#1
Hey guys I want to start this thread to help any of you guys out there studying OCR chemistry at A level. Hopefully this will turn out to be a really good revision thread!

I am simply an enthusiastic chemistry student who is going on to study it at university. I love helping others, and I am interested in teaching at some stage!
1) Try and relate your questions, e.g, page references, a past paper question, an example etc
2) It may be easier to post pictures of our working, so we'll see how that goes

Just want to help as many people as I can! Let's get as many people on this as possible, link your other classmates the more chemists the more revision!
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7 years ago
#2
I have been researching to try and find the correct way to convert a number from mol/dm^3 to g/dm^3 but I am really struggling. Which is the correct way to do it accurately?
Could you provide an exam so I can see how you do it.
thanks
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#3
Hi locket, I will do that for you, I'll reply shortly with some detailed explanations!
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7 years ago
#4
Hi, I was wondering if you could help me with this question.
When doing an enthalpy change of combustion experiment and you increase the mass of water that you are heating up, would this cause the resulting enthalpy change to be more or less exothermic?
Thanks.
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#5
Hi GCSE, I will reply shortly, just going to respond to my first question, bear with me!

Okay locket, I have an explanation and a question for you to practise. There are actually a couple of ways you can do it, but this way is probably the easiest way to describe it:

1) n = c x v.
2) c = n/v
3) n = m/M
4) Because we have two equations equal to n, we can sub them into eachother, which becomes m/M = c x v
5) divide by v, and multiply by M, there you have it, you now have m/v = c x M.
6) Use the periodic table to work out M, and use part 2) to work out c, and you will have gdm-3 (m/v)
7) remember its per dm3 so if you have say a 25cm3 solution you have to multiply by 40 to get 1 dm3, which is the other way of doing it

Try question 7b) on june 2010 F325, I have done the question so try your best and I will put up working if you're stuck, the method in the mark scheme involves part 7) but I personally think it's harder that way. In this question, using m/v = c x M does work, I've just done it! Let me know how it goes! I'll help if you get stuck it is a hard question but I'll help you ace it
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#6
Okay GCSE, when you increase the mass of the water, are you still burning the same mass of fuel? If you have a fixed amount of fuel being burnt, the answer may be different compared to a separate experiement, where the mass of water is simply different. Let me know, my answer will differ with your answer. Is this a question in a paper? My answer would be more accurate if I read the question.
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7 years ago
#7
(Original post by ChemistryOCR)
Okay GCSE, when you increase the mass of the water, are you still burning the same mass of fuel? If you have a fixed amount of fuel being burnt, the answer may be different compared to a separate experiement, where the mass of water is simply different. Let me know, my answer will differ with your answer. Is this a question in a paper? My answer would be more accurate if I read the question.
Hi, well I am preparing for a practical that I have tomorrow and I think a question like this may come up but I can't find any example questions on the internet! I think that it might be keeping the mass of the fuel the same but increasing the mass of the water, would it be more, less exothermic or no change.
And also, if you increased the starting temperature of the water (to say 60 degrees) would it have an affect on the enthalpy change?
Thanks
0
7 years ago
#8
Would anyone be able to help me with this question on the F325 mock (June 2014), no idea how you can get the gas just from density and form the equation. I know that the precipitate is Fe(OH)3 as it says is has an orange precipitate. thanks

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0
7 years ago
#9
Here is the question:

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0
7 years ago
#10
(Original post by gcse1998)
Hi, I was wondering if you could help me with this question.
When doing an enthalpy change of combustion experiment and you increase the mass of water that you are heating up, would this cause the resulting enthalpy change to be more or less exothermic?
Thanks.
It depends if everything else (i.e the other values) were scaled up also as well as increasing the temperature.

If the experiment did complete combustion of the same mass of fuel in the burner in both experiments or If it was timed incomplete combustion of the same mass of fuel in the burner in both experiments then either way it would be fine because the mass of the fuel burned needs to be controlled (i.e the same in both).

As long as that was controlled and nothing else was changed about the experiment. I would say that because there is more water to heat up with the same amount of fuel. The energy change would be less exothermic because there would be more water to heat up (so less of a temperature change.) Interested to see what others think.
0
7 years ago
#11
(Original post by ChemistryOCR)
Hey guys I want to start this thread to help any of you guys out there studying OCR chemistry at A level. Hopefully this will turn out to be a really good revision thread!

1) Try and relate your questions, e.g, page references, a past paper question, an example etc
2) It may be easier to post pictures of our working, so we'll see how that goes

Just want to help as many people as I can! Let's get as many people on this as possible, link your other classmates the more chemists the more revision!

There are other threads, specific to each exam too. So check them out.
0
#12
(Original post by gcse1998)
Hi, well I am preparing for a practical that I have tomorrow and I think a question like this may come up but I can't find any example questions on the internet! I think that it might be keeping the mass of the fuel the same but increasing the mass of the water, would it be more, less exothermic or no change.
And also, if you increased the starting temperature of the water (to say 60 degrees) would it have an affect on the enthalpy change?
Thanks
If we think of the definition of enthalpy change of combustion, it is the energy change which occurs when one mole of a compound is burnt completely with oxygen. Changing the mass of the water cannot change the value of this enthalpy change, because the energy is being given off. What does change, is how much the water is heated up by. More water is heated up, so the temperature change will be less, and without numbers it's hard to disprove, but mass increases, delta T decreases, so the answer should be the same. Do you understand? The same amount of energy is heating a higher volume of water, and because of the specific heat capacity, 4.18 J of energy is required to raise the temperature of 1 g of water by 1 degrees, so you can see that more mass will equal a smaller temperature change.

Regarding your second question, initial temperature does not really matter. It is the temperature change that is required. If the temperature rises by 20 degrees, it will always increase by 20 degrees no matter what the initial temperature is, because the amount of energy given off was enough to raise that volume of water by 20 degrees. Again look back at the definition of specific heat capacity, understanding the definition makes this clearer. If the mass of fuel being burnt is constant, the energy heating the water is constant, and so the temperature will raise by the same amount, regardless of initial temperature. Only when mass of water changes does it get more complicated as you can probably see in my first answer for you, but you are unlikely to be asked that, I never encountered it and working out the answer is a mathematical skill, not chemistry as such, because it's a ratio question. Mass goes up, change in temp goes down, Q should be the same, because the same amount of heat is being given off, to the best of my knowledge this is all correct, but due to lack of data, real questions and the fact that I've never encountered it before makes it a slightly dodgy area, but I'm sure, if they were asking what would happen, you know the same amount of energy is being given off, so no the enthalpy change cannot differ. How you record the enthalpy change however, will.
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#13
(Original post by _NMcC_)
It depends if everything else (i.e the other values) were scaled up also as well as increasing the temperature.

If the experiment did complete combustion of the same mass of fuel in the burner in both experiments or If it was timed incomplete combustion of the same mass of fuel in the burner in both experiments then either way it would be fine because the mass of the fuel burned needs to be controlled (i.e the same in both).

As long as that was controlled and nothing else was changed about the experiment. I would say that because there is more water to heat up with the same amount of fuel. The energy change would be less exothermic because there would be more water to heat up (so less of a temperature change.) Interested to see what others think.
NmcC yeah less of a temperature change, but then the mass has gone up, so read my answer and see what you think, do you agree? Enthalpy cannot change, the temperature of the water can, but it has gone up in the first place. mc delta T. Mass up, T down, do you think the answer would be the same? I think it should be. Very interesting though
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7 years ago
#14
(Original post by ChemistryOCR)
If we think of the definition of enthalpy change of combustion, it is the energy change which occurs when one mole of a compound is burnt completely with oxygen. Changing the mass of the water cannot change the value of this enthalpy change, because the energy is being given off. What does change, is how much the water is heated up by. More water is heated up, so the temperature change will be less, and without numbers it's hard to disprove, but mass increases, delta T decreases, so the answer should be the same. Do you understand? The same amount of energy is heating a higher volume of water, and because of the specific heat capacity, 4.18 J of energy is required to raise the temperature of 1 g of water by 1 degrees, so you can see that more mass will equal a smaller temperature change.

