[Exam Cram] The Ultimate Chemistry Revision Thread 2019

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CheeseIsVeg
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Chemistry Revision Thread 2019 :yay:

yo yo yo :bhangra: Welcome to the thread of your dreams. :blow:

Here you can ask your Chemistry questions and get help for all your chemistry needs. :ahee:

Post below with...
Your study level
The topics you're revising
How you're feeling about it
Any questions you're stuck on?

Cheese will hover around the thread later being able to help you wonderful lot revise for the best science (Chemistry) :woohoo:

Please also feel free to ask/answer each others questions here. Let's all revise together :woo:
Last edited by 04MR17; 2 years ago
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username4515420
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A- level Chem Ocr A
Im just recapping the gaps in my knowledge for paper 1 so electrochemical series and transition metal colors
I feel all in all ready for paper 1 physical chemistry is my strong point
Im not a fan of paper 2 organic chem but i know majority of my synthetic routes and all my mechanisms so hopefully everything will be all good
Not a fan off polyesters and poly amides.
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RazzzBerries
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GCSE AQA Chem :'(
I have to go to school today because they are running a revision session. I'll probably let you know what we studied when I come back :yes:

Edit: Cheese, this is my worst science. Send help. :cry:
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CheeseIsVeg
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(Original post by RazzzBerries)
GCSE AQA Chem :'(
I have to go to school today because they are running a revision session. I'll probably let you know what we studied when I come back :yes:

Edit: Cheese, this is my worst science. Send help. :cry:
help is being sent
we can go over any of Ur least favourite topics when u get back
happy to help
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CheeseIsVeg
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(Original post by MidLad78)
A- level Chem Ocr A
Im just recapping the gaps in my knowledge for paper 1 so electrochemical series and transition metal colors
I feel all in all ready for paper 1 physical chemistry is my strong point
Im not a fan of paper 2 organic chem but i know majority of my synthetic routes and all my mechanisms so hopefully everything will be all good
Not a fan off polyesters and poly amides.
nice, my favourite stuff! :yay: sounds like Ur revision is going well! all the best!
organic and amides not so much my favourite
but we can take a look at any of these things together if u like! anything welcome!
Last edited by CheeseIsVeg; 2 years ago
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04MR17
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How many rude words can you spell using Chemistry Equations?:teehee:
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CheeseIsVeg
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(Original post by 04MR17)
How many rude words can you spell using Chemistry Equations?:teehee:
Fluorine Uranium Carbon K(Potassium)
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username4515420
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hey thank you! , could we run over the pre exponential factor and Arrhenius equation if that's okay?
(Original post by CheeseIsVeg)
nice, my favourite stuff! :yay: sounds like Ur revision is going well! all the best!
organic and amides not so much my favourite
but we can take a look at any of these things together if u like! anything welcome!
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CheeseIsVeg
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(Original post by MidLad78)
hey thank you! , could we run over the pre exponential factor and Arrhenius equation if that's okay?
Sure!
So the Arrhenius Equation is k = A e^(-Ea/RT) It's a very useful equation that describes the temperature dependence of the rate constant k
k is the rate constant
A is the pre-exponential factor
both k and A have the same units :youbetcha:
Ea is the activation energy usually measured in kJmol^-1 or Jmol^-1
It represents the minimum energy required for a reaction to proceed. (i.e. molecules have to overcome this energy barrier to have enough energy to react)
R is the gas constant (8.314 JK^-1)
T is the temperature measured in Kelvin.

Is there something about this or the equation you'd like to go over specifically?
really hope this helps
Last edited by CheeseIsVeg; 2 years ago
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username4515420
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That does help a lot wow thank you!
but theres this one question that I thoroughly hate its number 17b if you could explain it that would be amazing thank you so much for the help
https://www.google.co.uk/url?sa=t&rc...oUcUYV0LAAa_rt
(Original post by CheeseIsVeg)
Sure!
So the Arrhenius Equation is k = A e^(-Ea/RT) It's a very useful equation that describes the temperature dependence of the rate constant k
k is the rate constant
A is the pre-exponential factor
both k and A have the same units :youbetcha:
Ea is the activation energy usually measured in kJmol^-1 or Jmol^-1
It represents the minimum energy required for a reaction to proceed. (i.e. molecules have to overcome this energy barrier to have enough energy to react)
R is the gas constant (8.314 JK^-1)
T is the temperature measured in Kelvin.

