Post by: ~ Miss Relina ~ on November 29, 2010, 10:42:51 pm
http://www.amazon.co.uk/Edexcel-Chemistry-Student-Level-Sciences/dp/1405896353
plz plz plz need it asap help as much as u can I have exams soon :'( :'( for edexcel
Post by: Deadly_king on November 30, 2010, 04:58:05 am
guys anybody who have answrs to chemistry book inthis link plz help
http://www.amazon.co.uk/Edexcel-Chemistry-Student-Level-Sciences/dp/1405896353
plz plz plz need it asap help as much as u can I have exams soon :'( :'( for edexcel
Am sorry but I don't have the answers. :-[
Nevertheless, if you need them so much, you could post the questions here and i'll work them out such that we can compare answers. ;)
Post by: elemis on November 30, 2010, 11:36:33 am
There should be answers in it, I think.
Post by: The Golden Girl =D on November 30, 2010, 11:38:47 am
guys anybody who have answrs to chemistry book inthis link plz help
http://www.amazon.co.uk/Edexcel-Chemistry-Student-Level-Sciences/dp/1405896353
plz plz plz need it asap help as much as u can I have exams soon :'( :'( for edexcel
I have the exact book but Unfortunately I never found the Answers -.- tho the CGP Guide for it is incredible it has Qs and Answers at the back ;D
I assume you're solving the Questions in the book , it's Okay take a screen shot of your doubt *from the CD* and Post it over here , We'll help ya :D
Good Luck
Post by: ~ Miss Relina ~ on November 30, 2010, 09:49:09 pm
I have the exact book but Unfortunately I never found the Answers -.- tho the CGP Guide for it is incredible it has Qs and Answers at the back ;D
I assume you're solving the Questions in the book , it's Okay take a screen shot of your doubt *from the CD* and Post it over here , We'll help ya :D
Good Luck
well I have questions but cando the screen shot so I will give u the page 23 Q3 if u could do it telll me ? if u donot tell mwe to post it here
many Thanks :)
Post by: The Golden Girl =D on December 01, 2010, 07:34:02 am
well I have questions but cando the screen shot so I will give u the page 23 Q3 if u could do it telll me ? if u donot tell mwe to post it here
many Thanks :)
I'll working on it :)
I got a) 0.03211 for C , 0.003699 for N , 0.2247 for H , and 0.06033 for O
b) as 9 : 1 : 61 : 16
c) as C9NH61O
d) I didn't know how to solve it :-X *srry I'm distracted by the TV -.- *
here is the pic ppl :)
(http://sphotos.ak.fbcdn.net/hphotos-ak-ash2/hs613.ash2/156537_468655218026_600433026_5829445_1875510_n.jpg)
Post by: Deadly_king on December 02, 2010, 05:29:44 am
a) Make use of the formula ---> No of moles = Mass/Mr
Oxygen : 0.66712/16 = 0.041695
Carbon : 0.3853/12 = 0.032108
Hydrogen : 0.05178/1 = 0.05178
Nitrogen : 0.2247/14 = 0.01605
Phosphorus : 0.2981/31 = 0.0096161
b) To get the ratio, we divide everything by the smallest number which in this case is 0.0096161.
Example : Oxygen -----> 0.041695/0.0096161 = 13/3(nearest)
So the ratio is O : C : H : N : P
13/3 : 10/3 : 16/3 : 5/3 : 1
Now we just have to multiply by 3 to get C10O13N5H16P3. This is the empirical formula. ;)
d) The empirical formula gives us an Mr of 494.
494n = 507 which is almost n=1.
Hence the molecular formula is equal to the empirical one. :D
Molecular formula : C10O13N5H16P3
Post by: ~ Miss Relina ~ on December 02, 2010, 03:23:53 pm
I'll working on it :)
I got a) 0.03211 for C , 0.003699 for N , 0.2247 for H , and 0.06033 for O
b) as 9 : 1 : 61 : 16
c) as C9NH61O
d) I didn't know how to solve it :-X *srry I'm distracted by the TV -.- *
here is the pic ppl :)
(http://sphotos.ak.fbcdn.net/hphotos-ak-ash2/hs613.ash2/156537_468655218026_600433026_5829445_1875510_n.jpg)
Thanks that is great :D
how did u took the screen shot pm me ;) other wise I wonot be able to post my questions as they are too long to type :(
Post by: ~ Miss Relina ~ on December 02, 2010, 03:26:12 pm
Here's my working.Thanks amillion Thanks :) :)
a) Make use of the formula ---> No of moles = Mass/Mr
Oxygen : 0.66712/16 = 0.041695
Carbon : 0.3853/12 = 0.032108
Hydrogen : 0.05178/1 = 0.05178
Nitrogen : 0.2247/14 = 0.01605
Phosphorus : 0.2981/31 = 0.0096161
b) To get the ratio, we divide everything by the smallest number which in this case is 0.0096161.
Example : Oxygen -----> 0.041695/0.0096161 = 13/3(nearest)
So the ratio is O : C : H : N : P
13/3 : 10/3 : 16/3 : 5/3 : 1
Now we just have to multiply by 3 to get C10O13N5H16P3. This is the empirical formula. ;)
d) The empirical formula gives us an Mr of 494.
494n = 507 which is almost n=1.
Hence the molecular formula is equal to the empirical one. :D
Molecular formula : C10O13N5H16P3
but I really had too many questions there hope that u could help
ur ansers are really great I will try to do it by meslf now ;)
Post by: Deadly_king on December 02, 2010, 05:02:13 pm
Thanks amillion Thanks :) :)
but I really had too many questions there hope that u could help
ur ansers are really great I will try to do it by meslf now ;)
No problem. :)
Do post your doubts and i'll try to help you as much as possible. ;)
Post by: The Golden Girl =D on December 02, 2010, 05:24:32 pm
Thanks that is great :D
how did u took the screen shot pm me ;) other wise I wonot be able to post my questions as they are too long to type :(
u open ur CD
then open the page and zoom in , how to do the zoom in is to click on the interactive and I guess that's it .
then press on a button that says *print screen* or *screen shot*
then paste it on the Paint thingy ;)
then save it as a JPG. file
then when u wrie a reply there is smthn at the end that says Additional Options u press that and then Attachments and then u upload it
that's it =D
Post by: ~ Miss Relina ~ on December 03, 2010, 04:45:37 pm
u open ur CD
then open the page and zoom in , how to do the zoom in is to click on the interactive and I guess that's it .
then press on a button that says *print screen* or *screen shot*
then paste it on the Paint thingy ;)
then save it as a JPG. file
then when u wrie a reply there is smthn at the end that says Additional Options u press that and then Attachments and then u upload it
that's it =D
Thanks vvvv much u have made for me favour :-*
Post by: The Golden Girl =D on December 03, 2010, 04:46:32 pm
Thanks vvvv much u have made for me favour :-*
Anytime :)
Post by: ~ Miss Relina ~ on December 03, 2010, 05:05:23 pm
Q2: i have attatched the table and the question
if u know what are the studies that would be needed to determine if food labelling has an impact on dietary choices?? ??? ???
Post by: The Golden Girl =D on December 03, 2010, 05:09:11 pm
my second question is is this Q1:definition write or should it include 1/12th the mass of carbon-12 atom
Q2: i have attatched the table and the question
if u know what are the studies that would be needed to determine if food labelling has an impact on dietary choices?? ??? ???
first Q , my teacher said that it has to be compared to 1/12 of a carbon12 atom .
Second Q , sorry can't help u I'm doing something else :-X
Post by: Deadly_king on December 06, 2010, 10:57:49 am
my second question is is this Q1:definition write or should it include 1/12th the mass of carbon-12 atom
Q2: i have attatched the table and the question
if u know what are the studies that would be needed to determine if food labelling has an impact on dietary choices?? ??? ???
1. The definition is sometimes said to be enough but I'll advice you to always include the 1/12th. This will make your answer more precise and you can be sure not to lose marks in case they take it as a key point. ;)
2. (a) According to the table, the very high mass of CO in air is about 20 or more ppm within an average of 8 hours.
Since you're asked minimum, you should take 20. And since you need for 24 hours, juts multiply by 3. ;)
Hence answer is (20 x 3) = 60 ppm
(b) As you can note from the table, each pollutant has its own lapse of time or is measured in either ppm or ppb. This pose a problem when you're going to compare the data.
Therefore you should convert all of them to the same lapse of time (Example : 24hr) and same unit.(Example : ppb)
Now to your question.
According to me food labelling indeed has an impact on dietary choices. But how can we determine that?
