When heated, CaCO₃ decomposes into CaO and CO₂ according to the following equation: CaCO₃(s) ⇌ CaO(s) + CO₂(g) This is an endothermic reaction in which ∆H° = 178.3 kJ/mol and in which entropy increases, ∆S° = 159 J/mol･K. To what temperature (in °C) must CaCO₃ be heated in a closed container in order to produce CO₂ at an equilibrium pressure of 0.100 atm? (Assume ∆H° and ∆S° do not change appreciably with temperature).

T= 1,121K

Explanation:

At equilibrium,

∆H°= T∆S°

∆H°= 178.3, ∆S°= 159J/mol= 0.159kJ/mol

Substitute into above formula

178.3 = T× 0.159

T= 1,121K

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i  

c-14 is carbon-14, or radiocarbon.   it is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons.