Chemistry: Post your doubts here!

[Mohammed salik]

but barium chloride is also a ppt. it is given as a standard test for Cl-.
And also how can we see solubility when ppt of sulfate is left?

Barium Chloride is Solubleeeee...!!!

 

[Suchal Riaz]

but barium chloride is also a ppt. it is given as a standard test for Cl-.
And also how can we see solubility when ppt of sulfate is left?

among all salts in our syllabus the only insoluble salts of Barium are Barium Carbonate, BaSO4 and BaS03
we will first add HCl to remove carbonate and then add BaCl. if we get ppt then we add dilute acid in excess to make sure it is SO4 not SO3.
BaSO4 is insoluble in acidic solution.

 

[Browny]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_31.pdf

2)a)ii)

How do you know from whose number of moles to divide from when calculating enthalpy change per mole?
Like in question if 2 compounds are reacting at a ratio of 1:2 and neither compounds are in excess.

 

[mehria]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_31.pdf

2)a)ii)

How do you know from whose number of moles to divide from when calculating enthalpy change per mole?
Like in question if 2 compounds are reacting at a ratio of 1:2 and neither compounds are in excess.

it is clearly stated that Mg is in excess... so we will use the moles of H2SO4 to find enthalpy chnge per mole

 

[Mohammed salik]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_31.pdf

2)a)ii)

How do you know from whose number of moles to divide from when calculating enthalpy change per mole?
Like in question if 2 compounds are reacting at a ratio of 1:2 and neither compounds are in excess.

I just told You in Convo!!

 

[princeali97]

(ii) Use the Data Booklet to suggest a value for the energy required to form one mole of
the gaseous radium cation you have given in (i) from one mole of gaseous radium
atoms. Explain your answer.
Mark scheme :
less than (502 + 966)
allow answers in the range 1000–1400 kJ mol–1 (1)
ionisation energies decrease down the Group
or must be less than IE for Ba → Ba2+
or size of atom increases down Group/
electrons are further away from nucleus
or there is increased shielding down Group

 

[mehria]

(ii) Use the Data Booklet to suggest a value for the energy required to form one mole of
the gaseous radium cation you have given in (i) from one mole of gaseous radium
atoms. Explain your answer.

u have to write the value for the first ionisation energy of radium... as Ionisation energy is the energy required to remove one mole of valence electron from one mole of gaseous atom to form one mole of a unipositive ion...
here one electron is removed so u'll write the value for the first ionisation energy of the radium

 

[princeali97]

u have to write the value for the first ionisation energy of radium... as Ionisation energy is the energy required to remove one mole of valence electron from one mole of gaseous atom to form one mole of a unipositive ion...
here one electron is removed so u'll write the value for the first ionisation energy of the radium

But u wont find the 1st ionisation energy for Ra in the data booklet my friend ==" .If it was so easy i wud not have posted this question.I know the IE will be less than that of Ba but why did they add up the 1st and 2nd IE of Ba ? i am confused

 

[kitkat <3 :P]

what is bond polarity :/?

 

[mehria]

But u wont find the 1st ionisation energy for Ra in the data booklet my friend ==" .If it was so easy i wud not have posted this question.I know the IE will be less than that of Ba but why did they add up the 1st and 2nd IE of Ba ? i am confused

which paper is ths?

 

[roxylara]

For question B (i) the answer is reaction number 4 and the addition product is CH3COCH3. Can someone explain to me how?

 

[princeali97]

which paper is ths?

w09 qp 21 q1(d)

 

[princeali97]

For question B (i) the answer is reaction number 4 and the addition product is CH3COCH3. Can someone explain to me how?

I just did this question.And the answer is CH3C(OH)(CN)CH3.Its a nucleophilic addition reaction.

 

[kitkat <3 :P]

For question B (i) the answer is reaction number 4 and the addition product is CH3COCH3. Can someone explain to me how?

well reaction 4 shoows an addition reaction and in an addition reaction always one product forms so we can say its the addition product

 

[mehria]

w09 qp 21 q1(d)

it's so cnfusing :/

 

[princeali97]

it's so cnfusing :/

I know right!

 

[Jelleh Belleh]

well reaction 4 shoows an addition reaction and in an addition reaction always one product forms so we can say its the addition product

But the product in reaction 4 is CH3C(OH)(CN)CH3, not CH3COCH3.

 

[mehria]

w09 qp 21 q1(d)

oh ...the thing is that here we r removing 2 electrons at one time so the first ionisation energy + the second ionisation energy will give us the total energy required to remove 2 electrons from a barium atom... as down the group ionisation energy is decreasing and I.E of radium will be less than Ba so we will write a value that is lower than the value for Ba...

 

[princeali97]

oh ...the thing is that here we r removing 2 electrons at one time so the first ionisation energy + the second ionisation energy will give us the total energy required to remove 2 electrons from a barium atom... as down the group ionisation energy is decreasing and I.E of radium will be less than Ba so we will write a value that is lower than the value for Ba...

hmmmm..i got it! yaaay! thank you.I wasted 30 mins on this question.i didnt see that 2 electrons are being removed from Ra .Good thinking.
Thanks!

 

[mehria]

hmmmm..i got it! yaaay! thank you.I wasted 30 mins on this question.i didnt see that 2 electrons are being removed from Ra .Good thinking.
Thanks!

lol... no problm