First you should have a certain number of persons who are on diet. Then you note all the substances that they buy after having consulted a doctor and considering his advice with utmost importance.
After that you list all the ingredients of these substances and compare what are present in high numbers and what is absent. ;)
NOTE : The last answer is just my opinion about it. There may be various ways of proceeding. ;D
Post by: ~ Miss Relina ~ on December 07, 2010, 11:57:07 am
1. The definition is sometimes said to be enough but I'll advice you to always include the 1/12th. This will make your answer more precise and you can be sure not to lose marks in case they take it as a key point. ;)
2. (a) According to the table, the very high mass of CO in air is about 20 or more ppm within an average of 8 hours.
Since you're asked minimum, you should take 20. And since you need for 24 hours, juts multiply by 3. ;)
Hence answer is (20 x 3) = 60 ppm
(b) As you can note from the table, each pollutant has its own lapse of time or is measured in either ppm or ppb. This pose a problem when you're going to compare the data.
Therefore you should convert all of them to the same lapse of time (Example : 24hr) and same unit.(Example : ppb)
Now to your question.
According to me food labelling indeed has an impact on dietary choices. But how can we determine that?
First you should have a certain number of persons who are on diet. Then you note all the substances that they buy after having consulted a doctor and considering his advice with utmost importance.
After that you list all the ingredients of these substances and compare what are present in high numbers and what is absent. ;)
NOTE : The last answer is just my opinion about it. There may be various ways of proceeding. ;D
Thanks very much I really appreciate ur help pal Thanks and I'll be waitng always for ur replies to me ;)
Post by: ~ Miss Relina ~ on December 07, 2010, 11:58:08 am
first Q , my teacher said that it has to be compared to 1/12 of a carbon12 atom .
Second Q , sorry can't help u I'm doing something else :-X
ohno robs Thanks anyway :D
Post by: Deadly_king on December 07, 2010, 12:00:11 pm
Thanks very much I really appreciate ur help pal Thanks and I'll be waitng always for ur replies to me ;)
You're welcome dear. :D
I'll be glad to help you as much as I can. ;)
Post by: The Golden Girl =D on December 07, 2010, 12:55:45 pm
ohno robs Thanks anyway :D
Hmm np =]
Post by: ~ Miss Relina ~ on December 07, 2010, 10:21:05 pm
Post by: Deadly_king on December 09, 2010, 12:07:36 pm
me agaun that my questions and for question concerning calorimeter i have attatched the picture for (fig1.2.10)
1st attachment. No 3
Temperature change for A - exothermic : (30 - 10) = 20oC
Temperature change for B - endothermic : (40 - 20) = 20oC
4th attachment
You should use the two equations provided below to form the required equation which is :
H2SO4 + 2KCl -----> 2HCl + K2SO4
Now we need to re-arrange the information provided to get this. You need to inverse the 2nd equation provided and multiply it by 2.
H2SO4 + 2KOH ----> K2SO4 + 2H2O ^H = -324 KJ
2KCl + 2H2O ----> 2HCl + 2KOH ^H = +408 KJ
NOTE : When you inverse an equation and multiply it by 2, you should inverse the enthalpy change as well and multiply it too.
Now you mix the two equations.
H2SO4 + 2KOH + 2KCl + 2H2O ----> K2SO4 + 2H2O + 2HCl + 2KOH
From this resulting equation you can note that 2KOH and 2H2O cancel each other from both sides of the equation.
So now you get your equation. For its enthalpy change you just need to add them tegether.
^ H = -324 + 408 = +84 KJ
Post by: Deadly_king on December 09, 2010, 12:42:04 pm
This is the equation representing the formation of propanoic acid.
3C + 3H2 + 1/2O2 ----> C3H5OH
Since the enthalpy changes of combustion has been given to you, you also need to write the equation relating each substance and its respective combustion.
1. C + O2 ----> CO2 (Enthalpy change of formation of carbon dioxide)
2. H2 + 0.5O2 ----> H2O (Enthalpy change of formation of water)
3. C3H5OH + 9/2O2 ----> 3CO2 + 3H2O (Enthalpy change of combustion of propanoic acid)
As you may observe, the combustion of both hydrogen and carbon leads to carbon dioxide and hydrogen as well as the combustion of butane.
Hence, according to Hess's law, we may opt for another route in the conversion of carbon and hydrogen to butane.
