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# AS Chemistry- helping each other out! watch

1. Green chemistry bores me so much
2. I really don't understand hess's law or born harber cycles too so I would also appreciate any help on this topic thanks
3. (Original post by SubwayLover1)
Guys...
If CCl2 has a bond angle of 118 then why does it change so dramatically when you have two lone pairs... I thought if it was V-shaped then it would be 115
If it has 2 lone pairs and 2 bonding pairs then it ought to have bond angle of 104.5o and it'll be a bent shape.
4. (Original post by TeaAndTextbooks)
I really don't understand hess's law or born harber cycles too so I would also appreciate any help on this topic thanks
5. (Original post by TeaAndTextbooks)
I really don't understand hess's law or born harber cycles too so I would also appreciate any help on this topic thanks
Hey guys I'm really struggling.
I think O understand the Hess's cycle but when I'm solving past papers I never seem to solve them. So could someone please explain the hess's cycle? Thanks much appreciated.

Just to mention, the cycle for Combustion and the route is ALWAYS the same for every combustion-related question. The values just differ from question to question.
1) Begin by forming an equation to form the products.
2) Substitute numbers into the cycle and calculate the Sum of Reactants.
3) As the cycle suggests to form the combustion products you can directly combust the products, we go against this arrow as suggested by the route which needs to be taken - and so, the value of the product changes from NEGATIVE to POSITIVE. Enthalpy change of reactants + products = answer.

Hope that helped, I'll try and find a question for Enthalpy change of Formation.
Attached Images

6. (Original post by Disney0702)
If it has 2 lone pairs and 2 bonding pairs then it ought to have bond angle of 104.5o and it'll be a bent shape.
I don't understand why it goes from 118 to 104.5 though?

Just to mention, the cycle for Combustion and the route is ALWAYS the same for every combustion-related question. The values just differ from question to question.
1) Begin by forming an equation to form the products.
2) Substitute numbers into the cycle and calculate the Sum of Reactants.
3) As the cycle suggests to form the combustion products you can directly combust the products, we go against this arrow as suggested by the route which needs to be taken - and so, the value of the product changes from NEGATIVE to POSITIVE. Enthalpy change of reactants + products = answer.

Hope that helped, I'll try and find a question for Enthalpy change of Formation.

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Thankyou that helps alot I finally get the hess's law stuff, born haber cycles are still killing me though but thanks anyway
8. (Original post by TeaAndTextbooks)
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Thankyou that helps alot I finally get the hess's law stuff, born haber cycles are still killing me though but thanks anyway
What specification are you doing?
9. (Original post by SubwayLover1)
I don't understand why it goes from 118 to 104.5 though?
the lone pair-lone pair repulsion is bigger and greater than the electron-electron bond
10. (Original post by sara_ara98)
the lone pair-lone pair repulsion is bigger and greater than the electron-electron bond
But for each electron pair they are repeled by 2.5 so I dont get why it goes from 118 to 104.5
What specification are you doing?
Edexcel you?

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12. (Original post by haemo)
I don't know if this will help you, but this works for me:

If given formation data: enthalpy = sum of products - sum of reactants

If given combustion data: enthalpy = sum of reactants - sum of products

I know that quite a few students use energy cycles, but I find that the formulas are easier to use.
This makes life alot easier thank you

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13. (Original post by TeaAndTextbooks)
Edexcel you?

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OCR, can't help you with Haber cycle - sorry! Although, I can help with alternative overlaps.
14. Can someone help me out here, The spec says this:
a) Describe the reactions of alkanes in terms of combustion and substitution by chlorine showing the mechanism of free radical substitution in terms of initiation, propagation and termination, and using curly half-arrows in the mechanism to show the formation of free radicals in the initiation step using a single dot to represent the unpaired electron.

To me this makes no sense, Alkanes don't undergo substitution reaction?? Plus the book has nothing on it, somone help.
15. (Original post by Rstlss)
Can someone help me out here, The spec says this:
a) Describe the reactions of alkanes in terms of combustion and substitution by chlorine showing the mechanism of free radical substitution in terms of initiation, propagation and termination, and using curly half-arrows in the mechanism to show the formation of free radicals in the initiation step using a single dot to represent the unpaired electron.

To me this makes no sense, Alkanes don't undergo substitution reaction?? Plus the book has nothing on it, somone help.

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Alkanes do undergo substitution reactions.UV radiation breaks a C-h bond and makes radicals.
The pics below should help.
Attached Images

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Alkanes do undergo substitution reactions.UV radiation breaks a C-h bond and makes radicals.
The pics below should help.
Ahh yes, I jst came here to say I found it, it's jsut me being an iodiot i was busy trying to find it in the big pictures, it's not a picture in my textbook xD
17. (Original post by Rstlss)
Can someone help me out here, The spec says this:
a) Describe the reactions of alkanes in terms of combustion and substitution by chlorine showing the mechanism of free radical substitution in terms of initiation, propagation and termination, and using curly half-arrows in the mechanism to show the formation of free radicals in the initiation step using a single dot to represent the unpaired electron.

To me this makes no sense, Alkanes don't undergo substitution reaction?? Plus the book has nothing on it, somone help.
Initiation:
Cl2 -> Cl + Cl
Propagation:
CH4 + Cl -> CH3 + HCl
CH3 + Cl2 -> CH3Cl + Cl
Termination/Overall:
Cl + Cl -> Cl2
CH3 + CH3 -> C2H6
Cl + CH3 -> CH3Cl
18. Can anyone summarise mass spectrometry especially the fragment ion part?
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Alkanes do undergo substitution reactions.UV radiation breaks a C-h bond and makes radicals.
The pics below should help.
You mean UVbreaks Cl-Cl bond

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20. (Original post by rm_27)
Green chemistry bores me so much
literally
why does it exist in the spec

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