Find ΔH for: BaCO3 (s) → BaO (s)+ CO2 (g) given 2 Ba (s) + O2 (g) → 2 BaO (s) ΔH = – 1107.0 kJ Ba (s) + CO2 (g) + ½ O2 (g) → BaCO3 (s) ΔH = – 822.5 kJ

The answer to your question is ΔH = 269 kJ

Explanation:

Change the sense of the third reaction

           BaCO₃(s)  ⇒  Ba(s)  + CO₂(g)  + 1/2O₂(g)       ΔH = 822.5 kJ

Divide the second reaction by 2

           Ba(s)   +  1/2O₂(g)  ⇒  BaO(s)                         ΔH = -553.5 KJ

Add both reactions and cancel reactants and products

            BaCO₃(s) ⇒  BaO (s)  +  CO₂ (g)                  ΔH = 269  (822.5 - 553.5)

i know insulin is primary but i dont know the others im sorry

explanation:

under a microscope, cancer cells may look very different from normal cells. they often have different sizes and some may be larger than normal while others are smaller. cancer cells are often abnormally shaped and the control centre of the cell (the nucleus) may have an abnormal appearance.

i think? ?