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Q1- 3*10^-21 * 6.02*10^23 = 1806Chemistry oct 2003 q 1,2,3.
Thanks I understand it nowA and B have H-bonds as well. In D Vander Waals forces are becoming stronger as its solidifiacation. In C there are only Vander Waals forces and its melting so the bonds need to be overcome.
Q14- A because max height means it covered half the distance so the energy also gets halved.
Q16- C. first find input power. output/input *100=efficiency. so 400000000/80=5MW
then find I using P=IV, 5MW/25kV=200A
Q21- find the area under the graph. Make it a triangle 1st. A=0.5*100*(2*10^-3)=0.1J but the area is not exactly a triangle. It is slightly more than a triangle so we assume it to be 0.11J
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s03_qp_1.pdf
no 26, 27, 35, 38
your help will be highly appreciated....
Thanks
Q9- momentum=mass*velocity= 2mvPhysics October November 2010 Paper 12
9, 14, 21, 27
well PV=nRT
well PV=nRT
we know number of moles wont change before and after the gases are mixed. T is also constant R is always constant so
P1V1 for helium P2V2 for Neon P3V3 for mixed gases.
P1V1+P2V2=P3V3
(5*12)+(6*10)=P3*15
solving this we get P3 as 8
Q9- momentum=mass*velocity= 2mv
if its elastic, momentum after collision will also be 2mv so the total collision would be 4mv. here its inelastic so it is not 4mv. Its not 2mv as well because this would mean it did not rebound. so it is 3mv
Q14- find the component of the force. it will be sin30=200/force. force=400N then multiply 1.5 to get the work = 600J add frictional force and you get 750
Q21- E=FL/A*extension. E is the same and F is same so Lp/Ap*extensionp=Lq/Aq*extensionq
2L/4A*extensionp=L/A*extensionq
extensionp/extensionq= 2LA/L4A = 0.5
October november 2010/ paper 11
Question 9 and 37
Please help !
well PV=nRT
we know number of moles wont change before and after the gases are mixed. T is also constant R is always constant so
P1V1 for helium P2V2 for Neon P3V3 for mixed gases.
P1V1+P2V2=P3V3
(5*12)+(6*10)=P3*15
solving this we get P3 as 8
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