Let's say we'll convert 3 moles of carbon and 3 moles of Hydrogen gas to carbon dioxide and water. Then these two products are converted to propanoic acid.
In practise these reactions are not feasible. But in theory we may opt for that method to find standard enthalpy changes of a particular reaction.
Let ^C be the enthalpy change of combustion of carbon.
So Enthalpy change of formation of propanoic acid = 3(^C) + 3(^H) - (^C3H5OH) = 3(-393.5) + 3(-285.8) - (-1527.2) = -510.7 KJ
NOTE : We need to subtract the enthalpy change of combustion since we are converting carbon dioxide and water back to propanoic acid whereas the enthalpy change of combustion of propanoic acid actually is the energy evolved when 1 mole of propanoic acid is completely burnt to carbon dioxide and water. In other words we are doing just the contrary, that's why we need to subtract it.
Post by: ~ Miss Relina ~ on December 16, 2010, 11:01:59 pm
5th attachment
This is the equation representing the formation of propanoic acid.
3C + 3H2 + 1/2O2 ----> C3H5OH
Since the enthalpy changes of combustion has been given to you, you also need to write the equation relating each substance and its respective combustion.
1. C + O2 ----> CO2 (Enthalpy change of formation of carbon dioxide)
2. H2 + 0.5O2 ----> H2O (Enthalpy change of formation of water)
3. C3H5OH + 9/2O2 ----> 3CO2 + 3H2O (Enthalpy change of combustion of propanoic acid)
As you may observe, the combustion of both hydrogen and carbon leads to carbon dioxide and hydrogen as well as the combustion of butane.
Hence, according to Hess's law, we may opt for another route in the conversion of carbon and hydrogen to butane.
Let's say we'll convert 3 moles of carbon and 3 moles of Hydrogen gas to carbon dioxide and water. Then these two products are converted to propanoic acid.
In practise these reactions are not feasible. But in theory we may opt for that method to find standard enthalpy changes of a particular reaction.
Let ^C be the enthalpy change of combustion of carbon.
So Enthalpy change of formation of propanoic acid = 3(^C) + 3(^H) - (^C3H5OH) = 3(-393.5) + 3(-285.8) - (-1527.2) = -510.7 KJ
NOTE : We need to subtract the enthalpy change of combustion since we are converting carbon dioxide and water back to propanoic acid whereas the enthalpy change of combustion of propanoic acid actually is the energy evolved when 1 mole of propanoic acid is completely burnt to carbon dioxide and water. In other words we are doing just the contrary, that's why we need to subtract it.
Thanks alot I'll post other questions later
Post by: Deadly_king on December 17, 2010, 12:29:17 pm
Thanks alot I'll post other questions later
You're welcome. :D
Post by: Vampire-Love4ever on December 27, 2010, 12:42:19 pm
guys anybody who have answrs to chemistry book inthis link plz help
http://www.amazon.co.uk/Edexcel-Chemistry-Student-Level-Sciences/dp/1405896353
plz plz plz need it asap help as much as u can I have exams soon :'( :'( for edexcel
Answers are there in the CD.
Check the CD properly.
You'll find it :)
Post by: ~ Miss Relina ~ on December 27, 2010, 10:02:39 pm
Answers are there in the CD.
Check the CD properly.
You'll find it :)
i'm sure if u know how tell me i need them v.. urgently... :'(
Post by: Helium on January 15, 2011, 01:09:10 pm
just ask your teacher, they're found in the teacher guide of the book , so he
brought a copy of the anwsers, too bad he didn't give us a copy, so we ask him every time
he'll tell. simply just ask your teacher ;D, by the way i think that book kinda sucks i give it 5.9/10. :P
Post by: ~ Miss Relina ~ on January 15, 2011, 10:39:51 pm
Our class also wanted the answer's for that book,
just ask your teacher, they're found in the teacher guide of the book , so he
brought a copy of the anwsers, too bad he didn't give us a copy, so we ask him every time
he'll tell. simply just ask your teacher ;D, by the way i think that book kinda sucks i give it 5.9/10. :P
unfortunately, i am a private student so plz if u got them post them to me but also if u have got another book which is better tell me about it.
Post by: The Golden Girl =D on January 16, 2011, 08:17:32 am
here :
AS-Level Chemistry
Exam board :Edexcel
The Revision Guide
ISBN 978 1 84762 124 5
It helped me A LOT =D