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
B, they then interpret that data to find their answers
Answer:(3)
Explanation: 2Al+3H2SO4----->Al(2SO4)3+3H2
Answer:
It can be removed by acidic chemicals
Explanation:
