OCR Chemistry AS Depth Thursday 23rd 2019 Unofficial Markscheme

I got 8.16g

1) weighted mean mass of an atom compared to the mass of 1/12 a carbon 12 atom
2) something like 86?
3) some volumes were 60, 35.8 (or 38.5)? cm3?

Sr + 2H2O --> Sr(OH)2 + H2
If you use calcium (less reactive), it's less vigorous (it'll take more time for the metal to dissolve and more times for the bubbling assuming same mass used) but also pH will be less (as you go down group 2, they get more alkaline).

For the hydrolysis, you use a timer and see how long it takes for a precipitate to form. C-I bond enthalpy is lower so reacts faster.

For catalyst, reduces activation energy by providing alternative reaction pathway so more particles have energy >= activation increasing rate.

As the reaction progresses, rate decreases as rectants are used up (conc goes down) so the number of unreacted particles colliding decreases per unit time ==>> rate goes down over time.

With the new graph, it hits the same volume at the end (as number of moles is the same) but it's done faster.



First 6 marker:
I found the mass needed, 25.63g or something. So I said get this mass, weighing boat, 100cm3 beaker and stirr, add distilled water, put into volumetric flask (through funnel), rise the beaker and transfer the rinse and then add distilled water until bottom of miniscus is at 250 then invert for uniform concentration then you've got it?

Second 6 marker:
Mass 8.16g? So first you remove water by getting organic and aqueous layer, separating funnel, invert, allow to settle, add water, the layer that expands is water and water is denser than both so in bottom then remove the water onto a flask and same for conical flask for organic layer and then you have both the organic layer and aqueous then you wanna separate hexan-1-ol and the alkene by redistilling it and colleting off the liquid hex-1-ene.

If you use hexan-2-ol, you reduce yield as you also make hexan-2-ene.


Metallic bonding - strong electrostatic attraction between sea of delocalised electrons and positive metal cations. A high temperature is needed to provide sufficient energy to overcome these strong electrostatic forces, hence high elting point. These delocalised electrons can move and conduct.

Polymer - I said fertiliser? Is this even a mark?

For drawing, I did 2 repeat units for it.

Nickel catalyst needed, solution turns from orange to colorless.

For drawing the 3, one was just an alkane, one had a H at the end and chlorine at penultimate one (for major product) and one had 2 bromines.


X=5.8
He did it many times to heat to constant mass and ensure no water of crystallisation left.

Modifications:
1) more accurate mass balance to more dp
2) use larger mass.

Glad I got basically all the same answers as you. For that mass of Mg(no3)2 where you got 23.63, I got the same answer or something very similar, but I got confused and I feel like we got it wrong, because 23.63 was the mass we calculated from using Mg(no3)2 • 6h2o, but the standard we were supposed to make was 250cm3 of 0.4moldm-3 of Mg(no3)2. So using the Mr of the hydrated salt would’ve given us the wrong mass to make up that solution wouldn’t it?

Question 1:

1) The weighted mean mass of an atom compared to 1/12th the mass of a carbon-12 atom
2) Something like 86?

I did a diagram and said


Metallic bonding - the strong electrostatic attraction between the sea of delocalised electrons and positive metal cations. A high temperature is needed to provide sufficient energy to overcome these strong electrostatic forces, hence a high melting point. These delocalised electrons can move and conduct, hence a good electrical conductor.


Sr + 2H2O --> Sr(OH)2 + H2
If you use calcium (less reactive), it's less vigorous (it'll take more time for the metal to dissolve and more times for the bubbling) but also pH will be less (as you go down group 2, they get more alkaline) so if you use calcium, pH won't be as high (as alkaline).


Question 2:


1) The water of crystallisation: x=5.8 to 2sf?
2 i) He did it 4 times to heat to constant mass and ensure all the water of crystallisation has evaporated
2 ii) He can use a mass balance accurate to more decimal points or increase the mass used - both of these reduce % uncertainty

Question 3:

I found the mass needed, 25.63g or something.

Calculation:
I'm a bit foggy on the numbers but I think what was required was 250cm3 at 0.4M so I found the number of moles by N=CV and used the molar ratio then multiplied by the Mr of hydrated magnesium nitrate to get 25.63 grams or something.

Method:

- get the mass and use weighing boat and method of differences to accurately measure a mass of this value
- add to 100cm3 beaker and add distilled water and stir thoroughly
- pour it onto a glass rod through a funnel into a volumetric flask
- rinse the beaker and transfer the rinse onto the volumetric flask
- add distilled water so the bottom of meniscus hits the graduation mark (250cm3 volumetric flask)
- invert the flask several times to ensure uniform concentration



Question 4 (I think - rates):

Calculations were 60cm3 and 35.8cm3 (or 38.5cm3).

Then you plot it and put the curve, time/s is the x-axis.

As the reaction progresses, rate decreases as reactants are used up (concentration goes down) so the number of unreacted particles colliding decreases per unit time ==>> rate goes down over time.

For catalyst, reduces activation energy by providing an alternative reaction pathway so more particles have energy >= activation increasing rate increasing the frequency of successful collisions, increasing rate.

With the new graph, it hits the same volume at the end (as the number of moles is the same) but it's done faster.

For the hydrolysis, you use a timer and see how long it takes for a precipitate to form.

C-I bond enthalpy is lower so energy is required to break it and therefore reaction occurs faster.


Question 5 (I think - organic chemistry):

First was drawing the 3 compounds which were fine I think

Nickel catalyst
Colour change from orange to colourless

6 marker:
Mass 8.16g?

Calculation:
So the student wants 4.20g of C6H12 so I found the actual number of moles he wants to get. I divided this by 0.625 to find the theoretical yield of moles and as it's 1-1,I multiplied by the Mr of hexan-1-ol to get 8.16g needed in the beginning.

Method:
- pour the mixture into separating funnel
- invert it for the layers to mix
- allow layers to settle
- add water; the layer that expands in volume = aqueous layer
- water is denser than both of these organic molecules so it's at the bottom
- once clearly separated, get a conical flask and open tap and let it out
- then pour the remaining mixture of organic liquids into another conical flask (ensure these are labelled)
- redistill it and collect your alkene


If you use hexan-2-ol, you reduce yield as you also make hexan-2-ene.


For repeating unit, I did the 2, put brackets, n and stuff

For 2 ways to dispose of - I said fertiliser - is this even a mark? I was unsure about this.

For that last one i spoke about it being biodegradable and photodegradable

Alcohol 8.16
Separating funnel and distill
Standard solution
60cm3
35.8cm2 or somerhing
86 for isotope
5. 8 for x
Mass of zinc like 26g I can't remember just plug formula this is the standard solution q
Nickel catalyst
Decolorises from orange to colorless
Major is when the H goes TO c connected to most h
Metallic bonding explanation
Observation is lower pH and takes more time to get the fizzing and metal dissolves (less vigorous)

Less H2 given off?

Wouldnt that depend on mass of calcium used eg perhaps he used enough so same amount is produced so we cannot assume this

But like slower rate of it being produced soz that’s what I meant

Do you think the grade boundaries will be higher than last year?

Uhh anyone else get 11?

I got 11 too

nah its defo 5.8 i think

Did you calculate the mean there were 4 values and 2 were the same so I used that value. Moles worked out at 0.008 and 0.09 then divided by the smallest and whole number ratio was 1:11