Is there something about this or the equation you'd like to go over specifically?
really hope this helps
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CheeseIsVeg
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(Original post by MidLad78)
That does help a lot wow thank you!
but theres this one question that I thoroughly hate its number 17b if you could explain it that would be amazing thank you so much for the help
https://www.google.co.uk/url?sa=t&rc...oUcUYV0LAAa_rt
hey, yeah I'll take a look for you when I get home! I'll be about 30mins
so glad it helped !
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username4515420
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That's fine have a safe one. youre a legend thank you for the help I really appreciate it!
(Original post by CheeseIsVeg)
hey, yeah I'll take a look for you when I get home! I'll be about 30mins
so glad it helped !
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Stiff Little Fingers
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(Original post by MidLad78)
That does help a lot wow thank you!
but theres this one question that I thoroughly hate its number 17b if you could explain it that would be amazing thank you so much for the help
https://www.google.co.uk/url?sa=t&rc...oUcUYV0LAAa_rt

For part one, draw your line of best fit. You're working on the natural log of the Arrhenius equation (so now ln k = -Ea/R(1/T) + ln A) - if the reaction constant obeys the Arrhenius equation then you should get a linear fit and so the gradient of the line (-Ea/R) can be used to calculate the activation energy. So, rearrange from:

Gradient = -Ea/R

To

-Gradient * R = Ea

Check all your units match up (you're going to be wanting to eliminate Kelvin from it) and plug in the gradient and gas constant to get your activation energy.

Then for part 2, take the activation energy you've calculated and plug it in with an ln K and it's corresponding (1/T) value, then solve for A.
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username4515420
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Thank you very much I get all of it now ,apart from when you have too calculate the gradient so when youre doing ea/r how do you work that out in the mark scheme it says 800-1000 but what numbers have they used too get them apart from that the rest of its gold thank you
(Original post by Stiff Little Fingers)
For part one, draw your line of best fit. You're working on the natural log of the Arrhenius equation (so now ln k = -Ea/R(1/T) + ln A) - if the reaction constant obeys the Arrhenius equation then you should get a linear fit and so the gradient of the line (-Ea/R) can be used to calculate the activation energy. So, rearrange from:

Gradient = -Ea/R

To

-Gradient * R = Ea

Check all your units match up (you're going to be wanting to eliminate Kelvin from it) and plug in the gradient and gas constant to get your activation energy.

Then for part 2, take the activation energy you've calculated and plug it in with an ln K and it's corresponding (1/T) value, then solve for A.
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username4515420
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Dw bout it ive finally figured it out , thank you so much for youre help and you too cheese! thank you guys


(Original post by Stiff Little Fingers)
For part one, draw your line of best fit. You're working on the natural log of the Arrhenius equation (so now ln k = -Ea/R(1/T) + ln A) - if the reaction constant obeys the Arrhenius equation then you should get a linear fit and so the gradient of the line (-Ea/R) can be used to calculate the activation energy. So, rearrange from:

Gradient = -Ea/R

To

-Gradient * R = Ea

Check all your units match up (you're going to be wanting to eliminate Kelvin from it) and plug in the gradient and gas constant to get your activation energy.

Then for part 2, take the activation energy you've calculated and plug it in with an ln K and it's corresponding (1/T) value, then solve for A.
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Stiff Little Fingers
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(Original post by MidLad78)
Thank you very much I get all of it now ,apart from when you have too calculate the gradient so when youre doing ea/r how do you work that out in the mark scheme it says 800-1000 but what numbers have they used too get them apart from that the rest of its gold thank you
So I got the gradient as -Ea/R because every linear equation fits the format y= mx+c. So, with the natural log of the Arrhenius equation:

ln k = -Ea/R(1/T) + ln A

You can see on your graph y is ln k and x is (1/T) so:

y = -Ea/R x + ln A

-Ea/R is m, your gradient, ln A is c, your intercept.
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CheeseIsVeg
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(Original post by MidLad78)
Dw bout it ive finally figured it out , thank you so much for youre help and you too cheese! thank you guys
Awesome stuff!
If you wanted any help or clarification on any topic - feel free to ask :hugs:
(Original post by Stiff Little Fingers)
So I got the gradient as -Ea/R because every linear equation fits the format y= mx+c. So, with the natural log of the Arrhenius equation:

ln k = -Ea/R(1/T) + ln A

You can see on your graph y is ln k and x is (1/T) so:

y = -Ea/R x + ln A

-Ea/R is m, your gradient, ln A is c, your intercept.
beat me to it thanks for helping out!
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username4515420
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I sure will!
(Original post by CheeseIsVeg)
Awesome stuff!
If you wanted any help or clarification on any topic - feel free to ask :hugs:


beat me to it thanks for helping out!
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AH101
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Any tips on how to answer them long 6-8 marks titration questions that come up in paper 1? (AQA A-Level)
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rehman15
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Any help would be much appreciated
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