Author Topic: Moles Again...  (Read 4484 times)

Offline astarmathsandphysics

  • SF Overlord
  • *********
  • Posts: 11271
  • Reputation: 65534
  • Gender: Male
  • Free the exam papers!
Re: Moles Again...
« Reply #15 on: September 10, 2009, 03:41:03 pm »
NaHCO3 has molar mass 23+1+12+3*16=84

Offline Ghost Of Highbury

  • O_o_O lala!
  • SF Farseer
  • *******
  • Posts: 4096
  • Reputation: 41428
  • Gender: Male
  • Namaskaram!
Re: Moles Again...
« Reply #16 on: September 10, 2009, 03:45:28 pm »
NaHCO3 has molar mass 23+1+12+3*16=84
sir the equation hs 2NaHCO3
divine intervention!

Offline ~~~~shreyapril~~~~

  • SF Master
  • ******
  • Posts: 2606
  • Reputation: 2640
  • Gender: Male
  • "Prototype"
Re: Moles Again...
« Reply #17 on: September 10, 2009, 03:46:00 pm »
This is my original solution hpe u understand it.
i think u may
find it in attachment
Friendship is like peeing on yourself. Everyone can see it but only you get the warm feeling that it gives :D :P :)

Offline Ghost Of Highbury

  • O_o_O lala!
  • SF Farseer
  • *******
  • Posts: 4096
  • Reputation: 41428
  • Gender: Male
  • Namaskaram!
Re: Moles Again...
« Reply #18 on: September 10, 2009, 03:58:12 pm »
This is my original solution hpe u understand it.
i think u may
find it in attachment
OKK..FIRST OF ALL

I DIDNT GET Y U APPLIED RATIO PROPORTION WITH 168:9.3::18:x

y 18?
they have talked abt CO2 not H2O

ok..heres the sol.--------------------------

take only NAHCO3 and CO2 into consideration

168g NAHCO3 - 44g CO2 (by molar masseS)

this means that pure NAHCO3 168g will give out 44g of CO2

however here..2.24g of CO2 is given out ..

by ratio and proportion find out how much grams of NAHCO3 should be used to give out 2.24g of CO2

44g - 168g
2.24 - 8.55g

but...here 9.3g of NAHCO3 impure is used

this means that there is 9.3 - 8.55g of impurity = 0.75g of impurity

they've asked to calculate % purity

so what v do is

=> 8.55/9.3 * 100 = 92%

divine intervention!

Offline Ghost Of Highbury

  • O_o_O lala!
  • SF Farseer
  • *******
  • Posts: 4096
  • Reputation: 41428
  • Gender: Male
  • Namaskaram!
Re: Moles Again...
« Reply #19 on: September 10, 2009, 04:00:04 pm »
This is my original solution hpe u understand it.
i think u may
find it in attachment
OKK..FIRST OF ALL

I DIDNT GET Y U APPLIED RATIO PROPORTION WITH 168:9.3::18:x

y 18?
they have talked abt CO2 not H2O

ok..heres the sol.--------------------------

take only NAHCO3 and CO2 into consideration

168g NAHCO3 - 44g CO2 (by molar masseS)

this means that pure NAHCO3 168g will give out 44g of CO2

however here..2.24g of CO2 is given out ..

by ratio and proportion find out how much grams of NAHCO3 should be used to give out 2.24g of CO2

44g - 168g
2.24 - 8.55g

but...here 9.3g of NAHCO3 impure is used

this means that there is 9.3 - 8.55g of impurity = 0.75g of impurity

they've asked to calculate % purity

so what v do is

=> 8.55/9.3 * 100 = 92%


what do u think slvri and nid404?
divine intervention!

Offline slvri

  • SF Master
  • ******
  • Posts: 1773
  • Reputation: 1735
  • Gender: Male
Re: Moles Again...
« Reply #20 on: September 10, 2009, 04:02:20 pm »
This is my original solution hpe u understand it.
i think u may
find it in attachment
OKK..FIRST OF ALL

I DIDNT GET Y U APPLIED RATIO PROPORTION WITH 168:9.3::18:x

y 18?
they have talked abt CO2 not H2O

ok..heres the sol.--------------------------

take only NAHCO3 and CO2 into consideration

168g NAHCO3 - 44g CO2 (by molar masseS)

this means that pure NAHCO3 168g will give out 44g of CO2

however here..2.24g of CO2 is given out ..

by ratio and proportion find out how much grams of NAHCO3 should be used to give out 2.24g of CO2

44g - 168g
2.24 - 8.55g

but...here 9.3g of NAHCO3 impure is used

this means that there is 9.3 - 8.55g of impurity = 0.75g of impurity

they've asked to calculate % purity

so what v do is

=> 8.55/9.3 * 100 = 92%


what do u think slvri and nid404?
i solved the question too rite nw and i got da same ans as u adi....92%
i hate A level...........

Offline Ghost Of Highbury

  • O_o_O lala!
  • SF Farseer
  • *******
  • Posts: 4096
  • Reputation: 41428
  • Gender: Male
  • Namaskaram!
Re: Moles Again...
« Reply #21 on: September 10, 2009, 04:08:43 pm »
This is my original solution hpe u understand it.
i think u may
find it in attachment
OKK..FIRST OF ALL

I DIDNT GET Y U APPLIED RATIO PROPORTION WITH 168:9.3::18:x

y 18?
they have talked abt CO2 not H2O

ok..heres the sol.--------------------------

take only NAHCO3 and CO2 into consideration

168g NAHCO3 - 44g CO2 (by molar masseS)

this means that pure NAHCO3 168g will give out 44g of CO2

however here..2.24g of CO2 is given out ..

by ratio and proportion find out how much grams of NAHCO3 should be used to give out 2.24g of CO2

44g - 168g
2.24 - 8.55g

but...here 9.3g of NAHCO3 impure is used

this means that there is 9.3 - 8.55g of impurity = 0.75g of impurity

they've asked to calculate % purity

so what v do is

=> 8.55/9.3 * 100 = 92%


what do u think slvri and nid404?
i solved the question too rite nw and i got da same ans as u adi....92%
told him..

what mistake did u find in his sol.
i didnt understand the ratio proportion thingy.

shrey...answer check kar..:)
divine intervention!

Offline slvri

  • SF Master
  • ******
  • Posts: 1773
  • Reputation: 1735
  • Gender: Male
Re: Moles Again...
« Reply #22 on: September 10, 2009, 04:16:47 pm »
This is my original solution hpe u understand it.
i think u may
find it in attachment
OKK..FIRST OF ALL

I DIDNT GET Y U APPLIED RATIO PROPORTION WITH 168:9.3::18:x

y 18?
they have talked abt CO2 not H2O

ok..heres the sol.--------------------------

take only NAHCO3 and CO2 into consideration

168g NAHCO3 - 44g CO2 (by molar masseS)

this means that pure NAHCO3 168g will give out 44g of CO2

however here..2.24g of CO2 is given out ..

by ratio and proportion find out how much grams of NAHCO3 should be used to give out 2.24g of CO2

44g - 168g
2.24 - 8.55g

but...here 9.3g of NAHCO3 impure is used

this means that there is 9.3 - 8.55g of impurity = 0.75g of impurity

they've asked to calculate % purity

so what v do is

=> 8.55/9.3 * 100 = 92%


what do u think slvri and nid404?
i solved the question too rite nw and i got da same ans as u adi....92%
told him..

what mistake did u find in his sol.
i didnt understand the ratio proportion thingy.

shrey...answer check kar..:)
firstly he used the Mr of water instead of carbon dioxide and secondly 168 should be related to 44 (not 9.3) while x (mass of nahco3) should be related to 2.24
so....168->44
and..x->2.24
cross multipying we find x=8.55 g
then % purity = (8.55/9.3)*100=92% (approx)
i hate A level...........

Offline Ghost Of Highbury

  • O_o_O lala!
  • SF Farseer
  • *******
  • Posts: 4096
  • Reputation: 41428
  • Gender: Male
  • Namaskaram!
Re: Moles Again...
« Reply #23 on: September 10, 2009, 04:18:52 pm »
This is my original solution hpe u understand it.
i think u may
find it in attachment
OKK..FIRST OF ALL

I DIDNT GET Y U APPLIED RATIO PROPORTION WITH 168:9.3::18:x

y 18?
they have talked abt CO2 not H2O

ok..heres the sol.--------------------------

take only NAHCO3 and CO2 into consideration

168g NAHCO3 - 44g CO2 (by molar masseS)

this means that pure NAHCO3 168g will give out 44g of CO2

however here..2.24g of CO2 is given out ..

by ratio and proportion find out how much grams of NAHCO3 should be used to give out 2.24g of CO2

44g - 168g
2.24 - 8.55g

but...here 9.3g of NAHCO3 impure is used

this means that there is 9.3 - 8.55g of impurity = 0.75g of impurity

they've asked to calculate % purity

so what v do is

=> 8.55/9.3 * 100 = 92%


what do u think slvri and nid404?
i solved the question too rite nw and i got da same ans as u adi....92%
told him..

what mistake did u find in his sol.
i didnt understand the ratio proportion thingy.

shrey...answer check kar..:)
firstly he used the Mr of water instead of carbon dioxide and secondly 168 should be related to 44 (not 9.3) while x (mass of nahco3) should be related to 2.24
so....168->44
and..x->2.24
cross multipying we find x=8.55 g
then % purity = (8.55/9.3)*100=92% (approx)
true..

thanks for confirmation!
:)
divine intervention!

Offline slvri

  • SF Master
  • ******
  • Posts: 1773
  • Reputation: 1735
  • Gender: Male
Re: Moles Again...
« Reply #24 on: September 10, 2009, 04:26:08 pm »
This is my original solution hpe u understand it.
i think u may
find it in attachment
OKK..FIRST OF ALL

I DIDNT GET Y U APPLIED RATIO PROPORTION WITH 168:9.3::18:x

y 18?
they have talked abt CO2 not H2O

ok..heres the sol.--------------------------

take only NAHCO3 and CO2 into consideration

168g NAHCO3 - 44g CO2 (by molar masseS)

this means that pure NAHCO3 168g will give out 44g of CO2

however here..2.24g of CO2 is given out ..

by ratio and proportion find out how much grams of NAHCO3 should be used to give out 2.24g of CO2

44g - 168g
2.24 - 8.55g

but...here 9.3g of NAHCO3 impure is used

this means that there is 9.3 - 8.55g of impurity = 0.75g of impurity

they've asked to calculate % purity

so what v do is

=> 8.55/9.3 * 100 = 92%


what do u think slvri and nid404?
i solved the question too rite nw and i got da same ans as u adi....92%
told him..

what mistake did u find in his sol.
i didnt understand the ratio proportion thingy.

shrey...answer check kar..:)
firstly he used the Mr of water instead of carbon dioxide and secondly 168 should be related to 44 (not 9.3) while x (mass of nahco3) should be related to 2.24
so....168->44
and..x->2.24
cross multipying we find x=8.55 g
then % purity = (8.55/9.3)*100=92% (approx)
true..

thanks for confirmation!
:)
dont mention it ;) :D
and shrey......when r u giving igcse?
i hate A level...........

Offline Ghost Of Highbury

  • O_o_O lala!
  • SF Farseer
  • *******
  • Posts: 4096
  • Reputation: 41428
  • Gender: Male
  • Namaskaram!
Re: Moles Again...
« Reply #25 on: September 10, 2009, 04:27:38 pm »
we r giving it together...oct/nov

hez in my class.. :)
divine intervention!

Offline slvri

  • SF Master
  • ******
  • Posts: 1773
  • Reputation: 1735
  • Gender: Male
Re: Moles Again...
« Reply #26 on: September 10, 2009, 04:28:49 pm »
we r giving it together...oct/nov

hez in my class.. :)
coooool........so wut r ur subs adi?
i hate A level...........

Offline Ghost Of Highbury

  • O_o_O lala!
  • SF Farseer
  • *******
  • Posts: 4096
  • Reputation: 41428
  • Gender: Male
  • Namaskaram!
Re: Moles Again...
« Reply #27 on: September 10, 2009, 04:30:03 pm »
math
phy
chem
eng
computer studies
french
add. math
divine intervention!

Offline slvri

  • SF Master
  • ******
  • Posts: 1773
  • Reputation: 1735
  • Gender: Male
Re: Moles Again...
« Reply #28 on: September 10, 2009, 04:34:56 pm »
math
phy
chem
eng
computer studies
french
add. math

nice.....u take french too?
my mom knws both french and german but i havent learned either even though i have books in ma house of them.......
so is french hard? or easy?
i hate A level...........

Offline ~~~~shreyapril~~~~

  • SF Master
  • ******
  • Posts: 2606
  • Reputation: 2640
  • Gender: Male
  • "Prototype"
Re: Moles Again...
« Reply #29 on: September 11, 2009, 07:51:58 am »
haha ;D
a silly mistake and the whole question went wrong
tks slvri and aadi
Friendship is like peeing on yourself. Everyone can see it but only you get the warm feeling that it gives :D :P :)