IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => IGCSE/ GCSE => Sciences => Topic started by: ~~~~shreyapril~~~~ on September 03, 2009, 10:13:31 am
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I need chem notes on moles that is stoichiometry and carbon chemistry..
these would help me n my mocks..
any sites would also serve my needs
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find 'em attached
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thanx mann
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anytime...
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gr8 job adi .. keep it up !! ;)
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hey tks Q80
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wat is the empirical formula of a comoud weighing 60g
whic has 20g content oxygen
and other hydrogen..
i am gettign a weird ans
pls help
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maybe because its a wierd question...
H O
mass 40g 20g
molar mass 1 16
(x)mass/mm 40 1.25
x/least 40/1.25 1
ratio 32 1
H32O
---this compound cannot exist-----
plzz check the question
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maybe because its a wierd question...
H O
mass 40g 20g
molar mass 1 16
(x)mass/mm 40 1.25
x/least 40/1.25 1
ratio 32 1
H32O
---this compound cannot exist-----
plzz check the question
H32O........LOOOOOOOOOOOL :P :P :P
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wat answer r u getting..slvri?
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wat answer r u getting..slvri?
same as the one ur getting.........definitely some problem in da question
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ok so
40g H and 20g O
WHat you first do is find out % composition of each
so
H O
% composition 40/60*100 20/60*100
ratio of compostion 2 : 1
It's H2O
That's how you do empirical formula for this one i guess
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dont u do it by the mass method...???
can u plz spot the error in the steps i rote...
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ok so
40g H and 20g O
WHat you first do is find out % composition of each
so
H O
% composition 40/60*100 20/60*100
ratio of compostion 2 : 1
It's H2O
That's how you do empirical formula for this one i guess
but hows that possible???? i mean u found percentage composition by mass but thats not necessarily the same as the empirical formula(number of moles of atoms of each element)
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ok so
40g H and 20g O
WHat you first do is find out % composition of each
so
H O
% composition 40/60*100 20/60*100
ratio of compostion 2 : 1
It's H2O
That's how you do empirical formula for this one i guess
but hows that possible???? i mean u found percentage composition by mass but thats not necessarily the same as the empirical formula(number of moles of atoms of each element)
true..
u didnt divide it by the molar mass
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ok so
40g H and 20g O
WHat you first do is find out % composition of each
so
H O
% composition 40/60*100 20/60*100
ratio of compostion 2 : 1
It's H2O
That's how you do empirical formula for this one i guess
but hows that possible???? i mean u found percentage composition by mass but thats not necessarily the same as the empirical formula(number of moles of atoms of each element)
true..
u didnt divide it by the molar mass
and anyways hydrogen atoms are so much lighter than oxygen atoms that the mass of hydrogen must be less than that of oxygen e.g. 4g of hydogen combines with 32g of oxygen to produce 36g of water
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ok so
40g H and 20g O
WHat you first do is find out % composition of each
so
H O
% composition 40/60*100 20/60*100
ratio of compostion 2 : 1
It's H2O
That's how you do empirical formula for this one i guess
but hows that possible???? i mean u found percentage composition by mass but thats not necessarily the same as the empirical formula(number of moles of atoms of each element)
true..
u didnt divide it by the molar mass
and anyways hydrogen atoms are so much lighter than oxygen atoms that the mass of hydrogen must be less than that of oxygen e.g. 4g of hydogen combines with 32g of oxygen to produce 36g of water
very true///
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ok so
40g H and 20g O
WHat you first do is find out % composition of each
so
H O
% composition 40/60*100 20/60*100
ratio of compostion 2 : 1
It's H2O
That's how you do empirical formula for this one i guess
but hows that possible???? i mean u found percentage composition by mass but thats not necessarily the same as the empirical formula(number of moles of atoms of each element)
true..
u didnt divide it by the molar mass
and anyways hydrogen atoms are so much lighter than oxygen atoms that the mass of hydrogen must be less than that of oxygen e.g. 4g of hydogen combines with 32g of oxygen to produce 36g of water
I know...i just tried to derive the empirical formula...found it this way so put it up :P..I know this is theoritically incorrect....but i just gave it a try
@shreyapril-plz check the q again...im being screwed for trying some crazy method
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hahaha
nice one though.
lolzz
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hahaha
nice one though.
lolzz
very innovative no?? :P
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extremely.. ;D
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hey let me check the question i got it frm some weird site..
i think i made some mistakes wit the numerical... :P :P :P :P :P
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@shreyapril
check the question
its definately incorrect
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hey let me check the question i got it frm some weird site..
i think i made some mistakes wit the numerical... :P :P :P :P :P
sure u did..
haha
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hey someone knos the site frm whre i got this question..
i have lost it..
completely..
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hey someone knos the site frm whre i got this question..
i have lost it..
completely..
haha.
how can we know
lol..
u got that question na,.,
anyway,..,its an incorrect question i guess
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aadi do u hav ne notes on the oxidation state??/
2days class was interesting..
but y didnt she explain oxidation state..
or were those electrons only the oxydation sattes
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aadi do u hav ne notes on the oxidation state??/
2days class was interesting..
but y didnt she explain oxidation state..
or were those electrons only the oxydation sattes
oxidation state can be explained as gain and loss of electrons too..
the concept is clear now..
i dont have any notes..
but i can give u few sitese..
and the repository of knowledge...nid404..wil definately have awesome sites for it..
chk out the thread...chem q...she had posted an awesome site for chem..
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find redox notes from that site attached
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k..
tks by the way..
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k..
tks by the way..
anytime...