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

1. (Original post by soutioirsim)
I swear we didn't have to write an equation for that, just describe the interactions.
yeah, thats what i thought too! i didn't see anything about an equation
2. (Original post by Betsss)
-38 i think
can you show how you worked it out, im not sure....

did you multiply the N-H bond by 6? because theres three of them but two moles of NH3?
3. (Original post by A level Az)
Why would you get the mark for doing it wrong/writing the full equation? The question clearly asked for the ionic equations, and to be honest if you knew the full equations it was really easy to work out.
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4. (Original post by spocckka)
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what question is this?
5. (Original post by Betsss)
hehe what were the ionic equations?
2Al^3+ + 6Cl^- -----> Al2Cl3

2Al^3+ + 8OH^- ---------> 2Al(OH)4^-

Something like that.
6. (Original post by Zakir)
can you show how you worked it out, im not sure....

did you multiply the N-H bond by 6? because theres three of them but two moles of NH3?
I rewrote the equation as:
1/2N2 + 1.5H2 -> NH3

reactants: 0.5 x N2 bond (0.5 x 944) + 1.5 x H2 bond (1.5 x 436) = 1126
products: 3x NH bond = 3x388 = 1164
1126-1164 = -38 Kjmol-1
7. (Original post by Zakir)
can you show how you worked it out, im not sure....

did you multiply the N-H bond by 6? because theres three of them but two moles of NH3?
It asked for the enthalpy of formation for one mole of NH3, so you had to divide the equation by two.
8. (Original post by A level Az)
2Al^3+ + 6Cl^- -----&gt; Al2Cl3

2Al^3+ + 8OH^- ---------&gt; 2Al(OH)4^-

Something like that.
yeah, i got that too, apart from AlCl3 (as Cl has a 1- charge)
and divided the 2nd equation by 2, as its the simplified ratio
9. It wanted ionic equation of the reactions of Al2O3 and HCl, and the same for Al2O3 and NaOH. So instead of Al2O3 + HCl... blah blah, it was Al^3+ + 3Cl-....

I'm so stupid, I even considered doing that.
10. (Original post by spocckka)
It asked for the enthalpy of formation for one mole of NH3, so you had to divide the equation by two.
yeah, or alternatively, divide the overall answer by 2
11. (Original post by A level Az)
2Al^3+ + 6Cl^- -----&gt; Al2Cl3

2Al^3+ + 8OH^- ---------&gt; 2Al(OH)4^-

Something like that.
what if we put..

Al2O3 + 6HCL ----> 2AlCl3

and

Al2O3 + 2NaOH + 3H2O ----> 2NaAl(OH)4...

:s
12. (Original post by spocckka)
It asked for the enthalpy of formation for one mole of NH3, so you had to divide the equation by two.
was that for the first part of the question as well?

i only did that for the 4th part...
13. (Original post by Zakir)
what if we put..

Al2O3 + 6HCL ----> 2AlCl3

and

Al2O3 + 2NaOH + 3H2O ----> 2NaAl(OH)4...

:s
It asked specifically for the ionic equations, so I wouldn't expect them to give any marks for that. Sorry
14. (Original post by A level Az)
It asked specifically for the ionic equations, so I wouldn't expect them to give any marks for that. Sorry
hmm...ok thanks
15. (Original post by A level Az)
It asked specifically for the ionic equations, so I wouldn't expect them to give any marks for that. Sorry
DAAYUM.. Didn't even notice thats what they asked for.. two marks down (N)
16. What did people get for the value of Delta G then?
17. (Original post by MrLukee)
DAAYUM.. Didn't even notice thats what they asked for.. two marks down (N)
Yeah I actually began writing down the full equations first before I saw the word "ionic"
18. does anyone know the UMS marks for EACH module (i.e. empa's and exam)...at AS and A2..

just trying to work out what grade i might be expecting...

thanks!
19. (Original post by soutioirsim)
What did people put for the "How do the fluoride ions interact with the water?" Or something a long the lines.

I put that the fluoride ions act as a lewis base :s
I talked about how the charges on the ions allow them to dissolve in a polar solvent.. Don't expect a mark though, it was a guess... :L
20. (Original post by Betsss)
What did people get for the value of Delta G then?
It was 33600Jmol^-1 or 33.6kJmol^-1

If you used the value they gave you you got answer in the 40000s (I think).

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