A level Chemistry help
1) A sample of 14N were placed in a mass spectrometer.
a. Calculate the mass of one ion of nitrogen
b. Calculate the velocity of its 1+ ion traveling down a 5m flight tube in 1.6 x 10-6 s
c. Calculate its kinetic energy. Give your answer to 3 significant figures
(1.135 x 10-13 J)
how do I work this out can anyone help please?
a. Calculate the mass of one ion of nitrogen
b. Calculate the velocity of its 1+ ion traveling down a 5m flight tube in 1.6 x 10-6 s
c. Calculate its kinetic energy. Give your answer to 3 significant figures
(1.135 x 10-13 J)
how do I work this out can anyone help please?
a. mass of ion/atom= mass of one ion/atom(same as atomic mass)/6.022x10^23(Avogadros constant)
=14/6.022x10^23
=2.32480903×10^-23
b.velocity= distance travelled/time
= 5/1.6 x 10-6
=3125000 m/s
c. kinetic energy= 0.5XmassX(velocity^2)
=0.5X2.32480903×10^-23X3125000
=
I think that is it. What year are you in if you don't mind me asking?
=14/6.022x10^23
=2.32480903×10^-23
b.velocity= distance travelled/time
= 5/1.6 x 10-6
=3125000 m/s
c. kinetic energy= 0.5XmassX(velocity^2)
=0.5X2.32480903×10^-23X3125000
=
I think that is it. What year are you in if you don't mind me asking?
Original post by andrei00
a. mass of ion/atom= mass of one ion/atom(same as atomic mass)/6.022x10^23(Avogadros constant)
=14/6.022x10^23
=2.32480903×10^-23
b.velocity= distance travelled/time
= 5/1.6 x 10-6
=3125000 m/s
c. kinetic energy= 0.5XmassX(velocity^2)
=0.5X2.32480903×10^-23X3125000
=
I think that is it. What year are you in if you don't mind me asking?
=14/6.022x10^23
=2.32480903×10^-23
b.velocity= distance travelled/time
= 5/1.6 x 10-6
=3125000 m/s
c. kinetic energy= 0.5XmassX(velocity^2)
=0.5X2.32480903×10^-23X3125000
=
I think that is it. What year are you in if you don't mind me asking?
Year 12 but | haven't done chemistry in like over a year so I honestly don't remember anything at all
Original post by andrei00
a. mass of ion/atom= mass of one ion/atom(same as atomic mass)/6.022x10^23(Avogadros constant)
=14/6.022x10^23
=2.32480903×10^-23
b.velocity= distance travelled/time
= 5/1.6 x 10-6
=3125000 m/s
c. kinetic energy= 0.5XmassX(velocity^2)
=0.5X2.32480903×10^-23X3125000
=
I think that is it. What year are you in if you don't mind me asking?
=14/6.022x10^23
=2.32480903×10^-23
b.velocity= distance travelled/time
= 5/1.6 x 10-6
=3125000 m/s
c. kinetic energy= 0.5XmassX(velocity^2)
=0.5X2.32480903×10^-23X3125000
=
I think that is it. What year are you in if you don't mind me asking?
isn't it 2.32480903×10^-26 for part a and ecf for part c
Original post by andrei00
a. mass of ion/atom= mass of one ion/atom(same as atomic mass)/6.022x10^23(Avogadros constant)
=14/6.022x10^23
=2.32480903×10^-23
b.velocity= distance travelled/time
= 5/1.6 x 10-6
=3125000 m/s
c. kinetic energy= 0.5XmassX(velocity^2)
=0.5X2.32480903×10^-23X3125000
=
I think that is it. What year are you in if you don't mind me asking?
=14/6.022x10^23
=2.32480903×10^-23
b.velocity= distance travelled/time
= 5/1.6 x 10-6
=3125000 m/s
c. kinetic energy= 0.5XmassX(velocity^2)
=0.5X2.32480903×10^-23X3125000
=
I think that is it. What year are you in if you don't mind me asking?
and so c would be 1.135 x 10-13 J
Original post by limitlesspuffy
isn't it 2.32480903×10^-26 for part a and ecf for part c
This is from 3 years ago but how did you get 10^-26??
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