Chemistry: Post your doubts here!

[hendrixgirl]

ANYONE DOING A2 PLEASE HELP ME OUT HERE!

http://www.xtremepapers.com/community/threads/chemistry-a2-doubts.15688/

 

[roadtrip9o9]

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_s11_qp_22.pdf question one all the moles calculations wid explanations plz :/

 

[prannoy]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_21.pdf

can someone pls tell me how do you know the product in q5 b(iii) "identify J" and how to draw the structural formula in c

Thanks

CH3 C(OH)H COOH

 

[prannoy]

what are you confused about? Basically they are asking what are the products of a reaction between conc. sulfuric acid and sodium bromide, and the reaction between butanol and conc. sodium bromide.

2NaBr + 2H2SO4 -----> Na2SO4 + Br2 + 2H2O + SO2

and an alcohol can be dehydrated by a strong conc. acid into an alkene.

it can also be Nabr + H2so4= Nahso4+ Hbr

 

[shamil]

Please help me http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_s08_qp_2.pdf

question 3, d, i.
I have no idea how the answer is supposed to be to the RHS.

 

[raamish]

Please help me http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_2.pdf

question 3, d, i.
I have no idea how the answer is supposed to be to the RHS.

the pressure is halved or lowered so equilibrium will go in the direction of greater moles that is the right hand side which has 3 moles

 

[raamish]

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_s03_qp_2.pdf

how do we do qs 1b(i)???

 

[smartangel]

when 50cm^3 of a hydrocarbon Y is burnt it reacts with exactly 300 cm^3 of oxygen to form 200 cm^3 of C02. water is also formed. deduce the equation for this reaction. answer: C4H8 + 6O2 → 4CO2 + 4H2O..please tell the how to work this one out..its frm chemistry coursebook.

 

[DragonCub]

Im not adding it.. thats what happens

It's either the one with oxygen, or this one:
PbO2 + 2 H2SO4 = 2 H2O + Pb(SO4)2

This one is impossible. Lead is not reduced at all in this reaction.

 

[DragonCub]

I'm getting:
PbO2 + 4H^(+) + SO4^(2-) + 2e- ------> PbSO4 + 2H2O

This is a perfectly balanced half equation.
But the problem with this equation is that it doesn't make sense. Sulfuric acid is a dibasic acid. This means one mole of H2SO4 will give 2 moles of H+ and one moles of sulfate ions. But here, we have 4 moles of H+ and one moles of sulfate ions. Is such a reaction possible? I don't know.

This can be possible. The question only asks for ionic equation so you don't have to worry about where the ions come from. Perhaps the some or all of the hydrogen ions are provided by another kind of acid like HCl.
Here the sulphuric acid has two functions, one as an reactant for the reduction of lead, the other as an acidic environment for the reaction to take place. The more detailed order should be:
PbO2 + 4H(+) + 2e(-) ------> Pb(2+) + 2H2O [the reduction of Pb(IV) in PbO2 under acidic condition]
Then Pb(2+) + SO4(2-) ------> PbSO4 [ionic precipitation]

 

[DragonCub]

i am re-posting this question:

1. Construct ionic half equation for:
a) PbO2(s) being reduced to PbSO4(s) in presence of of H2SO4(aq)
b) PbSO4(s) being reduced to Pb(s) in the presence of water.
c) Mow write the balanced equation for the reaction between PbO2 and lead in the presence of dilute sulfuric acid.

a) is PbO2 + 4H(+) + SO(2-) + 4e(-) ------> PbSO4 + 2H2O
b) I guess it is PbSO4 + 2e(-) ------> Pb + SO4(2-) [water here just provides aqueous environment]
Then for (c) PbO2 + 4H(+) + 4e(-) ------> Pb + 2H2O

 

[jsg]

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_s11_qp_22.pdf

Help needed in Q1 part c

 

[USMAN Sheikh]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_22.pdf

Help needed in Q1 part c

I also just login to ask the same ques i think solution is wrong coz the ans by my mentod is 4/9 whats ur?

 

[JD REBORN]

I just wanted to know that when we are asked to draw a chiral compound,would the marks be deducted if I didnt draw it exactly as the way in marking scheme.Cuz almost all the times the position of 4 groups around chiral carbon in my structure are different from the structure given in marking scheme.And generally when writing structural formulae i cannot write them exactly as mentioned in marking scheme??

 

[RGBM211]

Please help me http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_2.pdf

question 3, d, i.
I have no idea how the answer is supposed to be to the RHS.

So this is the equation given 2NOCl (g) -> 2NO(g) + Cl2(g)

ok so mmm first understand this
we know that when pressure is increased of a reaction which is in equilibrium,the equilibrium shifts in such a direction which can reduce the pressure.So increase in pressure increases the no. of molecule per unit volume.So the reaction tries to reduce the no. of molecule in the system so it moves towards a less molecular side.

So looking at the mole ratio in this case its 2:3 reactant: product ,the question is asking What will be the effect on the equilibrium concentration of NOCl when The pressure of the system is halved at constant temperature, so equilibrium will shift toward the more molecular side that is RHS because decrease in pressure shifts the equilibrium towards a more molecular side

 

[hussamh10]

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_s09_qp_2.pdf Q4 B 2nd part it ask what is formed when a diol(having also a carboxylic group attached)is reacted with carboxylic ......esters are formed but there must be only one ester formed because there is only one carboxylic group that reacts !!why then two esters are formed??? please help!!

 

[hussamh10]

I just wanted to know that when we are asked to draw a chiral compound,would the marks be deducted if I didnt draw it exactly as the way in marking scheme.Cuz almost all the times the position of 4 groups around chiral carbon in my structure are different from the structure given in marking scheme.And generally when writing structural formulae i cannot write them exactly as mentioned in marking scheme??

no marks will not be deducted but while making a mirror image you should keep the position exactly opp

 

[Mad-Eye]

· What reagent(s) and solvent are normally used in a laboratory to reduce a >C=O group without reducing the >C=C< group present in the same molecule?

 

[leosco1995]

·What reagent(s) and solvent are normally used in a laboratory to reduce a >C=O group without reducing the >C=C< group present in the same molecule?

LiAlH4 in dry ether OR NaBH4 in water.

 

[DragonCub]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_22.pdf

Help needed in Q1 part c

From (b) there is 0.04 mole of ethanoic acid remaining in the equilibrium mixture.
Originally there is 0.1 mole of ethanoic acid, so 0.06 mole of it has reacted to form R ethanoate.
According to the equation 1 mole of ethanoic acid reacts with 1 mole of R hydroxide to form 1 mole of R ethanoate and 1 mole of water.
So there should be 0.06 mole of R hydroxide reacted, 0.04 mole left.
0.06 mole of R ethanoate and water formed.
You don't need to worry about the volume since these four substances share the same solution volume.
Just use the mole amount to calculate Kc: 0.06 × 0.06 / (0.04 × 0.04) = 9 / 4 = 2.25 (no unit)