Chemistry: Post your doubts here!

[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.

 

[mrnt3250]

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.

I'm confused because the question says at equilibrium, 70% of reactants (means all of them) gets converted to products, but I'm only taking no. of moles of one of the reactants.??

 

[izzahzainab]

I'm confused because the question says at equilibrium, 70% of reactants (means all of them) gets converted to products, but I'm only taking no. of moles of one of the reactants.??

The total number of moles of reactants at the beginning are 0.2 + 0.2 = 0.4 moles
70/100 * 0.4 = 0.28 moles of reactants. Now we know that ethanol and ethanoic acid react in 1 : 1 ratio so in 0.28 moles of reactants there are 0.14 moles of ethanoic acid and 0.14 moles of ethanol. (0.14 + 0.14 = 0.28) I hope you get it now.

 

[aalmuhannadi]

I uploaded two questions, need help with the answers please!

Answers: 13: B and 33: 1 and 2, but can someone please explain them? For 13, I did not find anything on chemguide or the coursebook about diagonal relationships so I'm not sure if it's in the new syllabus (because these are from the 2005 papers), and for 33, on the examiner report, it says that students were EXPECTED to know that glucose could be measured using a calorimeter but I didn't know that... and explain why 1 and 2 are correct? I know that 2 is correct because C and H cannot be directly reacted together but I can't prove why 1 is correct as well...

 

[Minato112]

I uploaded two questions, need help with the answers please!View attachment 19598View attachment 19599

Answers: 13: B and 33: 1 and 2, but can someone please explain them? For 13, I did not find anything on chemguide or the coursebook about diagonal relationships so I'm not sure if it's in the new syllabus (because these are from the 2005 papers), and for 33, on the examiner report, it says that students were EXPECTED to know that glucose could be measured using a calorimeter but I didn't know that... and explain why 1 and 2 are correct? I know that 2 is correct because C and H cannot be directly reacted together but I can't prove why 1 is correct as well...

For question 13, apply the diagonal relationship. I've learnt it also and I don't think it's new in the syllabus.

I've put an example below, I assume you already know it but still I've put it here for students who didnt know. (To note that diagonal relationship is a way to determine, using the periodic table, elements which have same electonegativities)

As for question 33, my reasoning is the same as yours. I will try to look at it deeply and will notify you.

Hope I helped.

 

[darknessinme]

I uploaded two questions, need help with the answers please!View attachment 19598View attachment 19599

Answers: 13: B and 33: 1 and 2, but can someone please explain them? For 13, I did not find anything on chemguide or the coursebook about diagonal relationships so I'm not sure if it's in the new syllabus (because these are from the 2005 papers), and for 33, on the examiner report, it says that students were EXPECTED to know that glucose could be measured using a calorimeter but I didn't know that... and explain why 1 and 2 are correct? I know that 2 is correct because C and H cannot be directly reacted together but I can't prove why 1 is correct as well...

Coursebook pg 104 q10 answer.

 

[scouserlfc]

For question 13, apply the diagonal relationship. I've learnt it also and I don't think it's new in the syllabus.

I've put an example below, I assume you already know it but still I've put it here for students who didnt know. (To note that diagonal relationship is a way to determine, using the periodic table, elements which have same electonegativities)

As for question 33, my reasoning is the same as yours. I will try to look at it deeply and will notify you.

Hope I helped.

When i did Question 13 myself i did it this way !
that the distance from fluorine of both aluminum and beryllium is same so they have the same electronegativity As u remember that these decrease as u go from grp 7 to 1 and down the groups then if we take each element as equal to 1 unit and count then we get exactly same units You might call it a pretty absurd method but it works

I uploaded two questions, need help with the answers please!View attachment 19598View attachment 19599

Answers: 13: B and 33: 1 and 2, but can someone please explain them? For 13, I did not find anything on chemguide or the coursebook about diagonal relationships so I'm not sure if it's in the new syllabus (because these are from the 2005 papers), and for 33, on the examiner report, it says that students were EXPECTED to know that glucose could be measured using a calorimeter but I didn't know that... and explain why 1 and 2 are correct? I know that 2 is correct because C and H cannot be directly reacted together but I can't prove why 1 is correct as well...

For 33 how can u find the enthalpy of hydration directly can u get CuSO4 in gaseous form ?? i dont think so ! But thats what the enthalpy of hydration states that the energy realeased when one mole of gaseous ions are dissolved in water !!!

