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
The average weight of an atom of an element, formerly based on the
weight of one hydrogen atom taken as a unit or on 1/16 (0.0625) the
weight of an oxygen atom, but after 1961 based on 1/12 the weight of the
carbon-12 atom.
Answer:
20 g/mol
Explanation:
We can use <em>Graham’s Law of diffusion</em>:
The rate of diffusion (<em>r</em>) of a gas is inversely proportional to the square root of its molar mass (<em>M</em>).

If you have two gases, the ratio of their rates of diffusion is

Squaring both sides, we get

Solve for <em>M</em>₂:


