Chemistry: Post your doubts here!

[The Sarcastic Retard]

http://maxpapers.com/syllabus-materials/chemistry-9701-a-level/attachment/9701_w09_qp_21/
Xylferion Please can you answer 3 part c?

3c) You'll have to draw a double hump diagram for this. It's essentially the same as the simpler energy diagram except we break the single hump into 2 to represent the intermediate steps. There should be two peaks and the first one should be higher than the second. If you recall free radical substitution mechanism you know the intermediate steps. The first hump represents the formation of methyl free radical and HCl and the second represents the chloromethane formation. The reason the second hump is lower is because the reaction can only proceed as long as the energy barrier for the following step is less then the preceding one. The activation energy is labeled for the first hump.
-lilcloud!

 

[Studydayandnight]

GUYS PLEASE HELP!!!!!!

Question 2 (b) May june 2014/23

I am struggling with this and my exams start in less than 10 days! :'(

http://maxpapers.com/wp-content/uploads/2012/11/9701_s14_qp_23.pdf

 

[Midnight dream]

3c) You'll have to draw a double hump diagram for this. It's essentially the same as the simpler energy diagram except we break the single hump into 2 to represent the intermediate steps. There should be two peaks and the first one should be higher than the second. If you recall free radical substitution mechanism you know the intermediate steps. The first hump represents the formation of methyl free radical and HCl and the second represents the chloromethane formation. The reason the second hump is lower is because the reaction can only proceed as long as the energy barrier for the following step is less then the preceding one. The activation energy is labeled for the first hump.
-lilcloud!

Thanks.....yaar.

 

[The Sarcastic Retard]

GUYS PLEASE HELP!!!!!!

Question 2 (b) May june 2014/23

I am struggling with this and my exams start in less than 10 days! :'(

http://maxpapers.com/wp-content/uploads/2012/11/9701_s14_qp_23.pdf

This simply means u haven't grasped your concept of moles properly. Physical chemistry needs calculation practice.
2bi) n = cv = 40*0.4*0.001 = 0.016mol.
ii) n = cv = 25*0.12*0.001 = 0.003mol.
iii) intitial - excess will be left to react with ammonia = 0.016-0.003 = 0.013mol.
iv) ratio is 1:1 hence no change. 0.013mol.
v) 0.413/63.5 = 0.0065mol of copper. 1:2 ratio hence x=2.
vi) 63.5 + (14*2) + (1*8)+(32.1*2)+(16*8)+(6*18) = 399.7

 

[Midnight dream]

3c) You'll have to draw a double hump diagram for this. It's essentially the same as the simpler energy diagram except we break the single hump into 2 to represent the intermediate steps. There should be two peaks and the first one should be higher than the second. If you recall free radical substitution mechanism you know the intermediate steps. The first hump represents the formation of methyl free radical and HCl and the second represents the chloromethane formation. The reason the second hump is lower is because the reaction can only proceed as long as the energy barrier for the following step is less then the preceding one. The activation energy is labeled for the first hump.
-lilcloud!

your getting good wid chem probs..

 

[Studydayandnight]

This simply means u haven't grasped your concept of moles properly. Physical chemistry needs calculation practice.
2bi) n = cv = 40*0.4*0.001 = 0.016mol.
ii) n = cv = 25*0.12*0.001 = 0.003mol.
iii) intitial - excess will be left to react with ammonia = 0.016-0.003 = 0.013mol.
iv) ratio is 1:1 hence no change. 0.013mol.
v) 0.413/63.5 = 0.0065mol of copper. 1:2 ratio hence x=2.
vi) 63.5 + (14*2) + (1*8)+(32.1*2)+(16*8)+(6*18) = 399.7

Thanks so much for this. I don't get one thing though, how can you tell the mole ratios in the (iv) part? By comparing the NH4+ ions to what in which equation?

(I know I may come across as stupid but moles and organic are my weakness :'( )

 

[Studydayandnight]

Question 1 (c) i!

Another hydrocarbon, W, with the formula C4H8, reacts with hydrogen bromide, HBr, to give two products X and Y. X and Y are structural isomers of molecular formula C4H9Br. Reaction of X with aqueous alkali produces an alcohol, Z, that has no reaction with acidifi ed dichromate(VI).

It asks for the names of the structures. What I came up with was.

W = CH2=CHCH2CH3 - Butene

X= CH3CH2CH2CH2Br - Bromobutane

Y= CH3CH(Br) CH2CH3 - 2- Bromobutane

but in the mark scheme W,X and Y are different than my answers. I don't understand how my answers can be wrong. :/ Can someone PLEASE PLEASE explain why my formulas aren't considered correct since they have the same molecular formula as C4H8 and C4H9Br.

