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AQA - Unit 5 - Energetics, Redox and Inorganic Chemistry

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yes i got all of those. execpt i forgot to put white solid for MgO dam. also did you put hydrogen fuel cell or standard hydrogen electrode for the emf one

Hydrogen fuel cell, that whole last part of the question was a little hazy but i might have been able to blag a few marks

Reply 1381

M | k e

Original post by Betsss

For the Cu with C2O42- complex, did people cut draw it with 2 lots of CrO42- and no water ligands? :s-smilie:

i think i saw on the question that you weren't supposed to draw the water ligands, but i cant remember now! :frown:

Yeah said you don't need to show the water ligands :smile:

Reply 1382

MoMatrix

Aluminium Oxide testing? Molten Cryolite, Electrolysis??

Reply 1383

Betsss

For the first question, where you had to explain the interactions between Flouride ions and water, was it hydrogen bonding? :s-smilie:

Reply 1384

lou12

Original post by jamest92

Hydrogen fuel cell, that whole last part of the question was a little hazy but i might have been able to blag a few marks

cool, me too :smile:

Reply 1385

jamest92

Original post by SnookerLoopy

what did people get as the percentage of iron? 8.74 or something ?

I got around 50 % :s-smilie:

Reply 1386

Betsss

Original post by M | k e

Yes

and in the CGP book it has the exact example but with their cell as 2.05V, and their answers -0.36. Ours was 2.15V so can be sure -0.46V is right :biggrin:

Yay thanks! :biggrin:

and the other one was +1.1? :smile:

Reply 1387

magicalelf

Original post by Betsss

For the first question, where you had to explain the interactions between Flouride ions and water, was it hydrogen bonding? :s-smilie:

ah yeah i didnt like that one, i put that water is polar so the polar hydrogon atoms in water have a slight positive charge and are attracted towards the fluoride ions and said something about forming like a sheath around it.. no idea if its right tho :/

Reply 1388

Scaudwell

For the ammonia question where we had to find the entropy, and many people forgot to divide by 2 (myself included), the question beforehand was the enthalpy change for the production of 1 mole of ammonia.

Would we have to divide that by two, also? Seeing as the equation creates 2 moles of ammonia.

Funny, I divided by two for the enthalpy, but not the entropy. How stupid of me.

Reply 1389

Betsss

Anyone get 53.7% for the iron one? :smile:

Reply 1390

al_habib

Original post by Van_der_Waals

Yeah, haha, an easy 6 marks (hopefully :wink:

). I don't see why they can't use consistent symbols throughout both physics and chemistry though, it's E=hf not

\Delta

E=hv!

can you remember the ans to the first que in that page

Reply 1391

M | k e

Original post by Betsss

Yay thanks! :biggrin:

and the other one was +1.1? :smile:

S'what I got :smile:

Reply 1392

lou12

Original post by Betsss

Did people get -0.46 for the E value for the negative terminal?

that value sounds familiar :smile:

Reply 1393

Smartie77

Original post by SnookerLoopy

what did people get as the percentage of iron? 8.74 or something ?

I got something like that but I think people are saying something about a ratio of 1:6. I don't remember seeing that in the question. I probably neglected to read it properly.

Reply 1394

M | k e

Original post by al_habib

can you remember the ans to the first que in that page

Something about why is Copper(II) ions blue?

I said the d subshell splits when ligands attach, forming levels of different energy. The electrons in it require energy = delta E to make the jump to higher energy level. They get this energy by absorbing light, in copper then absorb red light, and reflect the rest, because red is absorbed this mean blue gets reflected and is visible at this colour :smile:

.

Not 100% about my physicsy stuff but it's along those lines :tongue:

Reply 1395

jamest92

Original post by Betsss

Anyone get 53.7% for the iron one? :smile:

Thank god, someone else

Reply 1396

M | k e

Original post by Smartie77

I got something like that but I think people are saying something about a ratio of 1:6. I don't remember seeing that in the question. I probably neglected to read it properly.

You had to work out the ratio yourself. Cr2O7 2- + 6Fe2+ you ended up with. So when you work moles of Cr2O7 2- from the data given, you just times by 6 for moles of iron. Then can use n=mass/Mr to find the mass of iron, and the %.

I got like 53.7%

Reply 1397

SnookerLoopy

Original post by Smartie77

I got something like that but I think people are saying something about a ratio of 1:6. I don't remember seeing that in the question. I probably neglected to read it properly.

Oh right, I couldn't be bothered working out the ratio, I just said 'assuming it's a 1:1 reaction' ...We'll probably pick up about 3 marks for it though (out of 5)

Reply 1398

Scaudwell

Original post by Smartie77

I got something like that but I think people are saying something about a ratio of 1:6. I don't remember seeing that in the question. I probably neglected to read it properly.

Fe 2+ ---> Fe 3+ + e-

Cr2O7 2- + 14H+ + 6e- ---> 2Cr 3+ + 7H2O

THEREFORE: 6Fe 2+ + Cr2O7 2- + 14H+ ---> 2Cr 3+ + 6Fe 3+ + 7H2O

That's how you get your ratio.

Reply 1399

Betsss

For why copper (II) ions are blue, i put:
- Cu2+ is a transition metal ion, so it has an incomplete d-sublevel
- it can undergo d-to-d transitions
- the electrons absorb light, which excites them to a higher energy level
- they then emit light to fall back to ground state
- Cu2+ ions absorb red light from the visible spectrum; so they will appear the complementary colour (blue)

Is this ok, this was the section i was least confident with :frown:

and it had to come up!!

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