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Original post by frozo123
http://qualifications.pearson.com/content/dam/pdf/A%20Level/Chemistry/2013/Exam%20materials/6CH01_01_que_20110113.pdf

May someone explain to me why in Q6a) why Answer A can be completely ruled out?
and I don't get Q11)

thankyou


For Question 6, successive ionisation energy is the energy required to remove each electron in turn. Each successive ionisation energy is larger than the last, as more there is one less electron, so there is less electron repulsion, therefore the outer shell electrons are a tiny bit closer to the nucleus (meaning more energy is required to remove it). However, there is an even bigger jump in ionisation energies between shells, as electrons in inner shells are much closer to the nucleus than those in outer shells, and this can be seen on the graph where there are larger jumps than usual.

On the graph, there is at least 1 jump in energies which is between 3rd and 4th ionisation energies. Therefore, after the 3rd electron being removed, there is a new shell (I am ignoring the jump between 1st and 2nd ionisation energies, as that would imply that there would be 1 electron in the outer shell, then only 2 electrons in the next shell, which is not possible since it isn't the innermost shell)so it can't be in Group 1, as Group 1 elements only have 1 electron in the outer shell.

Hope this helped!
Original post by frozo123
http://qualifications.pearson.com/content/dam/pdf/A%20Level/Chemistry/2013/Exam%20materials/6CH01_01_que_20110113.pdf

May someone explain to me why in Q6a) why Answer A can be completely ruled out?
and I don't get Q11)

thankyou


For 6a) Answer A can be completely rules out because group 1 has only 1 electron in it's outer shell, the graph shows 10 electrons being removed.

Q11)It says that most compounds of lead are insoluble, basically inferring that if you add sulphuric acid to lead the lead will dissolve, but then it tells us that lead(II)nitrate is an exception to this. Then all the options but C include Lead(II) nitrate solution (1st clue) and all the others include adding sulphuric acid to other lead compounds (2nd clue ). There for the answer must be C.

Hope that helped!
Original post by lukejoshjames
For Question 6, successive ionisation energy is the energy required to remove each electron in turn. Each successive ionisation energy is larger than the last, as more there is one less electron, so there is less electron repulsion, therefore the outer shell electrons are a tiny bit closer to the nucleus (meaning more energy is required to remove it). However, there is an even bigger jump in ionisation energies between shells, as electrons in inner shells are much closer to the nucleus than those in outer shells, and this can be seen on the graph where there are larger jumps than usual.

On the graph, there is at least 1 jump in energies which is between 3rd and 4th ionisation energies. Therefore, after the 3rd electron being removed, there is a new shell (I am ignoring the jump between 1st and 2nd ionisation energies, as that would imply that there would be 1 electron in the outer shell, then only 2 electrons in the next shell, which is not possible since it isn't the innermost shell)so it can't be in Group 1, as Group 1 elements only have 1 electron in the outer shell.

Hope this helped!



I really don't understand why there is a jump between 1st and 2nd ionisation energies? stupid graph!
Yeah, I think they've just done that to catch you out. I wouldn't be able to explain why that happens (maybe something to do with electron shielding?) xD As long as you think logically about how many electrons are in each shell, though, it shouldn't be to difficult.
(edited 9 years ago)
Original post by TheonlyMrsHolmes
For 6a) Answer A can be completely rules out because group 1 has only 1 electron in it's outer shell, the graph shows 10 electrons being removed.

Q11)It says that most compounds of lead are insoluble, basically inferring that if you add sulphuric acid to lead the lead will dissolve, but then it tells us that lead(II)nitrate is an exception to this. Then all the options but C include Lead(II) nitrate solution (1st clue) and all the others include adding sulphuric acid to other lead compounds (2nd clue ). There for the answer must be C.

Hope that helped!


you mean soluble in the first line right?
but yeah thankyou it made sense:smile:
For your question about ionisation energies
I think the graph is of aluminium as the first electron being removed is from the 3p subshell, the consequent electrons are being removed from the 3s subshell, hence the significant jump, as there is less shielding, and the effective nuclear charge is greater :smile:
I'm an A2 student, just doing unit 1 papers to recall my knowledge haha which is slowly coming back
Original post by lukejoshjames
For Question 6, successive ionisation energy is the energy required to remove each electron in turn. Each successive ionisation energy is larger than the last, as more there is one less electron, so there is less electron repulsion, therefore the outer shell electrons are a tiny bit closer to the nucleus (meaning more energy is required to remove it). However, there is an even bigger jump in ionisation energies between shells, as electrons in inner shells are much closer to the nucleus than those in outer shells, and this can be seen on the graph where there are larger jumps than usual.

On the graph, there is at least 1 jump in energies which is between 3rd and 4th ionisation energies. Therefore, after the 3rd electron being removed, there is a new shell (I am ignoring the jump between 1st and 2nd ionisation energies, as that would imply that there would be 1 electron in the outer shell, then only 2 electrons in the next shell, which is not possible since it isn't the innermost shell)so it can't be in Group 1, as Group 1 elements only have 1 electron in the outer shell.

