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Chemistry: Post your doubts here!

daredevil

daredevil

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Raiyan3 said:
Im still struggling a bit :/ i didn't get the "find out the total energy released when 1.60 g of fuel is burnt through the enthlpy method", if you please give a little more explanation i will get the hang of it im sure.
Click to expand...
umm... i don't exactly remember the formula or stuff :p but i remember there was a way to do it with all the enthalpy changes like standard, combustion, hydration, etc. sab k different methods thay calculation k ... that's as far as i get this question and by the look of it i really don't wanna solve it :p if some1 gives the ryt answer and calculations plz tag me too ... lol :p
 
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PhyZac

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Raiyan3 daredevil

First find the gain in energy to the water using => mc(delta)T
therefore 200*4.2*(66-18) = 40320

As mentioned in the question this is 45% only therefore,
45% => 40320
100% => x
using cross multiplication find x
(100*40320)/45
u will get
89600

The question asked to find per gram...! and tht was for 1.6 gram...therefore
1.6 => 89600
1 => x

cross multiply and u get
(1*89600)/1.6 = 56000

And this is the answer !
 
daredevil

daredevil

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PhyZac said:
Raiyan3 daredevil

First find the gain in energy to the water using => mc(delta)T
therefore 200*4.2*(66-18) = 40320

As mentioned in the question this is 45% only therefore,
45% => 40320
100% => x
using cross multiplication find x
(100*40320)/45
u will get
89600

The question asked to find per gram...! and tht was for 1.6 gram...therefore
1.6 => 89600
1 => x

cross multiply and u get
(1*89600)/1.6 = 56000

And this is the answer !
Click to expand...
Ohhhhh.... thankuu soo much.. i c i got at a round about and starting turning round and round at the same point and lost my bearing somewhere in the questios :p lolx
thanks for posting the question Raiyan3 :)
 
Ahmedraza73

Ahmedraza73

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At a total pressure of 1.0 atm dinitrogen tetraoxide is 50% dissociated at a temperature of 60 degree Celcius. according to the following equation:
N2O4-------------2NO2( reversible reaction)
what is the value of the equilibrium constant Kp, for this reaction at 60 degree Celcius?
A:1/3 atm
B:2/3 atm
C: 4/3 atm
D:2 atm
:) Please reply some one
 
Raiyan3

Raiyan3

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PhyZac said:
Raiyan3 daredevil

First find the gain in energy to the water using => mc(delta)T
therefore 200*4.2*(66-18) = 40320

As mentioned in the question this is 45% only therefore,
45% => 40320
100% => x
using cross multiplication find x
(100*40320)/45
u will get
89600

The question asked to find per gram...! and tht was for 1.6 gram...therefore
1.6 => 89600
1 => x

cross multiply and u get
(1*89600)/1.6 = 56000

And this is the answer !
Click to expand...

Ahaa i knew there must be something with the percent! Thanks alott buddy!!
 
daredevil

daredevil

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ashiqbal said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdf
Q31 and Q35 please..
Click to expand...
for q 31 it wud be B??

as when the atoms become bigger the radii increase thus increasing bond length too because 2 radii make up the bond length as bond length is the measurement of length from the nuceus of one atom to the nucleus of the second atom.
and as the atom becomes bigger the attraction between the atoms decreases so the overlapping area of the radii of the atoms decreases. thus increasing the bond length... like this:
a-===-b

A---=---B

where a and b are smaller atoms
and A and B are larger atoms
and - is the part of radius not overlapping
and = is the part of radius overlapping

understood? :)
 
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PhyZac

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Raiyan3 said:
Please explain me question number 8
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
and question number 8 aswell from this on
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdf
Click to expand...
8 For which equation is the enthalpy change correctly described as an enthalpy change of
formation?
A 2NO(g) → N2(g) + O2(g) This is wrong since more than one compound formed
B 2CO(g) + O2(g) → 2CO2(g) Here the compound formed is not 1 mole..!
C H2O(l) + NaCl(s) → NaCl(aq) here the compound is not in its standard state tht is solid
D K(s) + Mn(s) + 2O2(g) → KMnO4(s) this is correct ! 1 mol, standard state !

8 Red lead oxide, Pb3O4, is used in metal priming paints. It can be made by heating PbO in air. 6PbO(s) + O2(g) → 2Pb3O4(s)

Which two values are needed to calculate the enthalpy change for this reaction?
I dont know how to explain this or whether my way is right, but i chose D since you have all values....formation of PbO and Pb3O4 and for the oxygen it is zero !
A enthalpy change of combustion of lead and enthalpy change of formation of Pb3O4
B enthalpy change of combustion of PbO and enthalpy change of formation of Pb3O4
C enthalpy change of formation of PbO and enthalpy change of atomisation of O2
D enthalpy change of formation of PbO and enthalpy change of formation of Pb3O4
 
Raiyan3

Raiyan3

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PhyZac said:
8 For which equation is the enthalpy change correctly described as an enthalpy change of
formation?
A 2NO(g) → N2(g) + O2(g) This is wrong since more than one compound formed
B 2CO(g) + O2(g) → 2CO2(g) Here the compound formed is not 1 mole..!
C H2O(l) + NaCl(s) → NaCl(aq) here the compound is not in its standard state tht is solid
D K(s) + Mn(s) + 2O2(g) → KMnO4(s) this is correct ! 1 mol, standard state !

8 Red lead oxide, Pb3O4, is used in metal priming paints. It can be made by heating PbO in air. 6PbO(s) + O2(g) → 2Pb3O4(s)

Which two values are needed to calculate the enthalpy change for this reaction?
I dont know how to explain this or whether my way is right, but i chose D since you have all values....formation of PbO and Pb3O4 and for the oxygen it is zero !
A enthalpy change of combustion of lead and enthalpy change of formation of Pb3O4
B enthalpy change of combustion of PbO and enthalpy change of formation of Pb3O4
C enthalpy change of formation of PbO and enthalpy change of atomisation of O2
D enthalpy change of formation of PbO and enthalpy change of formation of Pb3O4
Click to expand...
Thank you thank you so much my friend. Great explanation thanks! May Allah shower His blessings upon you!!
And if you have time can explain the question number 9 of the same year im just stuck there :/
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
 
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