Calculate the volume in mL of 0.100 of CaCl2 needed to produce 1.00 g of CaCO3. there is an excess of Na2CO3

100. mL

Explanation:

The computation of the volume in mL is shown below:

The balance equation of the double displacement reaction is

Now determine the moles that corresponding to 1.00 g of CaCO₃

As,

The molar mass of CaCO₃ is 100.09 g/mol.

So,

= 1.00 g × 1 mol/100.09 g

= 0.0100 mol

Now

the moles of CaCl₂ needed to generate 0.0100 moles of CaCO₃

As we know that

The molar ratio of CaCl₂ to CaCO₃ is 1:1.

So, The moles of CaCl₂ needed is

=  1 ÷ 1 × 0.0100 mol

= 0.0100 mol.

Now volume of 0.100 M CaCl₂ that contains 0.0100 mol is  

= 0.0100 mol × 1 L/0.100 mol × 1000 mL/1 L

= 100. mL

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.