• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

guys plz help!!!chem s09_qp3.plz help!!

Messages
622
Reaction score
48
Points
38
here's the link
page 8....question 5b...cant figure it out .. :unknown: :unknown: .how to you get the formula????????plz help guys!!!!!!!!!!

thanks in advance!
 
Messages
2,030
Reaction score
2,180
Points
173
i figured out the answer but it is sooo hard and i think ill try my best to explain it to u :D
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 :D
 
Messages
622
Reaction score
48
Points
38
haochen said:
look at my previous post :D

awesome explanation!!!!! :fool: (no offense)
quite honestly it didnt HELP AT ALL!!!!i also knew that first we have to.....than...than this,....then that....uuurrrggghhh!!!!!!
anyways u tried ur best dude...thanks....see ya around.. :wink:
 
Messages
622
Reaction score
48
Points
38
haochen said:
ok thx for telling :(

hey dude dnt b sad.....i didnt mean 2 hurt ya feelings.....at least u understood the question...i even couldnt do that.....nd u posted in this thread with ur best efforts...nobody did that.....chill out.. :)
 
Messages
499
Reaction score
100
Points
38
I have seen many people ask about Question 5.b)

So, I thought I would explain it to everyone here :p

From the graph, 8cm^3 of the Phosphate reacts fully with 12cm^3 of T Nitrate.

So, identifying only the atoms we need:

T : Phosphate

Taking a ratio of volumes:

8 : 12

Multiply to find the LCM of both sides.

3T : 2 Phosphate

=> Equal volumes:

24 : 24.

This implies that the formula is T3(PO4)2
 
Messages
499
Reaction score
100
Points
38
The way haochen said also applies, I just thought that this way was easier :p
 
Messages
622
Reaction score
48
Points
38
thanks...it helped a lot...nd u ""thought"' this way was easier(lol no offense again to haochen)
thank u both of u....keep up da gd work... :good:
 
Messages
7
Reaction score
0
Points
0
haochen said:
i figured out the answer but it is sooo hard and i think ill try my best to explain it to u :D
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 :D

Can you please do it on paper and upload that! :)
 
Messages
27
Reaction score
0
Points
11
For question paper 2009 Q9 b
Where did you get 8 from?
doesnt it react fully with 6cm^3? because from then on its constant?
 
Messages
5,877
Reaction score
4,244
Points
323
Unkownbunny said:
For question paper 2009 Q9 b
Where did you get 8 from?
doesnt it react fully with 6cm^3? because from then on its constant?

See the graph carefully.... :) Its on 8 cm^3
 
Top