Respuesta :
1) Calcium carbonate contains 40.0% calcium by weight.
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
w(Ca)=40.1/100.1=0.400 (40.0%)!
2) Mass fraction of this is excessive data.
3) The solution is:
m(Ca)=1.2 g
m(CaCO₃)=M(CaCO₃)*m(Ca)/M(Ca)
m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
w(Ca)=40.1/100.1=0.400 (40.0%)!
2) Mass fraction of this is excessive data.
3) The solution is:
m(Ca)=1.2 g
m(CaCO₃)=M(CaCO₃)*m(Ca)/M(Ca)
m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g
Answer:
10 grams of calcium carbonate are needed to provide the recommended daily allowance of calcium.
Explanation:
The recommended daily allowance (rda) of calcium = 1.2 g
Percentage of calcium by mass in calcium carbonate = 12.0%
Total mass of calcium carbonate be x
[tex]\%=\frac{\text{Mass of an element}}{\text{Total mass of compound}}\times 100[/tex]
[tex]12.0\%=\frac{1.2 g}{x}\times 100[/tex]
x = 10 g
10 grams of calcium carbonate are needed to provide the rda of calcium.
