.

[Fe(H2O)6]2+ +CO3(2-) --> FeCO3 + 6H2O
Pale green solution to green precipitate.

[Co(H2O)6]2+ +CO3(2-) --> CoCO3 + 6H2O
Pink solution to pink precipitate

And then

2[Al(H2O)6]3+ +3CO3(2-) --> 2Al(H2O)3(OH)3 + 3H2O + 3CO2
Colourless solution to white precipitate and effervescence

2[Cr(H2O)6]3+ +3CO3(2-) --> 2Cr(H2O)3(OH)3 + 3H2O + 3CO2
Ruby (or for the guaranteed mark green) solution to green precipitate and effervescence

2[Fe(H2O)6]3+ +3CO3(2-) --> 2Fe(H2O)3(OH)3 + 3H2O + 3CO2
Pale violet (or for guaranteed mark yellow) solution to brown precipitate and effervescence

Hope this has helped a smidge
Thankyou sooooo much!!!
6. if the chem5 exam is as hard as the bio5 ill probs do something bad
7. (Original post by jimmy303)
You multiply by 10 because the sample of H2O2 is 25cm3 from a solution of 250cm3.

you work out how many moles are in 25cm3, then multiply it by 10 to get the number of moles in the whole thing.

It's like working out how much fat it in a slice of cake which amounts to a sixth of the cake and multiplying it by 6 to finding out how much fat is in the whole thing.

You then have to work out the concentration of the original solution. Remember that concentration = moles / volume. Volume must be in dm3 but starts in cm3 so it becomes concentration = moles / (volume/1000).

Now for the tricky part. It doesn't ask for the concentration in 250cm3. It asks for the concentration from 5cm3. What you've got to realise is that the number of moles doesn't change when you dilute a substance.

So if you have 1.14 x 10^(-2 in 5cm3 and add 245cm3 to make 250cm3, the number of moles is still 1.14 x 10^(-2. Only the concentration is affected.

It's like having a spoonful of sugar in a glass of water compared to a spoonful of sugar in a bucket of water. The amount of sugar is the same but it is sweeter in the glass of water because it is more concentrated.

Then you just plug it into the calculation of concentration = moles / (volume / 1000)

= 1.14 x 10^(-2 / (5/1000)
= 2.28 moldm-3

Hope this helps
Firstly thanks again it makes so much sense now ...the MS was so confusing
Secondly, have u ever thought of going into teaching?
And thirdly , I now have cravings for cake lol
Thanks again, you explain things really well

Btw 1 more thing and I promise I will go away, what are the advantages and disadvantages of the different types of batteries?
8. (Original post by Silkielemon)
Aluminium and Sodium Carbonate forms

2[Al(H2O)^3(OH)^3] + CO2

What kind of reactions with HCl? With the transition metals or with Period 3?
I think you only have to know

Cu^+2 + 4Cl- ---> [CuCl4]^-2, forms a yellow solution

Co^+2 + 4Cl- ---> [CoCl4]^-2, forms a blue solution

and both from tetrahedral shapes

im pretty certain you dont have to know any period 3 equations with HCl
9. (Original post by Silkielemon)
Aluminium and Sodium Carbonate forms

2[Al(H2O)^3(OH)^3] + CO2

What kind of reactions with HCl? With the transition metals or with Period 3?
Thanks! How many co2 and h2o are formed? And with transition metals.
10. Does anyone know the equations for when Cu2+ acts as catalyst in the oxidation of I- by K2O72-?
11. Hey, has anyone done the Jan'2011 paper for CHEM5? I'm stuck on the last question...I managed to work out the moles of H2O2 but don't understand why mark scheme says to convert to 5cm^3 you have to multiply by 10.

What I don't understand is that *10 converts from 25cm^3 to 250cm^3 but isn't the 250cm^3 a dilution of the 5cm^3 they started off with?
12. (Original post by Harry Potter's sidekick)
Hey, has anyone done the Jan'2011 paper for CHEM5? I'm stuck on the last question...I managed to work out the moles of H2O2 but don't understand why mark scheme says to convert to 5cm^3 you have to multiply by 10.