Regarding your second question, initial temperature does not really matter. It is the temperature change that is required. If the temperature rises by 20 degrees, it will always increase by 20 degrees no matter what the initial temperature is, because the amount of energy given off was enough to raise that volume of water by 20 degrees. Again look back at the definition of specific heat capacity, understanding the definition makes this clearer. If the mass of fuel being burnt is constant, the energy heating the water is constant, and so the temperature will raise by the same amount, regardless of initial temperature. Only when mass of water changes does it get more complicated as you can probably see in my first answer for you, but you are unlikely to be asked that, I never encountered it and working out the answer is a mathematical skill, not chemistry as such, because it's a ratio question. Mass goes up, change in temp goes down, Q should be the same, because the same amount of heat is being given off, to the best of my knowledge this is all correct, but due to lack of data, real questions and the fact that I've never encountered it before makes it a slightly dodgy area, but I'm sure, if they were asking what would happen, you know the same amount of energy is being given off, so no the enthalpy change cannot differ. How you record the enthalpy change however, will.
Ahhh yes thank you so much that makes sense. Regarding my 2nd question, if the starting temperature was relatively high would it make a difference because the water could evaporate?
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#15
(Original post by gcse1998)
Ahhh yes thank you so much that makes sense. Regarding my 2nd question, if the starting temperature was relatively high would it make a difference because the water could evaporate?
Ooo nice question! If the initial temperature was higher than 100 degrees, then the mass would decrease constantly and the practical would be impossible at our level. As the mass decreased, temperature change would increase as the same amount of heat is heating less mass, but it would become exponential and it's just not something that will happen, nor be asked of you, I promise you that. Because of this, the initial and final temperature will be under 100 degrees for sure. Do not worry about evaporation they won't put it above 100.
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#16
(Original post by ChemistryOCR)
Ooo nice question! If the initial temperature was higher than 100 degrees, then the mass would decrease constantly and the practical would be impossible at our level. As the mass decreased, temperature change would increase as the same amount of heat is heating less mass, but it would become exponential and it's just not something that will happen, nor be asked of you, I promise you that. Because of this, the initial and final temperature will be under 100 degrees for sure. Do not worry about evaporation they won't put it above 100.
Also, if you do see slight evaporation occuring, worry not. This is not evaporation due to boiling, it is surface evaporation which occurs naturally, constantly. What is happening here is that the molecules of the surface of the water are moving around and can actually produce enough kinetic energy and thus heat energy to allow some molecules to escape the surface. Although yes mass is being lost this is seriously negligable because there isn't actually a state change happening. You will see this effect more and more as the temperature increases, but unless the temperature reads above 100, (which it definitely shouldn't) do not worry about it and ignore it, you are not losing mass, and the water isn't boiling!
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#17
(Original post by Jimmy20002012)
Would anyone be able to help me with this question on the F325 mock (June 2014), no idea how you can get the gas just from density and form the equation. I know that the precipitate is Fe(OH)3 as it says is has an orange precipitate. thanks

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I'll be right with you!
0
7 years ago
#18
Thanks

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0
7 years ago
#19
(Original post by ChemistryOCR)
Ooo nice question! If the initial temperature was higher than 100 degrees, then the mass would decrease constantly and the practical would be impossible at our level. As the mass decreased, temperature change would increase as the same amount of heat is heating less mass, but it would become exponential and it's just not something that will happen, nor be asked of you, I promise you that. Because of this, the initial and final temperature will be under 100 degrees for sure. Do not worry about evaporation they won't put it above 100.
Ahhh great, thank you so much for your help!
0
#20
(Original post by gcse1998)
Ahhh great, thank you so much for your help!
No worries hope I explained it properly for you, and good luck!!!
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