Can someone please help with this 4 mark question!

Chemistry

1 answer:

saul85 [17]2 years ago

4 0

Answer:

6.941 u

Explanation:

You might be interested in

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 o

lyudmila [28]

Answer:

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

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:

$atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu$

Thus, the atom percent turns out:

$\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%$

Best regards.

4 0

3 years ago

How many moles of iron are there in 55.85g of Fe3O4

soldier1979 [14.2K]

Answer:

• Molecular mass of Iron (III) tetraoxide

$\dashrightarrow \: { \tt{(56 \times 3) + (16 \times 4)}} \\ = { \tt{168 + 64}} \\ = { \tt{232\:g}}$

[ molar masses: Fe → 56, O → 16 ]

$\dashrightarrow \:{ \rm{232 \: g \: = 1 \: mole}} \\ \\ \dashrightarrow \: { \rm{55.85 \: g = ( \frac{55.85}{232}) \: moles }} \\ \\ \dashrightarrow \: { \boxed{ \tt{ = 0.24 \: moles}}}$