Chemistry: Post your doubts here!

[Student12]

Does anyone know where I can find the 2012 oct/nov paper 4 please?

 

[snowbrood]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s07_qp_31.pdf i diluted 33.7cm^3. in titration i got 35.8 as the suitable volume of FA3 as i go on i was able to solve part c(i) of question 1 but got stuck in c(ii) where it asks as to how many moles of hcl are contained in the volume of fa3.
i dont know what do here i got 0.017 in the first part examiner says multiply this with the titre/250 to get the second answer need help urgent

 

[XPFMember]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_31.pdf i diluted 33.7cm^3. in titration i got 35.8 as the suitable volume of FA3 as i go on i was able to solve part c(i) of question 1 but got stuck in c(ii) where it asks as to how many moles of hcl are contained in the volume of fa3.
i dont know what do here i got 0.017 in the first part examiner says multiply this with the titre/250 to get the second answer need help urgent

AoA!

They want you to calculate the moles of HCl that reacted with 25cm^3 FA1.

You know, a vol. of 35.8 FA3 reacted.

you also know 250 cm3 of FA3 has 0.017 moles. (first part)

now apply the math here, if 250 cm3 has 0.017 moles, how many moles will be there in 35.8cm3?

35.8 x 0.017/250

 

[izzahzainab]

Here's a confusing question even my teacher had a hard time with (and it was in the AS paper 1, not A2!)

1. 34 Silver chloride dissolves in aqueous ammonia.
   What happens in this process?
   1. 1 A co-ordinate bond is formed.
   2. 2 The oxidation number of nitrogen is unchanged.
   3. 3 Ammonia acts as a Brønsted-Lowry base.

The solubility product is a complex silver ion I believe, which is clearly on the A2 syllabus only not the AS, and yet this was on paper one. On chemguide.co.uk, it lists the equation which forms this ion as not being needed by the AS NOR the A2 syllabus! Can someone please explain how to solve this question with the knowledge that an AS student would already have? Thanks!!

There was a similar question in AS paper last year :/ left me confused

 

[snowbrood]

AoA!

They want you to calculate the moles of HCl that reacted with 25cm^3 FA1.

You know, a vol. of 35.8 FA3 reacted.

you also know 250 cm3 of FA3 has 0.017 moles. (first part)

now apply the math here, if 250 cm3 has 0.017 moles, how many moles will be there in 35.8cm3?

35.8 x 0.017/250

waliakumaslam oh man that was so easy lol thanks man u owe me

 

[snowbrood]

There was a similar question in AS paper last year :/ left me confused

which year and what is correct answer please

 

[snowbrood]

Here's a confusing question even my teacher had a hard time with (and it was in the AS paper 1, not A2!)

1. 34 Silver chloride dissolves in aqueous ammonia.
   What happens in this process?
   1. 1 A co-ordinate bond is formed.
   2. 2 The oxidation number of nitrogen is unchanged.
   3. 3 Ammonia acts as a Brønsted-Lowry base.

The solubility product is a complex silver ion I believe, which is clearly on the A2 syllabus only not the AS, and yet this was on paper one. On chemguide.co.uk, it lists the equation which forms this ion as not being needed by the AS NOR the A2 syllabus! Can someone please explain how to solve this question with the knowledge that an AS student would already have? Thanks!!

u dont really need a level concept according to me answer should be 1 and 3 because ammonia have one lone pair which would be use for bonding with silver whenever there is a coordinate bonding involved the donor of lone pair is considered bronsted lowry base according to their thoery . as this is a displacement reactin so there is nothing oxidized or reduced hope that answers your question .
bronsted lowry theory oxidation and redution and coordinate bnding is all as level hope i was helpful

 

[aalmuhannadi]

u dont really need a level concept according to me answer should be 1 and 3 because ammonia have one lone pair which would be use for bonding with silver whenever there is a coordinate bonding involved the donor of lone pair is considered bronsted lowry base according to their thoery . as this is a displacement reactin so there is nothing oxidized or reduced hope that answers your question .
bronsted lowry theory oxidation and redution and coordinate bnding is all as level hope i was helpful

You can never have an answer as 1 and 3 on a multiple choice AS test! that's never one of the combination options... and the answer is 1 and 2 but I don't really understand why.. I need to know where exactly is this coordinate bond and the product formed, because it's not in the CIE chemistry coursebook

 

[snowbrood]

You can never have an answer as 1 and 3 on a multiple choice AS test! that's never one of the combination options... and the answer is 1 and 2 but I don't really understand why.. I need to know where exactly is this coordinate bond and the product formed, because it's not in the CIE chemistry coursebook

go to the syllabus in as section u would see that coordinate bond is formed and i am afraid u are wrong when ever there is a coordinate bond formed the donor is always considered a base please have u studied the lowry theory

 

[scouserlfc]

u dont really need a level concept according to me answer should be 1 and 3 because ammonia have one lone pair which would be use for bonding with silver whenever there is a coordinate bonding involved the donor of lone pair is considered bronsted lowry base according to their thoery . as this is a displacement reactin so there is nothing oxidized or reduced hope that answers your question .
bronsted lowry theory oxidation and redution and coordinate bnding is all as level hope i was helpful

by the way can i have the year for this question please ??

