Edexcel CHEMISTRY DOUBTS!!!!

[~ Miss Relina ~]

Thanks GG for the notes 

Does anyone have the link to AS chem Unit 3B Jan 2011? I cant seem to find it anywhere.. ;/

merci ^^

here

[Malak]

Sorry..whats pplw?
:S
Q15 which part?

[NightHawkerr]

In some past papers I've seen that if we have to find the enthalpy change, they do delta H of reactants - delta H of products, when Ive studied that to find delta H its delta H of prod - delta H of reactants.. When is this not the case? I dont remember what past paper it is actually from, but help would be appreciated. 
& What's the difference between Mean bond enthalpy and just standard bond enthalpy? Thanks alot!

[~ Miss Relina ~]

Sorry..whats pplw?
:S
Q15 which part?

pplw was a mis typing
Q 15 part b

[Malak]

In some past papers I've seen that if we have to find the enthalpy change, they do delta H of reactants - delta H of products, when Ive studied that to find delta H its delta H of prod - delta H of reactants.. When is this not the case? I dont remember what past paper it is actually from, but help would be appreciated. 
& What's the difference between Mean bond enthalpy and just standard bond enthalpy? Thanks alot!

Its Products - reactants when you are finding it for Enthalpy of Formation
And for Enthalpy of combustion its reactants - products

Mean bond enthalpy is the Average value for that bond whereas the other one is particularly for that bond in that compound
For eg C-H bond in methane is different from in ethanol but mean value of C-H bond is used assuming they are all the same

Hope u get it?

[NightHawkerr]

Its Products - reactants when you are finding it for Enthalpy of Formation
And for Enthalpy of combustion its reactants - products

Mean bond enthalpy is the Average value for that bond whereas the other one is particularly for that bond in that compound
For eg C-H bond in methane is different from in ethanol but mean value of C-H bond is used assuming they are all the same

Hope u get it?

Hmm I get it thanks alot 

And for the Hess's Law, it's reactant-products?

[Malak]

pplw was a mis typing
Q 15 part b

b (i) - appropriate readings are the ones which are concordant so, (11.40 + 11.20)/2 = 11.30

b-ii no. of moles is simply n = cv, so 0.300 * 11.3 = 0.00339 mol

iii- the ratio is 1 --> 1 so same no. of moles of HCl are used i.e 0.00339

iv- in 10 cm3 moles are 0.00339 so in 100 there will be 0.0339 (just * 10)

M not sure if you need the rest of parts as well?

[The Golden Girl =D]

2010 Jan Question 3d (v) and (vi)

Why is the temperature change double? There are same no. moles in both :s

Thx

ang3l , was this doubt cleared out ?

if not send me a PM ,and I'll answer you tomorrow inshAllah

@NightHawkerr : You're Welcome

[~ Miss Relina ~]

2010 Jan Question 3d (v) and (vi)

Why is the temperature change double? There are same no. moles in both :s

Thx

ya i had same doubt too

[~ Miss Relina ~]

b (i) - appropriate readings are the ones which are concordant so, (11.40 + 11.20)/2 = 11.30

b-ii no. of moles is simply n = cv, so 0.300 * 11.3 = 0.00339 mol

iii- the ratio is 1 --> 1 so same no. of moles of HCl are used i.e 0.00339

iv- in 10 cm3 moles are 0.00339 so in 100 there will be 0.0339 (just * 10)

M not sure if you need the rest of parts as well?

well what i understood for iii is that they said left is that means remainig or whT soory for annoying u but this was my doubt Thanks

[studz]

Anyone doing chemistry unit 6?? its on the 7th of june right
i wanna know what are the key things to remember and where can i get papers for practicing
we did not do unit 6 at school 

[Blizz_rb93]

A student prepared a sample of the fertilizer ammonium sulfate by adding ammonia solution to sulfuric acid :

2NH3 (aq) + H2SO4 (aq) -> (NH4)SO4 (aq)

a) Calculate the theoretical maximum yield that can be obtained by reacting 25.0 cm3 of 2.0mol dm-3 ammonia solution with an excess of sulfuric acid.

b) If the actual mass obtained was 3.12g, calculate the percentage yield and suggest reasons why the yield is less than 100%


can anyone help me with this asap please and id just like to know what are the exact methods of solving a question regarding this topic

Thank you

[Malak]

A student prepared a sample of the fertilizer ammonium sulfate by adding ammonia solution to sulfuric acid :

2NH3 (aq) + H2SO4 (aq) -> (NH4)SO4 (aq)

a) Calculate the theoretical maximum yield that can be obtained by reacting 25.0 cm3 of 2.0mol dm-3 ammonia solution with an excess of sulfuric acid.

b) If the actual mass obtained was 3.12g, calculate the percentage yield and suggest reasons why the yield is less than 100%


can anyone help me with this asap please and id just like to know what are the exact methods of solving a question regarding this topic

Thank you

No. of moles of ammonia : 25/1000 * 2.0 = 0.05 mol

eq. (should be (NH4)₂SO4) shows that 2 moles of NH3 form 1 mole of the product

So 0.025 moles of product are formed
molar mass of ammonium sulfate = 132.1

therefore mass formed will be 132.1 * 0.025 = 3.3025

(b) % yield = actual mass / maximum possible

so %yield = (3.12 / 3.3025) * 100
= 94.5 %

And reasons could be :
Reactants are impure
Product left behind on apparatus
Human error

I just did the calculations, if u have any doubt then ask:)

[Malak]

Q:
20 cm3 of sulfuric acid, concentration 0.25 mol dm–3, was neutralized in a titration with barium hydroxide, concentration 0.50 mol dm–3. The equation for the reaction is
Ba(OH)2(aq) + H2SO4(aq) ? BaSO4(s) + 2H2O(l)

During the titration, the barium hydroxide was added until it was present in excess.
The electrical conductivity of the titration mixture
A increased steadily.
B decreased steadily.
C increased and then decreased.
D decreased and then increased.

its 2010 Jan Q10 unit 2

[The Golden Girl =D]

2010 Jan Question 3d (v) and (vi)

Why is the temperature change double? There are same no. moles in both :s

Thx

in (v)
Q = m*c*delta T
115 = 25 *4.18 * delta T

so delta T = 11.0

in (iv) delta T = 5.5 => that's given to you at the beginning.

(v) 11.0 , because the same amount of energy present in Half the volume.

(vi) in this one measure the % error for each ;

for (iii) [0.1 * 2]/(5.5) * 100 = 3.64%    (two readings hence multiplied by two)

in (vi) = [ 0.2/11 ]* 100 = 1.82%

1.82% in (vi) is LESS THAN 3.64% in (iii)


Answer ; (v) will have LESS error because delta T is TWICE bigger.

I hope you got it girls