Question

What is the composition, in atom percent, of an alloy that contains a) 45.5 lbm of silver, b) 83.7 lbm of gold, and c) 6.3 lbm of Cu? The atomic weights for silver, gold, and copper are, respectively, 107.87, 196.97, and 63.55 g/mol.

Answer 1

`\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%`

Step-by-step explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

`atomsAg=45.5lbm*(453.59g)/(1lbm)*(1molAg)/(107.87gAg)*(6.022x10^(23)atomsAg)/(1molAg)=1.15x10^(26)atomsAg\\atomsAu=83.7lbm*(453.59g)/(1lbm)*(1molAu)/(196.97gAu)*(6.022x10^(23)atomsAu)/(1molAu)=1.16x10^(26)atomsAu\\atomsCu=6.3lbm*(453.59g)/(1lbm)*(1molAg)/(63.55gCu)*(6.022x10^(23)atomsCu)/(1molCu)=2.71x10^(25)atomsCu`

Thus, the atom percent turns out:

`\% atAg=(1.15x10^(26))/(1.15x10^(26)+1.16x10^(26)+2.71x10^(25))*100\% =44.6\%\\\% atAu=(1.16x10^(26))/(1.15x10^(26)+1.16x10^(26)+2.71x10^(25))*100\% =44.9\%\\\% atCu=(2.71x10^(25))/(1.15x10^(26)+1.16x10^(26)+2.71x10^(25))*100\% =10.5\%`

Best regards.