 

[Minato112]

When i did Question 13 myself i did it this way !
that the distance from fluorine of both aluminum and beryllium is same so they have the same electronegativity As u remember that these decrease as u go from grp 7 to 1 and down the groups then if we take each element as equal to 1 unit and count then we get exactly same units You might call it a pretty absurd method but it works

Sure is absurd but sounds interresting!

 

[aalmuhannadi]

minato112 scouserlfc Thanks you two your answers really helped much appreciated!

 

[Minato112]

minato112 scouserlfc Thanks you two your answers really helped much appreciated!

No problem!

 

[aalmuhannadi]

Need help with these questions as well

Q16 + 17 I do not understand at all... and aren't 'ligands' part of the A2 syllabus? That's what my teacher said
Q33 How would you deduce that it had an empirical formula of BN?

 

[scouserlfc]

Need help with these questions as wellView attachment 19629View attachment 19630

Q16 + 17 I do not understand at all... and aren't 'ligands' part of the A2 syllabus? That's what my teacher said
Q33 How would you deduce that it had an empirical formula of BN?

Dont do 17 its A2 syllabus part as we don't know the complex stuff forming when these reactions take place !

For 16 what i have till now understood is that since BaCO3 is more stable than any other Carbonate of grp 2 so when u heat it with bunsen flame which cant exceed a particular heat energy limit which is not enuf to decompose BaCO3 so BaCO3 remains as it is and then reacts with HCl while CaCO3 breaks and its oxide reacts with HCl too ! Let me know the answer

For 33 i dont think this is the correct logic but kind of looks correct to me ! that since BN has a structure like graphite then if it is supposed to have a layer structure like graphite it should have equal no. of B and N atoms in it otherwise you wont have graphite like bonding in it. Also a proof of this is seen when in 2 it says that it has a hexagonal pattern in which B and N atoms occupy alternate positions so in a hexagon u have six sides and each side is either occupied by B or N and since they are alternating so in each hexagon they will occupy 3 each corners and since hexagons only contain the B and N atoms and in them u have equal no. of B and N atoms so the emperical formula should be BN regardless of how many hexagons u attach to make the layer

 

[aalmuhannadi]

Dont do 17 its A2 syllabus part as we don't know the complex stuff forming when these reactions take place !

For 16 what i have till now understood is that since BaCO3 is more stable than any other Carbonate of grp 2 so when u heat it with bunsen flame which cant exceed a particular heat energy limit which is not enuf to decompose BaCO3 so BaCO3 remains as it is and then reacts with HCl while CaCO3 breaks and its oxide reacts with HCl too ! Let me know the answer

For 33 i dont think this is the correct logic but kind of looks correct to me ! that since BN has a structure like graphite then if it is supposed to have a layer structure like graphite it should have equal no. of B and N atoms in it otherwise you wont have graphite like bonding in it. Also a proof of this is seen when in 2 it says that it has a hexagonal pattern in which B and N atoms occupy alternate positions so in a hexagon u have six sides and each side is either occupied by B or N and since they are alternating so in each hexagon they will occupy 3 each corners and since hexagons only contain the B and N atoms and in them u have equal no. of B and N atoms so the emperical formula should be BN regardless of how many hexagons u attach to make the layer

But Q17 was part of the same paper, it was a paper 1 question... and I found a similar question about complex ions in the summer 2012 paper 1 as well! For Q16 you're right it's C. The question says that for all of them when heated a colourless gas was evolved, so how does barium carbonate not decompose if a gas is evolved from it on heating? Or does online the CaCO3 part of it get removed while the BaCO3 stays intact? How do you judge that Barium reacts most vigorously from the rest of the remaining solids like Calcium (answer A) or magnesium carbonate (answer B)?

Thanks for answering Q33 I understand your logic

 

[scouserlfc]

But Q17 was part of the same paper, it was a paper 1 question... and I found a similar question about complex ions in the summer 2012 paper 1 as well! For Q16 you're right it's C. The question says that for all of them when heated a colourless gas was evolved, so how does barium carbonate not decompose if a gas is evolved from it on heating? Or does online the CaCO3 part of it get removed while the BaCO3 stays intact? How do you judge that Barium reacts most vigorously from the rest of the remaining solids like Calcium (answer A) or magnesium carbonate (answer B)?

