For the following equilibrium system, which of the following changes will form more CaCO3? CO2(g) + Ca(OH)2(s) ⇌ CaCO3(s) + H2O(l) ΔH o rxn = −113 kJ a. Decrease temperature at constant pressure (no phase change). b. Increase volume at constant temperature. c. Increase partial pressure of CO2. d. Remove one-half of the initial CaCO3.

d. Remove one-half of the initial CaCO3.

Explanation: Le Chatelier's principle states that changes on the temperature, pressure, concentration and volume of a system will affect the reaction in an observable way. So in the reaction above:

A decrease in temperature will shift the equilibrium to the left because the reaction is exothermic, which means heat is released during the reaction. In other words, when you decrease temperature of a system, the equilibrium is towards the exothermic reaction;

A change in volume or pressure, will result in a production of more or less moles of gas. A increase in volume or in the partial pressure of CO2, the side which produces more moles of gas will be favored. In the equilibrium above, the shift will be to the left.

A change in concentration will tip the equilibrium towards the change: in this system, removing the product will shift the equilibrium towards the production of more CaCO3 to return to the equilibrium.

So, the correct answer is D. Remove one-half of the initial CaCO3.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.