A20.0 milliliter sample of 0.200 molar k₂co₃ solution is added to 30.0 milliliters of 0.400 molar ba(no₃)₂ solution. barium carbonate precipitates. the concentration of barium ion, ba₂, in solution after the reaction is

0.32M

Explanation:

Step 1:  Balance the reaction

K2CO3 + Ba(NO3)2 ⇔ KNO3 + BaCO3

We have a 20 mL 0.2 M K2CO3 and a 30mL 0.4M Ba(NO3)2 solution

SinceK2CO3 is the limiting reactant, there will remain Ba(NO3)2 after it's consumed and produced KNO3 + BaCO3

Step 2: Calculate concentration

To find the concentration of the barium cation we use the following equation:

Concentration = moles of the solute / volumen of the solution

[Ba2+] = (20 * 10^-3 * 0.2M + 30 * 10^-3 * 0.4M) / ( 20 + 30mL) *10^-3

[Ba2+] = 0.32 M

The concentration of Barium ion in solution is 0.32 M

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.

i do not get it   at all