I'll be forever grateful!

I've attached the mark scheme!

http://maxpapers.com/wp-content/uploads/2012/11/9701_s14_ms_23.pdf

 

[asadalam]

Question 1 (c) i!

Another hydrocarbon, W, with the formula C4H8, reacts with hydrogen bromide, HBr, to give two products X and Y. X and Y are structural isomers of molecular formula C4H9Br. Reaction of X with aqueous alkali produces an alcohol, Z, that has no reaction with acidifi ed dichromate(VI).

It asks for the names of the structures. What I came up with was.

W = CH2=CHCH2CH3 - Butene

X= CH3CH2CH2CH2Br - Bromobutane

Y= CH3CH(Br) CH2CH3 - 2- Bromobutane

but in the mark scheme W,X and Y are different than my answers. I don't understand how my answers can be wrong. :/ Can someone PLEASE PLEASE explain why my formulas aren't considered correct since they have the same molecular formula as C4H8 and C4H9Br.

I'll be forever grateful!

I've attached the mark scheme!

http://maxpapers.com/wp-content/uploads/2012/11/9701_s14_ms_23.pdf

Ok,firstly your own ans are wrong W is But-1-ene,at this level we have to use full names.
X forms an alcohol that gives no reaction with oxidising agent hence it must be a tertiary alcohol,and hence W must also have a structure where one carbon is surrounded by 3 others,as after electrophilic substitution wth Bromene it will again be nucleophilically substituted to form the TERTIARY ALCOHOL.

Keep that in mind and try to work out the structures via reverse processing each one,and work your way backwards.Try it keeping the tertiary alcohol in mind and then tell me if it doesnt work out for you.

 

[The Sarcastic Retard]

Thanks so much for this. I don't get one thing though, how can you tell the mole ratios in the (iv) part? By comparing the NH4+ ions to what in which equation?

(I know I may come across as stupid but moles and organic are my weakness :'( )

Consider NH3 moles in next part eqn with first part.

Question 1 (c) i!

Another hydrocarbon, W, with the formula C4H8, reacts with hydrogen bromide, HBr, to give two products X and Y. X and Y are structural isomers of molecular formula C4H9Br. Reaction of X with aqueous alkali produces an alcohol, Z, that has no reaction with acidifi ed dichromate(VI).

It asks for the names of the structures. What I came up with was.

W = CH2=CHCH2CH3 - Butene

X= CH3CH2CH2CH2Br - Bromobutane

Y= CH3CH(Br) CH2CH3 - 2- Bromobutane

but in the mark scheme W,X and Y are different than my answers. I don't understand how my answers can be wrong. :/ Can someone PLEASE PLEASE explain why my formulas aren't considered correct since they have the same molecular formula as C4H8 and C4H9Br.

I'll be forever grateful!

I've attached the mark scheme!

http://maxpapers.com/wp-content/uploads/2012/11/9701_s14_ms_23.pdf

Organic is yet not started. Sorry. Refer to this post.

 

[Studydayandnight]

Consider NH3 moles in next part eqn with first part.

Organic is yet not started. Sorry. Refer to this post.

Ok,firstly your own ans are wrong W is But-1-ene,at this level we have to use full names.
X forms an alcohol that gives no reaction with oxidising agent hence it must be a tertiary alcohol,and hence W must also have a structure where one carbon is surrounded by 3 others,as after electrophilic substitution wth Bromene it will again be nucleophilically substituted to form the TERTIARY ALCOHOL.

Keep that in mind and try to work out the structures via reverse processing each one,and work your way backwards.Try it keeping the tertiary alcohol in mind and then tell me if it doesnt work out for you.

Thank you so much guys for your help!

 

[Lola_sweet]

Define mole in terms of avogadros number

 

[The Sarcastic Retard]

Define mole in terms of avogadros number

the amount of substance containing 6(.02) x 1023 (fundamental) particles of that substance (or; the amount of substance containing as many particles as there are atoms in 12g of carbon-12)

 

[Sarosh Jameel]

PLZ someone explain me last two boxes and part B !!! ASAP
!

 

[Studydayandnight]

Could someone explain this equation?

The commonest form of iron(II) sulfate is the heptahydrate, FeSO4.7H2O. On heating at 90 C this loses some of its water of crystallisation to form a different hydrated form of iron(II) sulfate, FeSO4.xH2O. 3.40g of FeSO4.xH2O was dissolved in water to form 250 cm3 of solution. A 25.0cm3 sample of this solution was acidifi ed and titrated with 0.0200 mol dm–3 potassium manganate(VII). In this titration 20.0cm3 of this potassium manganate(VII) solution was required to react fully with the Fe2+ ions present in the sample.