Thanks for answering Q33 I understand your logic

For 17 can u please let me know its year because if it is not a recent one then the syllabus at that time might have been different,because ligands are a part of transition metals which are mostly in A2 except for the part where u have to state the catalysts

For 16 remember that reactivity of Grp 2 metals increases down the group like their solubility or the Grp 1 metals because their first and second IE decreases so Barium reacts more vigorously than any other metal in Grp 2,if u are lucky enuf to visit Youtube and your country hasnt banned it then type there reaction of Barium with water and compare it with calcium and observe the difference Also due to the high thermal stability of carbonates down the group and more enrgy is needed to break down their lattice structure ! try studying a bit of the A2 part of grp 2 metals it helps the understanding i did it because it was only based on 2 pages in my book when i did this part of syllabus

 

[aalmuhannadi]

For 17 can u please let me know its year because if it is not a recent one then the syllabus at that time might have been different,because ligands are a part of transition metals which are mostly in A2 except for the part where u have to state the catalysts

For 16 remember that reactivity of Grp 2 metals increases down the group like their solubility or the Grp 1 metals because their first and second IE decreases so Barium reacts more vigorously than any other metal in Grp 2,if u are lucky enuf to visit Youtube and your country hasnt banned it then type there reaction of Barium with water and compare it with calcium and observe the difference Also due to the high thermal stability of carbonates down the group and more enrgy is needed to break down their lattice structure ! try studying a bit of the A2 part of grp 2 metals it helps the understanding i did it because it was only based on 2 pages in my book when i did this part of syllabus

Alright thanks, and Q17 was from 2005

 

[scouserlfc]

Alright thanks, and Q17 was from 2005

Basically the answer to 17 just proves it that the syllabus then was a bit different and had ligands in AS so just forget this one

 

[aalmuhannadi]

Basically the answer to 17 just proves it that the syllabus then was a bit different and had ligands in AS so just forget this one

Oh alright, thanks. Well I stumbled onto an enthalpy change question which I couldn't do because Hf and Hc weren't labelled so it confused me, could you possible help do this one as well?

 

[scouserlfc]

Oh alright, thanks. Well I stumbled onto an enthalpy change question which I couldn't do because Hf and Hc weren't labelled so it confused me, could you possible help do this one as well?
View attachment 19698

It should be C thats what i think !
You could do it by using hess law i think but i didnt use it actually break the reaction in two parts,
1.where CO2 is converted to CO
2.where H2 is converted to H2O (gaseous)

for 1 the enthalpy is +283 as you use the opposite of the given enthalpy of combustion of CO !
for 2 it involves addition of two enthalpies, one in which H2 is combusted to form H2O liquid and also the atomisation enthalpy for converting H2O liquid to steam ! so the whole thing for 2 is (-286 + 44 ) we use +44 again as we are going in the opposite direction of the given equation !

So add up everything now
+283-286+44 =ΔHr
ΔHr = +41KJ/mol

 

[aalmuhannadi]

It should be C thats what i think !
You could do it by using hess law i think but i didnt use it actually break the reaction in two parts,
1.where CO2 is converted to CO
2.where H2 is converted to H2O (gaseous)

for 1 the enthalpy is +283 as you use the opposite of the given enthalpy of combustion of CO !
for 2 it involves addition of two enthalpies, one in which H2 is combusted to form H2O liquid and also the atomisation enthalpy for converting H2O liquid to steam ! so the whole thing for 2 is (-286 + 44 ) we use +44 again as we are going in the opposite direction of the given equation !

So add up everything now
+283-286+44 =ΔHr
ΔHr = +41KJ/mol

Wow never knew you could do it that way! I'm used to using Hess's law but I didn't know how to do that for this question. I wish you could be my tutor!!

 

[scouserlfc]

Wow never knew you could do it that way! I'm used to using Hess's law but I didn't know how to do that for this question. I wish you could be my tutor!!

Umm !!! i also started doing by hess law but since the equations given are such that if you use a bit of brain then u can easily comprehend about what you should do ! Basically my teacher he is a bit of a genius himself so he always keeps emphasizing that when u get to know a topic you should be able to apply otherwise you will be in the 99% of people who rote learn stuff so it really is his guidance and a bit of past papers help