Question #5c1f1

1 Answer
Aug 20, 2017

Answer:

#157.48" g"#

Explanation:

Given:

#(1.45xx10^24" atoms of Zn")/1#

Multiply by the conversion factor for the Avogadro constant :

#(1.45xx10^24" atoms of Zn")/1(1" mole of Zn")/(6.02xx10^23" atoms of Zn")#

Please observe how the conversion factor cancels the units "atoms of Zn" and we are left with the units "moles of Zn":

#(1.45xx10^24cancel(" atoms of Zn"))/1(1" mole of Zn")/(6.02xx10^23cancel(" atoms of Zn"))#

Multiply by the conversion factor for the Molar Mass of Zn :

#(1.45xx10^24cancel(" atoms of Zn"))/1(1" mole of Zn")/(6.02xx10^23cancel(" atoms of Zn"))(65.38" g of Zn")/(1" mole of Zn")#

Please observe that the units "mole of Zn" cancel and we are left with the units " g of Zn":

#(1.45xx10^24cancel(" atoms of Zn"))/1(1cancel(" mole of Zn"))/(6.02xx10^23cancel(" atoms of Zn"))(65.38" g of Zn")/(1cancel(" mole of Zn"))#

All that is left to do is the multiplication and division and we know that we have correct units:

#(1.45xx10^24)/1(1)/(6.02xx10^23)(65.38" g of Zn")/(1) = 157.48" g of Zn"#