Chemistey Doubts

[LimeReem]

I'm sorry but there's no NH3 in the question :/

OH WOOPS! mixed u up with some other person! sorry =/

 

[JmCullen]

why boron has high melting point and boiling point?
non metals have high melthing point and boiling point? :O

 

[shan5674]

why boron has high melting point and boiling point?
non metals have high melthing point and boiling point? :O

Boron is a semi metal!

 

[JmCullen]

why boron has high melting point and boiling point?
non metals have high melthing point and boiling point? :O

Boron is a semi metal!

ouhh.. ok!
so it has both metal and non metal's properties?

 

[shan5674]

why boron has high melting point and boiling point?
non metals have high melthing point and boiling point? :O

Boron is a semi metal!

ouhh.. ok!
so it has both metal and non metal's properties?

yea it has some metal properties and some non-metal properties like silicon or germanium They are also called metalloids

 

[shan5674]

Chemistry Paper 3 doubt!
MAY/JUNE 2009 Q)5 b) part
http://www.xtremepapers.me/CIE/Cambridg ... 9_qp_3.pdf
how to we get the formula for T? And what is the explanation!

i figured out the answer but it is sooo hard and i think ill try my best to explain it to u
first we write down the equation
the we solve for the mole (something related to math)
then we find the ratio of the mole of the reactant and product
then we substitute them in the equation.
after that balance the equation by putting number for the element (dont change the mole number)

hope u got it

Do u mind solving it here pleaseeeeeeeeeeeeeeeeeee i dont get it

 

[JmCullen]

yea it has some metal properties and some non-metal properties like silicon or germanium They are also called metalloids

oh! okok.. thanks alot!

 

[LimeReem]

http://www.xtremepapers.me/CIE/Cambridg ... _qp_03.pdf
q4b i , ii and iii [variant 1]
how to do it? D=

 

[JmCullen]

http://www.xtremepapers.me/CIE/Cambridge%20IGCSE/0620%20-%20Chemistry/0620_w08_qp_03.pdf
q4b i , ii and iii [variant 1]
how to do it? D=

(i) 100% - 92.3% = 7.7%

(ii) 92.3 (divide) 12(mr of 1 carbon) = 7.7
which is 1:1 (carbon : hydrogen)

(iii) empirical formula: CH
molecular formula: C6H6
why C6H6:
n(CH) = 78
n(12+1) = 78
13n = 78
n = 6

 

[JmCullen]

http://www.xtremepapers.me/CIE/Cambridge%20IGCSE/0620%20-%20Chemistry/0620_w08_qp_03.pdf
q4b i , ii and iii [variant 1]
how to do it? D=

(i) 100% - 92.3% = 7.7%

(ii) 92.3 (divide) 12(mr of 1 carbon) = 7.7
which is 1:1 (carbon : hydrogen)

(iii) empirical formula: CH
molecular formula: C6H6
why C6H6:
n(CH) = 78
n(12+1) = 78
13n = 78
n = 6

 

[LimeReem]

(i) 100% - 92.3% = 7.7%

(ii) 92.3 (divide) 12(mr of 1 carbon) = 7.7
which is 1:1 (carbon : hydrogen)

(iii) empirical formula: CH
molecular formula: C6H6
why C6H6:
n(CH) = 78
n(12+1) = 78
13n = 78
n = 6

Ouu thank you soo much! =D what is the formula of benzene btw for the empirical formula?

 

[shan5674]

http://www.xtremepapers.me/CIE/Cambridge%20IGCSE/0620%20-%20Chemistry/0620_w08_qp_03.pdf
q4b i , ii and iii [variant 1]
how to do it? D=

Q4 part b is all about calculating moles

(i)->100-92.3= 7.7%
(ii)->92.3/12(carbons Mr)= 7.69 and 7.7/1(hydrogens Mr) = 7.7 so therefore its 1:1
(iii)->since its a 1:1 ratio the empirical formula would be CH
and molecular formula would be 12+1=13 then divide the mass of the compound which is 78 but 13 = 6 so the molecular formula would be C6h6

Hope u got it

 

[JmCullen]

(i) 100% - 92.3% = 7.7%

(ii) 92.3 (divide) 12(mr of 1 carbon) = 7.7
which is 1:1 (carbon : hydrogen)

(iii) empirical formula: CH
molecular formula: C6H6
why C6H6:
n(CH) = 78
n(12+1) = 78
13n = 78
n = 6

Ouu thank you soo much! =D what is the formula of benzene btw for the empirical formula?

is it the B ii???
It's CH..
i wrote it in the previous post.

 

[SmartNour]

guys for chemistry paper 3 we all need to solve more past papers and focus on our mistakes!!!!!!!!! WISH u ALL THE BEST!

 

[JmCullen]

guys for chemistry paper 3 we all need to solve more past papers and focus on our mistakes!!!!!!!!! WISH u ALL THE BEST!

i've solve all past papers from 2002 to 2010!!!!!!
and now trying to read all of them once again..
hope it helps..
THANKS!!!!

 

[JmCullen]

Q(6)(a)(iii)
how do we know which type of polymerisation is which?

http://www.xtremepapers.me/CIE/Cambridg ... 4_qp_3.pdf

 

[shan5674]

Q(6)(a)(iii)
how do we know which type of polymerisation is which?

http://www.xtremepapers.me/CIE/Cambridg ... 4_qp_3.pdf

The polymerisation in a(i) is addition as it is the polyermisation of the above diagram right and the in a(ii) is condensation so theres one difference for u to state in for a(iii) and the other difference is that in an addition polymerisation there aren't any by products but in condensation water is produced

Hope u got it

 

[JmCullen]

Q(6)(a)(iii)
how do we know which type of polymerisation is which?

http://www.xtremepapers.me/CIE/Cambridg ... 4_qp_3.pdf

The polymerisation in a(i) is addition as it is the polyermisation of the above diagram right and the in a(ii) is condensation so theres one difference for u to state in for a(iii) and the other difference is that in an addition polymerisation there aren't any by products but in condensation water is produced

Hope u got it

oh..
ok.. i think i get it .
hahahahha..
thanks.

 

[ajmanutd]

hw do u solve q)5)b) on mj 09
this is d link fr d paper:- http://www.xtremepapers.me/CIE/Cambridg ... 9_qp_3.pdf

pls help i really dont understand this question..

 

[JmCullen]

hw do u solve q)5)b) on mj 09
this is d link fr d paper:- http://www.xtremepapers.me/CIE/Cambridg ... 9_qp_3.pdf

pls help i really dont understand this question..

that's the hardest one.
i cant do either