They are found in group 3 of the periodic table.
Answer:

Explanation:
Hello,
In this case, since the undergoing chemical reaction is:

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

Regards.
Answer:
Zn+H2O ------ ZnO + H2 is the correct answer
0.0015 kilometers is for sure the answer!