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# Combustion related help please watch

1. Does anybody know how to work this out? It was from the new Aqa Chem AS paper 2016
Thank You
2. (Original post by PBC Gyal)

Does anybody know how to work this out? It was from the new Aqa Chem AS paper 2016
Thank You
The working is right there in front of you!

Density = mass/vol

Thus

mass = density x volume

as the density is given in g cm-3 you have to use the volume in cm3

Once you have the mass you divide by the relative mass of ethanol to get the number of moles.

Then you multiply by the energy per mol.
3. (Original post by charco)
The working is right there in front of you!

Density = mass/vol

Thus

mass = density x volume

as the density is given in g cm-3 you have to use the volume in cm3

Once you have the mass you divide by the relative mass of ethanol to get the number of moles.

Then you multiply by the energy per mol.
Ohh okay thanks but why multiply at the end? 🤔
4. (Original post by PBC Gyal)
Ohh okay thanks but why multiply at the end? 🤔
You work out how many moles of ethanol you are burning. More ethanol = more energy release.

Every one mole of ethanol releases 1371 kJ

Therefore the total energy release is the energy released by 1 mol multiplied by the number of moles being burned.
5. (Original post by charco)
You work out how many moles of ethanol you are burning. More ethanol = more energy release.

Every one mole of ethanol releases 1371 kJ

Therefore the total energy release is the energy released by 1 mol multiplied by the number of moles being burned.
Oh thank you I understand where that's coming from 🤗
6. (Original post by charco)
You work out how many moles of ethanol you are burning. More ethanol = more energy release.

Every one mole of ethanol releases 1371 kJ

Therefore the total energy release is the energy released by 1 mol multiplied by the number of moles being burned.

Do you know how to work this out? Attachment 609424609426

The last image is the question for the graph above
Attached Images

7. (Original post by PBC Gyal)
Do you know how to work this out? Attachment 609424609426

The last image is the question for the graph above
Concentration of the second ethanoic acid is larger therefore the reaction will be faster (steeper slope)

Moles in the second case = 0.8 x 0.02 = 0.016 mol
Moles in the first case = 0.6 x 0.03 = 0.018 mol

Therefore in the second case the graph will not achieve the same height as the first case (less hydrogen produced)

So you need a steeper slope with a levelling off lower (8) than the first case.(9)
8. [QUOTE=charco;69438002]Concentration of the second ethanoic acid is larger therefore the reaction will be faster (steeper slope)

Moles in the second case = 0.8 x 0.02 = 0.016 mol
Moles in the first case = 0.6 x 0.03 = 0.018 mol

Therefore in the second case the graph will not achieve the same height as the first case (less hydrogen produced)

So you need a steeper slope with a levelling off lower (8) than the first case.(9)[/QUOTE

Aaah okay thank you so much for your help! I appreciate it 🤗

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