 

[SilverCrest]

Peace be upon you!
I need help with q1(c)(iv)
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_31.pdf

i dont understand why we have to multiply our answer to part(iii) into 40 ..

 

[snowbrood]

explain the shape of and the angles in the ethane and ethene in terms of sigma and alpha bonds can anyone please write an answer for both of these compounds

 

[scouserlfc]

explain the shape of and the angles in the ethane and ethene in terms of sigma and alpha bonds can anyone please write an answer for both of these compounds

umm well ill give u a website please check it from there because this particular syllabus point requires u to just remember the shape

For ethane : http://www.chemguide.co.uk/basicorg/bonding/methane.html#top
for ethene : http://www.chemguide.co.uk/basicorg/bonding/ethene.html#top

 

[aalmuhannadi]

go to the syllabus in as section u would see that coordinate bond is formed and i am afraid u are wrong when ever there is a coordinate bond formed the donor is always considered a base please have u studied the lowry theory

Yes I've studied the Bronsted-Lowry theory and if something donates a proton/H+ ion it's an acid not a base! Check for yourself xD And where in the syllabus do you find that the product has a coordinate bond? That's the thing that's troubling me I have no idea how/where the co-ordinate bond is formed here..

 

[aalmuhannadi]

by the way can i have the year for this question please ??

Summer 2012, Variant 21, paper 1, Q34

 

[snowbrood]

You can never have an answer as 1 and 3 on a multiple choice AS test! that's never one of the combination options... and the answer is 1 and 2 but I don't really understand why.. I need to know where exactly is this coordinate bond and the product formed, because it's not in the CIE chemistry coursebook

i dont know abt the book but coordinate bonding is in syllabus .
well there are two key concepts here i was wrong about that answer actually i blundered i thought it was changed rather than unchanged and u are right about the bronsted lowry theory.
i thought that it was lewis base lol according to lewis theory any compound that has a lone pair is capable of being a base. you know what ammonia is amphoteric( Able to react both as a base and as an acid.)
1. u should know from o level that metals do displacement reactions since there is displacement there would be no change in the oxidation number of nitrogen.
2.whenever a molecule has a lone pair it does coordinate bonding

 

[aalmuhannadi]

i dont know abt the book but coordinate bonding is in syllabus .
well there are two key concepts here i was wrong about that answer actually i blundered i thought it was changed rather than unchanged and u are right about the bronsted lowry theory.
i thought that it was lewis base lol according to lewis theory any compound that has a lone pair is capable of being a base. you know what ammonia is amphoteric(Able to react both as a base and as an acid.)
1. u should know from o level that metals do displacement reactions since there is displacement there would be no change in the oxidation number of nitrogen.
2.whenever a molecule has a lone pair it does coordinate bonding

Thanks now that explains it!

 

[Minato112]

Peace be upon you!
I need help with q1(c)(iv)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_31.pdf

i dont understand why we have to multiply our answer to part(iii) into 40 ..

You have calculated the number of moles in part (iii). Let's call it x

Therefore,
In 25.0 cm3, there are x moles of H2SO4.
In 1000 cm3, there are ((x/25)*1000) moles of H2SO4.

Note: 1 dm3 ----> 1000 cm3

Hope it helps.

 

[mrnt3250]

My confusion is that when it says 70% of the reactants (all ?) are converted to products, does it mean that I have to add the number of moles of both the reactants and then multiply it by 70/100 ? When I do this and try to subtract number of moles of a reactant from the final product, I get a negative value.

This is how I've done it:
<---
C2H5OH + CH3COOH ---> CH3COOC2H5 + H2O
initially: 12/60 = 0.2 mol 9.20/46 = 0.2 mol 0 mol 0 mol
concentration: 0.2/0.25 = 0.8 moldm^-3 0.8 moldm^-3

at equilibrium: x x 70/100 * 0.2 = 0.14 mol 0.14 mol
0.2-0.14 = 0.06 mol 0.2 - 0.14 = 0.06 mol

This is my problem, I don't understand how to get the number of moles of reactants and products separately once a percentage is given.

 

[izzahzainab]

My confusion is that when it says 70% of the reactants (all ?) are converted to products, does it mean that I have to add the number of moles of both the reactants and then multiply it by 70/100 ? When I do this and try to subtract number of moles of a reactant from the final product, I get a negative value.

This is how I've done it:
<---
C2H5OH + CH3COOH ---> CH3COOC2H5 + H2O
initially: 12/60 = 0.2 mol 9.20/46 = 0.2 mol 0 mol 0 mol
concentration: 0.2/0.25 = 0.8 moldm^-3 0.8 moldm^-3

at equilibrium: x x 70/100 * 0.2 = 0.14 mol 0.14 mol
0.2-0.14 = 0.06 mol 0.2 - 0.14 = 0.06 mol

This is my problem, I don't understand how to get the number of moles of reactants and products separately once a percentage is given.

You are doing it right; I don't see where the problem is :/ At equilibrium there will be 0.06 moles of ethanol, 0.06 moles of ethanoic acid, 0.14 moles of ethylethanoate and 0.14 moles of water.