OCR AS Salters May 23rd Exam Thread 2014

Thankyou very much !
What do we say for the source of unburnt hydrocarbons? Incomplete combustion of hydrocarbons?
Final question is do we just have to know that oxygenates increase the octane rating and therefore decrease chance of auto ignition and engine damage?

Unburnt hydrocarbons escape from the engine without combusting.
Incomplete combustion produces CO and C (and H2O).
And that is always useful to know, but probably not. You should know oxygenates and having more oxygen in a molecule means more complete combustion.

Ok thankyou
When we are asked to complete equations for incomplete combustion do we only include CO and H2O on the product side?

When it asks us why relative atomic mass is not a whole number the answer is "average of naturally occurring isotopes" by this I assume it means average masses of naturally occurring isotopes?

Yes. It could produce C in theory too, but I think CO and H2O is the main products. I think it would say if it was talking about CO or C. Sorry if I am confusing you.

No it's okay! You aren't confusing me, I know that incomplete combustion produces C as well just wasn't sure if we had to include it in the equation

Yes. This part is a little weird to explain.

Basically, thermal stability refers how stable the (group 2 elements in this context) are when heated. If they are more stable they will require more heat to undergo thermal decomposition

Why it increases down the group: carbonate ions (CO3-) can be distorted in the presence of a positively charged ion (e.g Mg2+). As we go down the group, there is less distortion (because larger atoms have a smaller charge density) which leads to a higher thermal stability.

I think you use the following equation: Moles=vol (in 24dm3) / 24

So rearrange it to : vol (in 24dm3) = Moles x 24

Does anyone know how to work out volume of air calculations?

My revision guide says n= volume (in dm^3) / 24 OR n= volume (in cm^3) /24 000

Can anybody tell me all practicals we need to know please!
I know about group 2 and comparing thermal stability, and about using spirit burner and known vol of water to calculate enthalpy change of combustion. Any other? Thanks

No I know that one.

If you look in the may 2012 paper question 3d, it's a little more different
http://www.ocr.org.uk/Images/131291-question-paper-unit-f331-chemistry-for-life.pdf

I thinks that's about it.

That ones a bit different but do-able

1) We have 60cm^3 of gaseous isooctane (which is 0.06dm^3). First work out the number of moles 0.06/24 (1 mol of a gas occupies 24dm^3) = 0.0025
2) We know that 12.5moles of oxygen reacts with 1 mol of isooctane so mol O = 0.0025 x 12.5 = 0.03125
3) We have 0.03125 moles of oxygen which is 0.03125 x 24 = 0.75dm^3 of oxygen
4) Oxygen makes up 21% of air so 0.75/0.21=3.57dm^3

Mark schemes are really cryptic (especially OCR B ) so it's usually hard to work out whats going on

or rather simpler, your told that the ratio is 1:12.5, so if 60cm3 of isooctane reacts with o2, then there must be 60x12.5 of o2

no need to go through moles if ur given the ratio and the volume of one thing already

Guys can I just confirm the products of each of the refining process please?

Isomerisation - Branched alkanes
Reforming - cycloalkes/Aromatic (mark scheme says cylcoalkanes but accepts aromatic)
Catalytic cracking - Shorter chain alkanes - Cycloalkanes - Alkenes (Benzene)

This was off the top off my head and I've seen different answers in different books so want to double check

Guys can I just confirm the products of each of the refining process please?

Isomerisation - Branched alkanes
Reforming - cycloalkes/Aromatic (mark scheme says cylcoalkanes but accepts aromatic)
Catalytic cracking - Shorter chain alkanes - Cycloalkanes - Alkenes (Benzene)

This was off the top off my head and I've seen different answers in different books so want to double check