Question #2a70b
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"If the momentum of an object increases by #20%#, what will be the percent increase in its kinetic energy?"
Assuming that you are saying #64g# of #O# (single oxygen) atoms:
Oxygen has a molar mass of #16g"/"mol#.
#:.64g# of oxygen atoms is #(64g)/(16g"/"mol)=4mol#
One mole is #6.02*10^23# particles/atoms.
#:. 4mol=4*6.02*10^23~~2.41*10^24# oxygen atoms.
To find the number of particles (in this case atoms), we need to find the number of moles of oxygen atoms.
#n(O)=(m(O))/(M_r(O))=64/16=4mol#
#"Number of particles"=n*N_a#, where #N_a# is Avogadros' constant (#6.02*10^23# #mol^(-1)#)
#"Number of oxygen atoms"=n(O)*N_a#
#color(white)("Number of oxygen atoms")=4(6.02*10^23)~~2.41*10^24# atoms
It's #2.4092 xx 10^24# particles in #64# g of Oxygen atom.
Since we have #O# = #16# g = #1# mole= #6.023 xx 10^23# particles, so on this basis, when we have 64g of O, which is the #4 xx16#, so it will be #4# moles, and #4# moles= #4 xx 6.023 xx 10^23# .. which is eventually #2.4092 xx 10^23# atoms in #O#.
Hope this helps.. :)
