Edexcel ial chemistry unit 1 june 2018 discussion & predictions Watch

E.143.G
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Hi all,

does anyone have any predictions for the upcoming unit 1 exam?

we can also have questions/topics discussions on here
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Ammaar99
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(Original post by E.143.G)
Hi all,

does anyone have any predictions for the upcoming unit 1 exam?

we can also have questions/topics discussions on here
I have no clue what's going to come but I'm **** scared dude
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Linda.ah
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(Original post by E.143.G)
Hi all,

does anyone have any predictions for the upcoming unit 1 exam?

we can also have questions/topics discussions on here

I don't know what's coming, but from what I solved, I think: •Mechanisms of either Free Radical substitution with its steps, or The alkene addition reactions with curly arrows (Usually they ask for names of reactions, catalysts and conditions and all)
•How to prepare pure crystals
•Enthalpy change question that has, calculating enthalpy change, Hess Law, labelling born haber cycle...
•Question about isotopes, ionization energy, radius length of atoms/ions....
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Kevin.US
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(Original post by Linda.ah)
I don't know what's coming, but from what I solved, I think: •Mechanisms of either Free Radical substitution with its steps, or The alkene addition reactions with curly arrows (Usually they ask for names of reactions, catalysts and conditions and all)
•How to prepare pure crystals
•Enthalpy change question that has, calculating enthalpy change, Hess Law, labelling born haber cycle...
•Question about isotopes, ionization energy, radius length of atoms/ions....
2017 october predictions huh?
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viktorbrown100
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can anyone here write down in detail the rends in ionization energies across and down a periodic table. with reasons
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Simplysubah
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Phosphorus reacts with iodine to produce phosphorus(III) iodide:P4 6I2 → 4PI3 What is the minimum mass of iodine required to produce 1 kg of phosphorus(III) iodide when the phosphorus is in excess? Data: molar mass of phosphorus(III) iodide = 411.7 g mol−1
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Linda.ah
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(Original post by Kevin.US)
2017 october predictions huh?
No, actually I have solved a good number of papers to be able to actually know what usually comes.

I suggest you solve some other papers to see the pattern.
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Linda.ah
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(Original post by viktorbrown100)
can anyone here write down in detail the rends in ionization energies across and down a periodic table. with reasons

As you go down a group, the atomic radius gets larger and so the attraction between the outermost negatively charged electron and the positively charged nucleus is less. So, the as you go down the group, the ionization energy decreases.


However, as you go across a group, it increases; because they all have the same shielding but the number of protons increase as well as the number of electrons. So the atomic radius is smaller, ending up with needing more energy to remove the outermost electron.
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Linda.ah
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(Original post by Simplysubah)
Phosphorus reacts with iodine to produce phosphorus(III) iodide:P4 6I2 → 4PI3 What is the minimum mass of iodine required to produce 1 kg of phosphorus(III) iodide when the phosphorus is in excess? Data: molar mass of phosphorus(III) iodide = 411.7 g mol−1

4P + 6I2 -----> 4PI3
Okay so here, they're asking for the mass of iodine for 1000 g (1 kg) pf Phosphorus iodide.
You have the mass, the molar mass, so you get out the number of moles using number of mol = mass/molar mass. mass should be in grams of course.
1000/411.7 = 2.43 mol
but in the equation they have 4.
PI3: 4 ----> 2.43
I2: 6 ------> ?
? = (6 x 2.43)/4 = 3.64 mol
Now you have Iodine's number of mol, you need the molar mass to get its mass, its molar mass is 254(since it's I2 molecule, not one atom)
3.64 x 254 = 924.56 grams
if you want it in kg, divide by 1000 you'll get 0.9245.
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M. Alrashid
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The boundaries for this unit are ridiculously high!
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viktorbrown100
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(Original post by Linda.ah)
As you go down a group, the atomic radius gets larger and so the attraction between the outermost negatively charged electron and the positively charged nucleus is less. So, the as you go down the group, the ionization energy decreases.


However, as you go across a group, it increases; because they all have the same shielding but the number of protons increase as well as the number of electrons. So the atomic radius is smaller, ending up with needing more energy to remove the outermost electron.
Thats a more simplified version of. I'm looking the one for the long 5-6 mark questions. But thanks anyway
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Mịřäžūľ Īšľâm
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(Original post by M. Alrashid)
The boundaries for this unit are ridiculously high!
Chemistry paper ain't that hard bro.
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Mịřäžūľ Īšľâm
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(Original post by viktorbrown100)
can anyone here write down in detail the rends in ionization energies across and down a periodic table. with reasons
Across the period Ionization energy increases generally. This is because nuclear charge increases (charge density). Therefore the atmic radius become smaller in size. As a result need more energy to remove electron from outermost orbit. But there are two exception group 3 because removing electron from from p subshell is easier than s subshell as p subshell is further away from nucleus than s subshell, required less enegy to remove electron from p subshell. And the other one is for group 6 this is because group 6 has pair of electron in p subshell there fore they repel compare to group 5 which has no pair of electron in p subshell . As a result removing electron is easier in group 6 comapre to group 5 so requires less energy.
As for down the group energy requires decreases as one extra shell is added therefore the electron become further away from nucleus and required less energy to be removed
.
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viktorbrown100
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Why does obtained value differ from the value calculated by using mean bond enthalpies. 2 REASONS
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IALs
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(Original post by viktorbrown100)
Why does obtained value differ from the value calculated by using mean bond enthalpies. 2 REASONS
1. The ionic compound has some COVALENT character in it because there isn't enough electronegativity difference between the two to allow for complete transfer of electrons.
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Presto
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How to do this?
Ans 290
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IALs
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(Original post by Presto)

How to do this?
Ans 290
The question states that only 10% of Oxygen gas reacts to form Ozone. 10% of it is 30 cm cubes. If 30 cm cube reacts, then we will get only 20cm^3 of Ozone gas. 3 moles of oxygen give 2 moles of ozone so 30cm will give 20cm. The remaining gas is the unreacted oxygen which is 300-30 = 270cm^3.
Total volume of gases present at the end of reaction: 20cm of ozone and 270cm of oxygen. = 290cm^3 of gases.
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Presto
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(Original post by IALs)
The question states that only 10% of Oxygen gas reacts to form Ozone. 10% of it is 30 cm cubes. If 30 cm cube reacts, then we will get only 20cm^3 of Ozone gas. 3 moles of oxygen give 2 moles of ozone so 30cm will give 20cm. The remaining gas is the unreacted oxygen which is 300-30 = 270cm^3.
Total volume of gases present at the end of reaction: 20cm of ozone and 270cm of oxygen. = 290cm^3 of gases.
Thanks so much
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IALs
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(Original post by Presto)
Thanks so much
My pleasure
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Mịřäžūľ Īšľâm
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