What I don't understand is that *10 converts from 25cm^3 to 250cm^3 but isn't the 250cm^3 a dilution of the 5cm^3 they started off with?
This helps to explain it really well
13. (Original post by Harry Potter's sidekick)
Hey, has anyone done the Jan'2011 paper for CHEM5? I'm stuck on the last question...I managed to work out the moles of H2O2 but don't understand why mark scheme says to convert to 5cm^3 you have to multiply by 10.

What I don't understand is that *10 converts from 25cm^3 to 250cm^3 but isn't the 250cm^3 a dilution of the 5cm^3 they started off with?
The original sample was diluted from 5cm^3 to 250cm^3, so this will contain the same number of moles as the original hydrogen peroxide solution. However when they titrate it they only use 25cm^3 of it, so they are only reacting one tenth of the moles of the original solution. Hence you have to multiply it by 10.
Hope this makes sense!
14. I have 2 questions which I was hoping somebody could help me with

The melting point of aluminium oxide is lower than that of the other metal oxides because of the degree of covalent character it has (I don't really see how this means the bonding is weaker but I have given up trying to get my head around it!). However, further on it the textbook it says that one of the reasons aluminium oxide is insoluble is because of the additional covalent bonding which results in a stronger lattice. Isn't this contradictory?

Also, is it the magnesium oxide or magnesium hydroxide that is sparingly soluble, or both?
15. (Original post by englishman129)
Just wondering how to work out the cell representation of the hydrogen fuel cell in the Feb 10 paper? Hardest one ive done.. and i didnt get the right anwser as well..
Many thanks and + rep for anyone who can help!
Since it's an alkaline fuel cell you need to work with the two electrode potentials that have OH- in them. The more negative one will be on the left and as always you have to put | between every change of state, and there are 3 of them in the first reaction. You also have to work out which way around the reaction goes. The one on the left is the reducing agent, so the product will be electrons, hence the more negative electrode potential comes first.
16. has anyone done 3b)i. on the specimen paper? anyone else find that vanadium species at the end of the reaction question abit weird? how did would you get (VO2)+
17. (Original post by OhNo_23)
I have 2 questions which I was hoping somebody could help me with

The melting point of aluminium oxide is lower than that of the other metal oxides because of the degree of covalent character it has (I don't really see how this means the bonding is weaker but I have given up trying to get my head around it!). However, further on it the textbook it says that one of the reasons aluminium oxide is insoluble is because of the additional covalent bonding which results in a stronger lattice. Isn't this contradictory?

Also, is it the magnesium oxide or magnesium hydroxide that is sparingly soluble, or both?
If you worked out the Lattice dissociation for Al2O3 using the perfect ionic model, it would be lower than the experimental value, but it still has strong ionic bonds, and the degree of covalent bonding makes it a stronger lattice. The melting point isn't weaker BECAUSE it has covalent bonding as well, but in spite of this. I guess the ionic bonds are weaker?

MgO is sparingly soluble in water and so makes a solution pH around 9/10, as there wont be much Mg(OH)2 i think!
18. (Original post by teddyWS)
has anyone done 3b)i. on the specimen paper? anyone else find that vanadium species at the end of the reaction question abit weird? how did would you get (VO2)+
I got confused over this also. Its because it works a bit like a chain reaction. The V2+ becomes the V3+ first, then that is oxidised again to form the VO(2+), and then this in turn is further oxidised to form the final (VO2)+.
It's a bit of nasty question but once you see it its easy. Hope this makes sense anyways!
19. (Original post by Luke0011)
If you worked out the Lattice dissociation for Al2O3 using the perfect ionic model, it would be lower than the experimental value, but it still has strong ionic bonds, and the degree of covalent bonding makes it a stronger lattice. The melting point isn't weaker BECAUSE it has covalent bonding as well, but in spite of this. I guess the ionic bonds are weaker?

MgO is sparingly soluble in water and so makes a solution pH around 9/10, as there wont be much Mg(OH)2 i think!
Thanks for answering I see what you are saying, but surely the ionic bonds should be stronger in aluminium oxide because the aluminium ion is smaller and more highly charged, and so should attract the oxide ion to a greater extent
20. (Original post by OhNo_23)
Thanks for answering I see what you are saying, but surely the ionic bonds should be stronger in aluminium oxide because the aluminium ion is smaller and more highly charged, and so should attract the oxide ion to a greater extent
I think this is due to the arrangement of ions, but we dont need to know that at our level

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