Respuesta :
The answers are the following:
1. 3.01 x 10^23 atoms B
solution: (5.40/10.81)(6.022x10^23)
2. 1.51 x 10^23 atoms S
solution: (.250)(6.022x10^23)
3. 2.31 x 10^22 atoms K
solution: (.0384)(6.022x10^23)
4. 7.872 x 10^19 atoms Pt
solution: (.02550/195.08)(6.022x10^23)
5. 3.06 x 10^11 atoms Au
solution: (1.00x10^-10/196.97)(6.022x10^23)
1. 3.01 x 10^23 atoms B
solution: (5.40/10.81)(6.022x10^23)
2. 1.51 x 10^23 atoms S
solution: (.250)(6.022x10^23)
3. 2.31 x 10^22 atoms K
solution: (.0384)(6.022x10^23)
4. 7.872 x 10^19 atoms Pt
solution: (.02550/195.08)(6.022x10^23)
5. 3.06 x 10^11 atoms Au
solution: (1.00x10^-10/196.97)(6.022x10^23)
Answer:
a) 5.40 g B : [tex]3.012\times 10^{23}[/tex] atoms
b) 0.250 mol K : [tex]1.51\times 10^{23}[/tex] atoms
c) 0.0384 mol K : [tex]0.23\times 10^{23}[/tex] atoms
d) 0.02550 g Pt: [tex]7.8\times 10^{19}[/tex] atoms
e) [tex]1.00\times 10^-{10} g[/tex] Au: [tex]0.03\times 10^{13}[/tex] atoms
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
a) 5.40 g B
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{5.40g}{11g/mol}=0.5moles[/tex]
1 mole of boron contains = [tex]6.023\times 10^{23}[/tex] atoms
Thus 0.5 moles of boron contain =[tex]\frac{6.023\times 10^{23}}{1}\times 0.5=3.012\times 10^{23}[/tex] atoms
b) 0.250 mol K
1 mole of potassium (K) contains = [tex]6.023\times 10^{23}[/tex] atoms
Thus 0.250 moles of potassium contain =[tex]\frac{6.023\times 10^{23}}{1}\times 0.250=1.51\times 10^{23}[/tex] atoms
c) 0.0384 mol K
1 mole of potassium (K) contains = [tex]6.023\times 10^{23}[/tex] atoms
Thus 0.0384 moles of potassium (K) contain =[tex]\frac{6.023\times 10^{23}}{1}\times 0.0384=0.23\times 10^{23}[/tex] atoms
d) 0.02550 g Pt
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{0.02550 g}{195g/mol}=1.3\times 10^{-4}moles[/tex]
1 mole of platinum contains = [tex]6.023\times 10^{23}[/tex] atoms
Thus [tex]1.3\times 10^{-4}moles[/tex] of platinum contain=[tex]\frac{6.023\times 10^{23}}{1}\times 1.3\times 10^{-4}=7.8\times 10^{19}[/tex] atoms
e) [tex]1.00\times 10^-{10}g[/tex] Au,
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{1.00\times 10^{-10}g}{197g/mol}=0.005\times 10^{-10}moles[/tex]
1 mole of gold contains = [tex]6.023\times 10^{23}[/tex] atoms
Thus [tex]0.005\times 10^{-10}moles[/tex] of platinum contain=[tex]\frac{6.023\times 10^{23}}{1}\times 0.005\times 10^{-10}=0.03\times 10^{13}[/tex] atoms