Hope this helped!


great answer thankyou :smile:
Original post by lukejoshjames
Yeah, I think they've just done that to catch you out. I wouldn't be able to explain why that happens (maybe something to do with electron shielding?) xD As long as you think logically about how many electrons are in each shell, though, it shouldn't be to difficult.


read above
Original post by frozo123
you mean soluble in the first line right?
but yeah thankyou it made sense:smile:
For your question about ionisation energies
I think the graph is of aluminium as the first electron being removed is from the 3p subshell, the consequent electrons are being removed from the 3s subshell, hence the significant jump, as there is less shielding, and the effective nuclear charge is greater :smile:
I'm an A2 student, just doing unit 1 papers to recall my knowledge haha which is slowly coming back


By bad :facepalm: Basically my explanation but flipped around!

(I also can't take away the bold writing so excuse that please)

Q11)It says that most compounds of lead are insoluble, basically inferring that if you add sulphuric acid to lead the lead will not dissolve, but then it tells us that lead(II)nitrate is an exception to this and it will dissolve. Then all the options but C include Lead(II) nitrate solution (
1st clue) and all the others include adding sulphuric acid to other lead compounds (2nd clue ). There for the answer must be C.

I had to write the correct explanation sorry if you already understood! haha

and thanks for explaining the other bit!
Reply 1088
Original post by TheonlyMrsHolmes
I am going over Hess' law cycles and I can't seem to remember why standard enthalpy change of formation always needs 1 mole of product on the right hand side and why combustion always has 1 mole of elements and compounds on the left side of the equation.

All I know is that it does! But whyyy?

Thanks

Becaue thats what the definition of standard enthalpy of formation is.. The amount of energgy needed for ONE mole of substance to be formed. If the moles of product on the right side of the reaction were different for every reaction, it would be hard to compare enthalpy changes, its just the standard for all reactions. Kinda like relative atomic mass is the mass of an element compared to carbon 12 which is seen as the standard.
Obvioulsy the enthalpy change for three moles of a compound to be formed will be gfeater than one mole, but we just want to know the basic standard.
I'm doing OCR B Salters.

I was just wondering, what would you say is the best way to revise chemistry?
Original post by TheonlyMrsHolmes
By bad :facepalm: Basically my explanation but flipped around!

(I also can't take away the bold writing so excuse that please)

Q11)It says that most compounds of lead are insoluble, basically inferring that if you add sulphuric acid to lead the lead will not dissolve, but then it tells us that lead(II)nitrate is an exception to this and it will dissolve. Then all the options but C include Lead(II) nitrate solution (
1st clue) and all the others include adding sulphuric acid to other lead compounds (2nd clue ). There for the answer must be C.

I had to write the correct explanation sorry if you already understood! haha

and thanks for explaining the other bit!


If you have any questions, let me know as it helps with my revision :smile:
Anyone know the equations relating to moles, avogadro and number of atoms.
Original post by chzm
Becaue thats what the definition of standard enthalpy of formation is.. The amount of energgy needed for ONE mole of substance to be formed. If the moles of product on the right side of the reaction were different for every reaction, it would be hard to compare enthalpy changes, its just the standard for all reactions. Kinda like relative atomic mass is the mass of an element compared to carbon 12 which is seen as the standard.
Obvioulsy the enthalpy change for three moles of a compound to be formed will be gfeater than one mole, but we just want to know the basic standard.



Not doing alevels however, doing Applied science.. haven't revised.. but could u ask me a random alevel chem q? and any alevel biology question e.g. topics such as: cell membrane, biological molecules, cells e.t.c..
Original post by Uz25
Not doing alevels however, doing Applied science.. haven't revised.. but could u ask me a random alevel chem q? and any alevel biology question e.g. topics such as: cell membrane, biological molecules, cells e.t.c..


what's the chemistry between ronaldo and bale on fifa 15?
Need help with question 2. Don't understand how to do it?
Original post by Super199
Need help with question 2. Don't understand how to do it?


I guess, it is asked for molar mass, but I am not sure. If that is the case, it is M = m/n, where m is the mass and n the amount of substance.

The empirical formula of magnesium oxide must be 2 MgO.
Original post by Super199
Need help with question 2. Don't understand how to do it?


Do 6.448 - 3.888 and that leaves you with the amount of oxygen, then work out the number of moles for magnesium and oxygen by using n=m/mr for Mg do 3.888/24.3 = 0.16 and for oxygen use the value obtained when you subtract 3.888 from 6.448 (2.56) and then use that to find n for oxygen, 2.56/16 = 0.1625

then divide the number of moles by the smallest value, in this case its 0.16 so 0.16/0.16 is 1 and 0.1625/0.16 is 1.01... but round that up to 1

so you have the ratio 1:1

empirical is MgO
(edited 9 years ago)
Reply 1097
Original post by Uz25
Not doing alevels however, doing Applied science.. haven't revised.. but could u ask me a random alevel chem q? and any alevel biology question e.g. topics such as: cell membrane, biological molecules, cells e.t.c..


erm I do AQA so you may not have learnt this but
CHEM: What is the definition of electronegativity?
Describe the type of bonding in metals

BIO: Describe how substances can move across the cell surface membrane in different ways

How can a change in the primary structure of an enzyme cause it to become non-functioning.
Original post by ThatMadClown
(...)

empirical is MgO


Not 2 MgO? as far as I know its: 2 Mg + O2 -> 2 MgO. Or does this not count?
Original post by Kallisto
Not 2 MgO? as far as I know its: 2 Mg + O2 -> 2 MgO. Or does this not count?


thats the balanced equation to show you how MgO is formed, the 2 at the front just shows that theres 2 MgO's not the empirical formula :smile:

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