Complete and balance the ionic equation for the reaction between the manganate(VII) ions and the iron(II) ions.

MnO4 – (aq) + 5Fe2+(aq) + ........H+(aq)  ..... +....(aq) + 5Fe3+(aq) + .......H2O(l)


Also this question!

Another element, Z, in the same period of the Periodic Table as A, reacts with chlorine to form a compound with empirical formula ZCl 2. The percentage composition by mass of ZCl 2 is Z, 31.13; Cl, 68.87.

Calculate the relative atomic mass, Ar , of Z. Give your answer to three significant figures.

I've attached the mark scheme for reference! http://maxpapers.com/wp-content/uploads/2012/11/9701_s14_ms_21.pdf

 

[Mahmood Magdy]

Hey guys, I have a silly question. Which type of bond is this

 

[Studydayandnight]

Could someone explain this equation?

The commonest form of iron(II) sulfate is the heptahydrate, FeSO4.7H2O. On heating at 90 C this loses some of its water of crystallisation to form a different hydrated form of iron(II) sulfate, FeSO4.xH2O. 3.40g of FeSO4.xH2O was dissolved in water to form 250 cm3 of solution. A 25.0cm3 sample of this solution was acidifi ed and titrated with 0.0200 mol dm–3 potassium manganate(VII). In this titration 20.0cm3 of this potassium manganate(VII) solution was required to react fully with the Fe2+ ions present in the sample.

Complete and balance the ionic equation for the reaction between the manganate(VII) ions and the iron(II) ions.

MnO4 – (aq) + 5Fe2+(aq) + ........H+(aq)  ..... +....(aq) + 5Fe3+(aq) + .......H2O(l)


Also this question!

Another element, Z, in the same period of the Periodic Table as A, reacts with chlorine to form a compound with empirical formula ZCl 2. The percentage composition by mass of ZCl 2 is Z, 31.13; Cl, 68.87.

Calculate the relative atomic mass, Ar , of Z. Give your answer to three significant figures.

I've attached the mark scheme for reference! http://maxpapers.com/wp-content/uploads/2012/11/9701_s14_ms_21.pdf

HELPPP! I NEED HELPPPP! #DesperationAtItsPeak

 

[Metanoia]

And also in the same question, it was asked whether the compound exhibits cis-trans isomerism. And the answer was No. So does it mean that cyclic alkenes never exhibit isomerism?

No. Cycloalkenes are restricted from having cis-trans isomers, but those which have more than 8 carbons DO have cis-trans isomerism.
Metanoia Please confirm this is true right??

Yes. Cis isomers would be very unlikely in small rings (less than 8 carbons) due to the strain on the bonds.

 

[Metanoia]

Hello. Can anyone please explain to me these questions? Thank you

1. The density of this nitrogen to be 1.2572gdm-3 at stp. Chemically, pure nitrogen has a density of 1.2505gdm-3 at stp. Which gas was present in atmospheric nitrogen to cause this discrepancy?
A. Argon B. Helium C. Methane D. Neon. ANS: Argon (Why?)

2. A sample of mg of an organic compound is vaporised in a gas syringe and occupies V cm3 at TK and p atm. What is the relative molecular mass of the compound, Mr?
A. Mr= (m.22400.T)/p.V.273 B. Mr= (m.22400.T+273)/p.V.273 C. Mr= (m.22400.273.p)/V.T D. (m.22400.273.p)/V.(T+273) ANS: A

Q1. Since the density of the mixture is higher than density of pure N2, it indicates that N2 is mixed with another gas that has a higher Mr than N2 (28).

Q2.
pV = nRT
pV = (mass/Mr)RT ------(1)

What is R?
When T = 273, n =1, p = 1, V = 22400
R= pV/nT = 22400/273

Sub into eqn (1) and rearrange the equation in terms of Mr

 

[Metanoia]

PLZ someone explain me last two boxes and part B !!! ASAP
!

The cold dilute MnO4- mildly oxidises the C=C bonds, adding 2 OH groups to each carbon to form a di-ol

The OH group subsequently oxidizes to form COOH and ketones respectively for the last box.


Reaction between C and D: C has an acid group and D has an alcohol group, create an ester bond between them.
Reaction between C and E: Again, use the COOH group on E to form an ester with any alcohol group won C.

 

[Metanoia]

Hey guys, I have a silly question. Which type of bond is this

Ester linkage.